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Provide tags and a correct Python 3 solution for this coding contest problem.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
|
instruction
| 0
| 1,679
| 3
| 3,358
|
Tags: binary search, geometry, ternary search
Correct Solution:
```
n , l , r = 0 , -100000000, 1000000000
x , y = [] , []
def cal(mid):
mx = 0
for i in range(n):
mx = max (mx ,(x[i] - mid) ** 2 / (2 * y[i]) + (y[i] / 2))
return mx;
n = int(input())
for i in range(n):
a , b = map(int , input().split())
x.append(a) , y.append(b)
for i in range(1 , n):
if(y[i] * y[0] < 0):
print(-1)
exit(0)
if(y[i] < 0):
y[i] *= -1
if(y[0] < 0):
y[0] *= -1
for it in range(100):
m1 , m2 = l + (r - l) / 3 , r - (r - l) / 3
v1 , v2 = cal(m1) , cal(m2)
if(v1 > v2):
l = m1
else:
r = m2
print(cal(m1))
```
|
output
| 1
| 1,679
| 3
| 3,359
|
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
|
instruction
| 0
| 1,680
| 3
| 3,360
|
Tags: binary search, geometry, ternary search
Correct Solution:
```
l , r =-100000000, 1000000000
def check(mid):
mx = 0
for i in range(n):
x,y = x1[i],y1[i]
mx = max (mx ,(x1[i] - mid) ** 2 / (2 * y1[i]) + (y1[i] / 2))
return mx
n = int(input())
count1 = 0
count2 = 0
x1 = []
y1 = []
for i in range(n):
a,b = map(int,input().split())
if b>=0:
count1+=1
else:
count2+=1
x1.append(a)
y1.append(abs(b))
if count1 and count2:
print(-1)
exit()
for i in range(100):
mid1 = l+(r-l)/3
mid2 = r-(r-l)/3
if check(mid1)>check(mid2):
l = mid1
else:
r = mid2
# print(l,r)
print(check(l))
```
|
output
| 1
| 1,680
| 3
| 3,361
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
Submitted Solution:
```
import math
n = int(input())
points = []
npt = 0
for i in range(n):
point = list(map(int, input().split()))
if point[1] < 0:
npt += 1
point[1] = -point[1]
points.append(point)
if npt < n and npt > 0:
print(-1)
else:
l, r = 0, 1e20
count = 200
while abs(r - l) > 1e-6 and count > 0:
count -= 1
ok = True
mid = (l + r) / 2
a, b = -float('inf'), float('inf')
for i in range(n):
x, y = points[i]
if y > 2 * mid:
l = mid
ok = False
break
delta = math.sqrt(y * (2*mid - y))
a, b = max(a, x - delta), min(b, x + delta)
#print("x: {}, y: {}, a: {}, b: {}, delta: {}".format(x, y, a, b, delta))
if not ok:
continue
#print(a, b, l, r, mid)
if a > b:
l = mid
else:
r = mid
print((l + r) / 2)
```
|
instruction
| 0
| 1,681
| 3
| 3,362
|
Yes
|
output
| 1
| 1,681
| 3
| 3,363
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
Submitted Solution:
```
import sys
input = sys.stdin.readline
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def get(x0, a, n):
r = 0
for i in range(n):
p = (x0 - a[i].x)*(x0 - a[i].x) + 1.0*a[i].y*a[i].y
p = p/2.0/a[i].y
if p < 0:
p = -p
r = max(r, p)
return r
def main():
n = int(input())
pos, neg = False, False
a = []
for i in range(n):
x, y = map(int, input().split())
t = Point(x, y)
if t.y > 0:
pos = True
else:
neg = True
a.append(t)
if pos and neg:
return -1
if neg:
for i in range(n):
a[i].y = -a[i].y
L, R = -1e8, 1e8
for i in range(120):
x1 = L + (R-L)/3
x2 = R - (R-L)/3
if get(x1, a, n) < get(x2, a, n):
R = x2
else:
L = x1
return get(L, a, n)
if __name__=="__main__":
print(main())
```
|
instruction
| 0
| 1,682
| 3
| 3,364
|
Yes
|
output
| 1
| 1,682
| 3
| 3,365
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def check(m):
r = -1
for j in range(n):
r = max(r, (x[j]-m)**2/(2*y[j])+(y[j]/2))
return r
n=int(input())
x=[]
y=[]
for j in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
p=0
q=0
for j in y:
if j<0:
p=1
else:
q=1
if p==1 and q==1:
print(-1)
else:
if p==1:
for j in range(n):
y[j]=abs(y[j])
l=-10**7
h=10**7
for i in range(100):
m1=(2*l+h)/3
m2=(2*h+l)/3
if check(m1)>check(m2):
l=m1
else:
h=m2
print(check(l))
```
|
instruction
| 0
| 1,683
| 3
| 3,366
|
Yes
|
output
| 1
| 1,683
| 3
| 3,367
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
Submitted Solution:
```
'''
-1 possible only when some are on both sides
'''
n = int(input())
X = []
Y = []
q1 = 0
q2 = 0
q3 = 0
q4 = 0
for i in range(n):
x, y = [int(i) for i in input().split()]
if y>0:
q1+=1
else:
q2+=1
X.append(x)
Y.append(y)
def dist(a, b, c, d):
return (a - c)**2 + (b - d)**2
def check(r):
global X, Y, finalx, n
for i in range(n):
if dist(finalx, r, X[i], Y[i]) > r*r:
return False
return True
if [q1,q2].count(0)!=1:
print(-1)
else:
hi = 0
lo = 0
finalx = 0
for i in range(n):
Y[i] = abs(Y[i])
hi = max(hi, Y[i])
finalx += X[i]
finalx = finalx/n
for i in range(60):
mid = (lo+hi)/2
#print(mid)
if check(mid):
hi = mid
else:
lo = mid
if check(mid):
print(mid)
elif check(lo):
print(lo)
elif check(hi):
print(hi)
```
|
instruction
| 0
| 1,684
| 3
| 3,368
|
No
|
output
| 1
| 1,684
| 3
| 3,369
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
Submitted Solution:
```
def print_return(func):
def wrapper(*args, **kwargs):
retval = func(*args, **kwargs)
print(retval)
return retval
return wrapper
@print_return
def solve_b(n=None, m=None, sign=None):
if n is None or m is None or sign is None:
n, m = (int(x) for x in input().split())
sign = [input() for _ in range(n)]
sheet = [list('.' * m) for _ in range(n)]
def can_write(r, c):
for i in range(r - 1, r + 2):
for j in range(c - 1, c + 2):
if (i, j) == (r, c):
continue
if sign[i][j] != '#':
return False
return True
def write(r, c):
for i in range(r - 1, r + 2):
for j in range(c - 1, c + 2):
if (i, j) == (r, c):
pass
else:
sheet[i][j] = '#'
for r in range(1, n - 1):
for c in range(1, m - 1):
if can_write(r, c):
write(r, c)
for r in range(n):
sheet[r] = ''.join(sheet[r])
if sheet[r] != sign[r]:
return 'NO'
return 'YES'
@print_return
def solve_c(n=None):
if n is None:
n = int(input())
ans = []
curr_pow = 1
while n > 3:
ans += [curr_pow] * ((n + 1) // 2)
n //= 2
curr_pow *= 2
if n == 1:
ans += [curr_pow]
if n == 2:
ans += [curr_pow, curr_pow * 2]
if n == 3:
ans += [curr_pow, curr_pow, curr_pow * 3]
ans = ' '.join((str(x) for x in ans))
return ans
@print_return
def solve_d(n=None, coors=None):
if n is None or coors is None:
n = int(input())
coors = [tuple(int(x) for x in input().split()) for _ in range(n)]
elif isinstance(coors, str):
coors = coors.split('\n')
coors = [tuple(int(x) for x in coor.split()) for coor in coors]
if any(x[1] > 0 for x in coors) and any(x[1] < 0 for x in coors):
return -1
def get_lcl__rcl(x, y, radius):
h = abs(y - radius)
d = (radius * radius - h * h) ** 0.5
return x - d, x + d
def is_valid_radius(radius):
x0, x1 = -10 ** 7, 10 ** 7
for x, y in coors:
l, r = get_lcl__rcl(x, y, radius)
if isinstance(l, complex) or isinstance(r, complex):
return False
x0 = max(x0, l)
x1 = min(x1, r)
if x1 < x0:
return False
return True
l_op, r_cl = 0, 10 ** 8
while r_cl - l_op > 10 ** (-7):
m = (l_op + r_cl) / 2
if is_valid_radius(m):
r_cl = m
else:
l_op = m
return r_cl
if __name__ == '__main__':
solve_c()
```
|
instruction
| 0
| 1,685
| 3
| 3,370
|
No
|
output
| 1
| 1,685
| 3
| 3,371
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
Submitted Solution:
```
n=int(input())
x=[]
y=[]
for j in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
p=0
q=0
for j in y:
if j<0:
p=1
else:
q=1
if p==1 and q==1:
print(-1)
else:
if p==1:
for j in range(n):
y[j]=abs(y[j])
c=sum(x)/n
r=-1
for j in range(n):
r=max(r,(((x[j]-c)**2+y[j]**2))/(2*y[j]))
print(min(r,max(y)/2))
```
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.
The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.
For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals.
Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.
Output
If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
Examples
Input
1
0 1
Output
0.5
Input
3
0 1
0 2
0 -3
Output
-1
Input
2
0 1
1 1
Output
0.625
Note
In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).
In the second sample it is impossible to build a reserve.
In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).
Submitted Solution:
```
from math import hypot
def dist(p1, p2):
return hypot(p1[0]-p2[0], p1[1]-p2[1])
def main():
n = int(input())
points = []
ySign = 0
for _ in range(n):
x, y = [int(x) for x in input().split()]
if ySign:
if y * ySign < 0:
print(-1)
return
else:
ySign = y
points.append((x,y))
# For every pair of points, find the center of the circle on the vertical line between them.
# The distance to the vertical line is x, the circle radius is r, and the height of the points is h.
# By geometry, r^2=x^2+d^2 when h-r=d by definition, so r=(x^2+h^2)/(2h)
# r is the height of the center of the circle, as it must be tangent to y=0.
if n == 1:
print(points[0][1]/2)
return
minR = 10**8
for i in range(len(points)-1):
for j in range(i+1, len(points)):
x0, y0 = points[i]
x1, y1 = points[j]
h = y0
x = abs(x1-x0)/2
r = (x**2+h**2)/2/h
#print(r)
if r < minR:
center = (x0+x, r)
bad = False
for pt in points:
#print(pt, center, dist(pt, center))
if dist(pt, center) > r:
bad = True
break
if not bad: minR = r
print(minR)
main()
```
|
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| 0
| 1,687
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No
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output
| 1
| 1,687
| 3
| 3,375
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A very unusual citizen lives in a far away kingdom — Dwarf Gracula. However, his unusual name is not the weirdest thing (besides, everyone long ago got used to calling him simply Dwarf Greg). What is special about Dwarf Greg — he's been living for over 200 years; besides, he lives in a crypt on an abandoned cemetery and nobody has ever seen him out in daytime. Moreover, nobody has ever seen Greg buy himself any food. That's why nobody got particularly surprised when after the infernal dragon's tragic death cattle continued to disappear from fields. The people in the neighborhood were long sure that the harmless dragon was never responsible for disappearing cattle (considering that the dragon used to be sincere about his vegetarian views). But even that's not the worst part of the whole story.
The worst part is that merely several minutes ago Dwarf Greg in some unintelligible way got inside your house and asked you to help him solve a problem. The point is that a short time ago Greg decided to order a new coffin (knowing his peculiar character, you are not surprised at all). But the problem is: a very long in both directions L-shaped corridor leads to Greg's crypt, and you can't drag just any coffin through that corridor. That's why he asked you to help.
<image>
You've formalized the task on a plane like this: let the corridor's width before and after the turn be equal to a and b correspondingly (see the picture). The corridor turns directly at a right angle, the coffin is a rectangle whose length and width are equal to l and w (l ≥ w) correspondingly. Dwarf Greg has already determined the coffin's length (l), which is based on his height; your task is to determine the coffin's maximally possible width (w), at which it can be brought to the crypt. Besides, due to its large mass (pure marble!) the coffin is equipped with rotating wheels; therefore it is impossible to lift it off the ground, however, arbitrary moves and rotations of the coffin in the plane become possible. The coffin may be rotated arbitrarily just before you drag it into crypt and move through the corridor.
Greg promised that if you help him, he will grant you immortality (I wonder how?). And if you don't, well... trust me, you don't want to know what happens if you don't help him...
Input
The first line contains three space-separated integers a, b and l from the problem's statement (1 ≤ a, b, l ≤ 104).
Output
Print the maximally possible width of a coffin with absolute or relative error no more than 10 - 7. If a coffin with the given length and positive width (the coffin that would meet the conditions from the problem's statement) does not exist, print "My poor head =(" (without quotes).
It is guaranteed that if the answer is positive, it will be not less than 10 - 7. All the hacks will also be checked to meet that condition.
Examples
Input
2 2 1
Output
1.0000000
Input
2 2 2
Output
2.0000000
Input
2 2 3
Output
1.3284271
Input
2 2 6
Output
My poor head =(
Note
In the first example the answer is restricted by the coffin's length (remember — coffin's widths should not be larger than it's length).
In the second example it is possible to drag the coffin through the corridor thanks to rotating wheels: firstly, drag it forward by one side while it will not be hampered by the wall, then move it forward by adjacent side perpendicularly to the initial movement direction (remember — arbitrary moves and rotations of the coffin are possible).
Submitted Solution:
```
a, b, l = map(int, input().split(' '))
d = (a+b)*(2**.5)
if l > d:
ans = "My poor head =("
else:
f = 0
e = (d-l)/2
if l < max(a, b):
f = max(a, b)
ans = str(float(min(l, max(f, min(l, e)))))
ans1 = ans.split('.')
if len(ans1[1]) == 7:
ans = str(float(ans))
elif len(ans1[1]) < 7:
while len(ans1[1]) != 7:
ans1[1] += '0'
elif len(ans1[1]) > 7:
print(ans1)
ans1[1] = ans1[1][0] + ans1[1][1] + ans1[1][2]+ ans1[1][3]+ ans1[1][4]+ ans1[1][5]+ans1[1][6]
ans = ans1[0] + '.' + ans1[1]
print(ans)
```
|
instruction
| 0
| 2,219
| 3
| 4,438
|
No
|
output
| 1
| 2,219
| 3
| 4,439
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A very unusual citizen lives in a far away kingdom — Dwarf Gracula. However, his unusual name is not the weirdest thing (besides, everyone long ago got used to calling him simply Dwarf Greg). What is special about Dwarf Greg — he's been living for over 200 years; besides, he lives in a crypt on an abandoned cemetery and nobody has ever seen him out in daytime. Moreover, nobody has ever seen Greg buy himself any food. That's why nobody got particularly surprised when after the infernal dragon's tragic death cattle continued to disappear from fields. The people in the neighborhood were long sure that the harmless dragon was never responsible for disappearing cattle (considering that the dragon used to be sincere about his vegetarian views). But even that's not the worst part of the whole story.
The worst part is that merely several minutes ago Dwarf Greg in some unintelligible way got inside your house and asked you to help him solve a problem. The point is that a short time ago Greg decided to order a new coffin (knowing his peculiar character, you are not surprised at all). But the problem is: a very long in both directions L-shaped corridor leads to Greg's crypt, and you can't drag just any coffin through that corridor. That's why he asked you to help.
<image>
You've formalized the task on a plane like this: let the corridor's width before and after the turn be equal to a and b correspondingly (see the picture). The corridor turns directly at a right angle, the coffin is a rectangle whose length and width are equal to l and w (l ≥ w) correspondingly. Dwarf Greg has already determined the coffin's length (l), which is based on his height; your task is to determine the coffin's maximally possible width (w), at which it can be brought to the crypt. Besides, due to its large mass (pure marble!) the coffin is equipped with rotating wheels; therefore it is impossible to lift it off the ground, however, arbitrary moves and rotations of the coffin in the plane become possible. The coffin may be rotated arbitrarily just before you drag it into crypt and move through the corridor.
Greg promised that if you help him, he will grant you immortality (I wonder how?). And if you don't, well... trust me, you don't want to know what happens if you don't help him...
Input
The first line contains three space-separated integers a, b and l from the problem's statement (1 ≤ a, b, l ≤ 104).
Output
Print the maximally possible width of a coffin with absolute or relative error no more than 10 - 7. If a coffin with the given length and positive width (the coffin that would meet the conditions from the problem's statement) does not exist, print "My poor head =(" (without quotes).
It is guaranteed that if the answer is positive, it will be not less than 10 - 7. All the hacks will also be checked to meet that condition.
Examples
Input
2 2 1
Output
1.0000000
Input
2 2 2
Output
2.0000000
Input
2 2 3
Output
1.3284271
Input
2 2 6
Output
My poor head =(
Note
In the first example the answer is restricted by the coffin's length (remember — coffin's widths should not be larger than it's length).
In the second example it is possible to drag the coffin through the corridor thanks to rotating wheels: firstly, drag it forward by one side while it will not be hampered by the wall, then move it forward by adjacent side perpendicularly to the initial movement direction (remember — arbitrary moves and rotations of the coffin are possible).
Submitted Solution:
```
a, b, l = map(int, input().split(' '))
d = (a+b)*(2**.5)
if l > d:
ans = "My poor head =("
else:
f = 0
e = (d-l)/2
if l <= max(a, b):
f = max(a, b)
ans = str(float(min(l, max(f, min(l, e)))))
ans1 = ans.split('.')
if len(ans1[1]) == 7:
ans = str(float(ans))
elif len(ans1[1]) < 7:
while len(ans1[1]) != 7:
ans1[1] += '0'
elif len(ans1[1]) > 7:
ans1[1] = ans1[1][0] + ans1[1][1] + ans1[1][2]+ ans1[1][3]+ ans1[1][4]+ ans1[1][5]+ans1[1][6]
ans = ans1[0] + '.' + ans1[1]
print(ans)
```
|
instruction
| 0
| 2,220
| 3
| 4,440
|
No
|
output
| 1
| 2,220
| 3
| 4,441
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A very unusual citizen lives in a far away kingdom — Dwarf Gracula. However, his unusual name is not the weirdest thing (besides, everyone long ago got used to calling him simply Dwarf Greg). What is special about Dwarf Greg — he's been living for over 200 years; besides, he lives in a crypt on an abandoned cemetery and nobody has ever seen him out in daytime. Moreover, nobody has ever seen Greg buy himself any food. That's why nobody got particularly surprised when after the infernal dragon's tragic death cattle continued to disappear from fields. The people in the neighborhood were long sure that the harmless dragon was never responsible for disappearing cattle (considering that the dragon used to be sincere about his vegetarian views). But even that's not the worst part of the whole story.
The worst part is that merely several minutes ago Dwarf Greg in some unintelligible way got inside your house and asked you to help him solve a problem. The point is that a short time ago Greg decided to order a new coffin (knowing his peculiar character, you are not surprised at all). But the problem is: a very long in both directions L-shaped corridor leads to Greg's crypt, and you can't drag just any coffin through that corridor. That's why he asked you to help.
<image>
You've formalized the task on a plane like this: let the corridor's width before and after the turn be equal to a and b correspondingly (see the picture). The corridor turns directly at a right angle, the coffin is a rectangle whose length and width are equal to l and w (l ≥ w) correspondingly. Dwarf Greg has already determined the coffin's length (l), which is based on his height; your task is to determine the coffin's maximally possible width (w), at which it can be brought to the crypt. Besides, due to its large mass (pure marble!) the coffin is equipped with rotating wheels; therefore it is impossible to lift it off the ground, however, arbitrary moves and rotations of the coffin in the plane become possible. The coffin may be rotated arbitrarily just before you drag it into crypt and move through the corridor.
Greg promised that if you help him, he will grant you immortality (I wonder how?). And if you don't, well... trust me, you don't want to know what happens if you don't help him...
Input
The first line contains three space-separated integers a, b and l from the problem's statement (1 ≤ a, b, l ≤ 104).
Output
Print the maximally possible width of a coffin with absolute or relative error no more than 10 - 7. If a coffin with the given length and positive width (the coffin that would meet the conditions from the problem's statement) does not exist, print "My poor head =(" (without quotes).
It is guaranteed that if the answer is positive, it will be not less than 10 - 7. All the hacks will also be checked to meet that condition.
Examples
Input
2 2 1
Output
1.0000000
Input
2 2 2
Output
2.0000000
Input
2 2 3
Output
1.3284271
Input
2 2 6
Output
My poor head =(
Note
In the first example the answer is restricted by the coffin's length (remember — coffin's widths should not be larger than it's length).
In the second example it is possible to drag the coffin through the corridor thanks to rotating wheels: firstly, drag it forward by one side while it will not be hampered by the wall, then move it forward by adjacent side perpendicularly to the initial movement direction (remember — arbitrary moves and rotations of the coffin are possible).
Submitted Solution:
```
a, b, l = map(int, input().split(' '))
d = (a+b)*(2**.5)
if l > d:
ans = "My poor head =("
else:
f = 0
e = (d-l)/2
if l < max(a, b):
f = max(a, b)
ans = str(float(max(f, min(l, e))))
ans1 = ans.split('.')
if len(ans1[1]) == 7:
ans = str(float(ans))
elif len(ans1[1]) < 7:
while len(ans1[1]) != 7:
ans1[1] += '0'
elif len(ans1[1]) > 7:
print(ans1)
ans1[1] = ans1[1][0] + ans1[1][1] + ans1[1][2]+ ans1[1][3]+ ans1[1][4]+ ans1[1][5]+ans1[1][6]
ans = ans1[0] + '.' + ans1[1]
print(ans)
```
|
instruction
| 0
| 2,221
| 3
| 4,442
|
No
|
output
| 1
| 2,221
| 3
| 4,443
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 2,617
| 3
| 5,234
|
Tags: dp
Correct Solution:
```
import sys
import math
INF = -100000000
memo = dict()
def func(line, r):
if line in memo and r in memo[line]:
return memo[line][r]
if len(line) == 1:
which = line[0] == 'T'
if r % 2 == 1:
which = not which
if which:
return [INF, INF, 0, 0]
else:
return [1, INF, INF, INF]
best = [INF, INF, INF, INF]
for i in range(r + 1):
a = func(line[:len(line) // 2], i)
b = func(line[len(line) // 2:], r - i)
for (j, k) in [(j, k) for j in range(4) for k in range(4)]:
D = j < 2
if a[j] == INF or b[k] == INF: continue
aa = -a[j] if j % 2 else a[j]
bb = -b[k] if k % 2 else b[k]
d1, d2 = 0, 1
if k < 2:
aa = aa + bb
if not D: d1, d2 = 2, 3
else:
aa = -aa + bb
if D: d1, d2 = 2, 3
if aa >= 0: best[d1] = max(best[d1], aa)
if aa <= 0: best[d2] = max(best[d2], -aa)
if not line in memo:
memo[line] = dict()
memo[line][r] = best
return best
line = input().strip()
k = int(input())
ans = max(func(line, k))
if ans == INF: print("0")
else: print(ans)
```
|
output
| 1
| 2,617
| 3
| 5,235
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A patient has been infected with an unknown disease. His body can be seen as an infinite grid of triangular cells which looks as follows:
<image>
Two cells are neighboring if they share a side. Therefore, each cell (x, y) has exactly three neighbors:
* (x+1, y)
* (x-1, y)
* (x+1, y-1) if x is even and (x-1, y+1) otherwise.
Initially some cells are infected, all the others are healthy. The process of recovery begins. Each second, for exactly one cell (even though there might be multiple cells that could change its state) one of the following happens:
* A healthy cell with at least 2 infected neighbors also becomes infected.
* An infected cell with at least 2 healthy neighbors also becomes healthy.
If no such cell exists, the process of recovery stops. Patient is considered recovered if the process of recovery has stopped and all the cells are healthy.
We're interested in a worst-case scenario: is it possible that the patient never recovers, or if it's not possible, what is the maximum possible duration of the recovery process?
Input
The first line contains one integer n (1 ≤ n ≤ 250000) — the number of infected cells.
The i-th of the next n lines contains two space-separated integers x_i and y_i (0 ≤ x_i, y_i < 500), meaning that cell (x_i, y_i) is infected. All cells (x_i, y_i) are distinct, and all other cells are considered healthy.
Output
If it is possible that the organism never fully recovers from the disease, print SICK. Otherwise, you should print RECOVERED and in the next line an integer k — the longest possible recovery period, modulo 998244353.
Examples
Input
4
0 0
1 0
2 0
0 1
Output
RECOVERED
4
Input
3
1 0
3 0
1 1
Output
SICK
Note
For the first testcase, the following drawings describe the longest possible recovery process. It can be proven that there are no recovery periods of length 5 or longer, and the organism always recovers in this testcase.
<image> \hspace{40pt} \downarrow <image> \hspace{40pt} \downarrow <image> \hspace{40pt} \downarrow <image> \hspace{40pt} \downarrow
\hspace{15pt} RECOVERED
For the second testcase, it is possible for the cells (2, 0), (2, 1), (0, 1) to become infected. After that, no cell can change its state, so the answer is SICK, as not all of the cells are healthy.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def sickdiamond(pos):
x=pos[0]
y=pos[1]
if grid[x][y]=="H":
return False
if grid[x][y+1]=="H":
return False
if grid[x-2][y+1]=="H":
return False
return True
def fn(pos):
x=pos[0]
y=pos[1]
ret=[]
if x>0:
ret.append([x-1,y])
if x<500:
ret.append([x+1,y])
if x%2==0:
if y>0 and x<500:
ret.append([x+1,y-1])
else:
if y<500 and x>0:
ret.append([x-1,y+1])
return ret
n=int(input())
grid=[0]*501
for i in range(501):
grid[i]=["H"]*501
toc=[]
inf=n
time=0
for i in range(n):
x,y=map(int,input().split())
toc.append([x,y])
grid[x][y]="S"
while len(toc)>0:
pos=toc.pop()
neig=fn(pos)
for n in neig:
if grid[n[0]][n[1]]=="H":
neig2=fn(n)
sick=0
for n2 in neig2:
if grid[n2[0]][n2[1]]=="S":
sick+=1
if sick>=2:
grid[n[0]][n[1]]="S"
inf+=1
time+=1
toc.append(n)
sick=False
for x in range(2,501):
if x%2==0:
for y in range(500):
if sickdiamond([x,y]):
sick=True
if sick:
print("SICK")
else:
print("RECOVERED")
print(time+inf)
```
|
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| 0
| 2,650
| 3
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|
No
|
output
| 1
| 2,650
| 3
| 5,301
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the galaxy of far, far away...
Darth Wader found out the location of a rebels' base. Now he is going to destroy the base (and the whole planet that the base is located at), using the Death Star.
When the rebels learnt that the Death Star was coming, they decided to use their new secret weapon — space mines. Let's describe a space mine's build.
Each space mine is shaped like a ball (we'll call it the mine body) of a certain radius r with the center in the point O. Several spikes protrude from the center. Each spike can be represented as a segment, connecting the center of the mine with some point P, such that <image> (transporting long-spiked mines is problematic), where |OP| is the length of the segment connecting O and P. It is convenient to describe the point P by a vector p such that P = O + p.
The Death Star is shaped like a ball with the radius of R (R exceeds any mine's radius). It moves at a constant speed along the v vector at the speed equal to |v|. At the moment the rebels noticed the Star of Death, it was located in the point A.
The rebels located n space mines along the Death Star's way. You may regard the mines as being idle. The Death Star does not know about the mines' existence and cannot notice them, which is why it doesn't change the direction of its movement. As soon as the Star of Death touched the mine (its body or one of the spikes), the mine bursts and destroys the Star of Death. A touching is the situation when there is a point in space which belongs both to the mine and to the Death Star. It is considered that Death Star will not be destroyed if it can move infinitely long time without touching the mines.
Help the rebels determine whether they will succeed in destroying the Death Star using space mines or not. If they will succeed, determine the moment of time when it will happen (starting from the moment the Death Star was noticed).
Input
The first input data line contains 7 integers Ax, Ay, Az, vx, vy, vz, R. They are the Death Star's initial position, the direction of its movement, and its radius ( - 10 ≤ vx, vy, vz ≤ 10, |v| > 0, 0 < R ≤ 100).
The second line contains an integer n, which is the number of mines (1 ≤ n ≤ 100). Then follow n data blocks, the i-th of them describes the i-th mine.
The first line of each block contains 5 integers Oix, Oiy, Oiz, ri, mi, which are the coordinates of the mine centre, the radius of its body and the number of spikes (0 < ri < 100, 0 ≤ mi ≤ 10). Then follow mi lines, describing the spikes of the i-th mine, where the j-th of them describes the i-th spike and contains 3 integers pijx, pijy, pijz — the coordinates of the vector where the given spike is directed (<image>).
The coordinates of the mines' centers and the center of the Death Star are integers, their absolute value does not exceed 10000. It is guaranteed that R > ri for any 1 ≤ i ≤ n. For any mines i ≠ j the following inequality if fulfilled: <image>. Initially the Death Star and the mines do not have common points.
Output
If the rebels will succeed in stopping the Death Star using space mines, print the time from the moment the Death Star was noticed to the blast.
If the Death Star will not touch a mine, print "-1" (without quotes).
For the answer the absolute or relative error of 10 - 6 is acceptable.
Examples
Input
0 0 0 1 0 0 5
2
10 8 0 2 2
0 -3 0
2 2 0
20 0 0 4 3
2 4 0
-4 3 0
1 -5 0
Output
10.0000000000
Input
8 8 4 4 4 2 6
1
-2 -2 -1 3 0
Output
-1
Input
30 30 2 1 2 1 20
3
0 0 40 5 1
1 4 4
-10 -40 -5 7 0
100 200 95 8 1
-10 0 0
Output
74.6757620881
|
instruction
| 0
| 3,051
| 3
| 6,102
|
Tags: geometry
Correct Solution:
```
import sys, math
def dist(a,b):
return pow((a[0] - b[0])**2 + (a[1] - b[1])**2 + (a[2] - b[2])**2,0.5)
def vxv(v1,v2):
x = -v1[2] * v2[1] + v1[1] * v2[2]
y = -v1[0] * v2[2] + v1[2] * v2[0]
z = -v1[1] * v2[0] + v1[0] * v2[1]
return (x,y,z)
readline = sys.stdin.readline
s1,s2,s3,v1,v2,v3,R = map(int,readline().split())
n = int(readline())
pm = []
p = []
for _ in range(n):
o1,o2,o3,r,m = map(int,readline().split())
p.append((o1,o2,o3,r))
for _ in range(m):
pm1,pm2,pm3 = map(int,readline().split())
pm.append((pm1 + o1, pm2 + o2, pm3 + o3))
nv = (v1,v2,v3)
m = (s1,s2,s3)
distance = []
for x in pm:
tp = (x[0] - m[0], x[1] - m[1], x[2] - m[2])
tp1 = vxv(tp,nv)
d = dist(tp1,(0,0,0))/dist(nv,(0,0,0))
if d <= R:
dd = pow(dist(x,m)**2-d**2,0.5)
dnv = dist(nv, (0, 0, 0))
nnv = (dd * nv[0] / dnv + m[0], dd * nv[1] / dnv + m[1], dd * nv[2] / dnv + m[2])
if dist(nnv, x) < dist(x, m):
distance.append(dd - pow(R**2 - d**2,0.5))
for i in range(len(p)):
pi =(p[i][0],p[i][1],p[i][2])
tp = (pi[0] - m[0], pi[1] - m[1], pi[2] - m[2])
tp1 = vxv(tp,nv)
d = dist(tp1,(0,0,0))/dist(nv,(0,0,0))
if d - p[i][3] <= R:
dd = pow(dist(pi,m)**2-d**2,0.5)
dnv = dist(nv,(0,0,0))
nnv = (dd * nv[0]/dnv + m[0], dd * nv[1]/dnv + m[1],dd * nv[2]/dnv + m[2])
if dist(nnv,p[i]) < dist(p[i],m):
dr = pow((R + p[i][3])**2 - d**2,0.5)
distance.append(dd - dr)
if len(distance):
distance.sort()
print(distance[0]/dist(nv,(0,0,0)))
else:
print(-1)
```
|
output
| 1
| 3,051
| 3
| 6,103
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the galaxy of far, far away...
Darth Wader found out the location of a rebels' base. Now he is going to destroy the base (and the whole planet that the base is located at), using the Death Star.
When the rebels learnt that the Death Star was coming, they decided to use their new secret weapon — space mines. Let's describe a space mine's build.
Each space mine is shaped like a ball (we'll call it the mine body) of a certain radius r with the center in the point O. Several spikes protrude from the center. Each spike can be represented as a segment, connecting the center of the mine with some point P, such that <image> (transporting long-spiked mines is problematic), where |OP| is the length of the segment connecting O and P. It is convenient to describe the point P by a vector p such that P = O + p.
The Death Star is shaped like a ball with the radius of R (R exceeds any mine's radius). It moves at a constant speed along the v vector at the speed equal to |v|. At the moment the rebels noticed the Star of Death, it was located in the point A.
The rebels located n space mines along the Death Star's way. You may regard the mines as being idle. The Death Star does not know about the mines' existence and cannot notice them, which is why it doesn't change the direction of its movement. As soon as the Star of Death touched the mine (its body or one of the spikes), the mine bursts and destroys the Star of Death. A touching is the situation when there is a point in space which belongs both to the mine and to the Death Star. It is considered that Death Star will not be destroyed if it can move infinitely long time without touching the mines.
Help the rebels determine whether they will succeed in destroying the Death Star using space mines or not. If they will succeed, determine the moment of time when it will happen (starting from the moment the Death Star was noticed).
Input
The first input data line contains 7 integers Ax, Ay, Az, vx, vy, vz, R. They are the Death Star's initial position, the direction of its movement, and its radius ( - 10 ≤ vx, vy, vz ≤ 10, |v| > 0, 0 < R ≤ 100).
The second line contains an integer n, which is the number of mines (1 ≤ n ≤ 100). Then follow n data blocks, the i-th of them describes the i-th mine.
The first line of each block contains 5 integers Oix, Oiy, Oiz, ri, mi, which are the coordinates of the mine centre, the radius of its body and the number of spikes (0 < ri < 100, 0 ≤ mi ≤ 10). Then follow mi lines, describing the spikes of the i-th mine, where the j-th of them describes the i-th spike and contains 3 integers pijx, pijy, pijz — the coordinates of the vector where the given spike is directed (<image>).
The coordinates of the mines' centers and the center of the Death Star are integers, their absolute value does not exceed 10000. It is guaranteed that R > ri for any 1 ≤ i ≤ n. For any mines i ≠ j the following inequality if fulfilled: <image>. Initially the Death Star and the mines do not have common points.
Output
If the rebels will succeed in stopping the Death Star using space mines, print the time from the moment the Death Star was noticed to the blast.
If the Death Star will not touch a mine, print "-1" (without quotes).
For the answer the absolute or relative error of 10 - 6 is acceptable.
Examples
Input
0 0 0 1 0 0 5
2
10 8 0 2 2
0 -3 0
2 2 0
20 0 0 4 3
2 4 0
-4 3 0
1 -5 0
Output
10.0000000000
Input
8 8 4 4 4 2 6
1
-2 -2 -1 3 0
Output
-1
Input
30 30 2 1 2 1 20
3
0 0 40 5 1
1 4 4
-10 -40 -5 7 0
100 200 95 8 1
-10 0 0
Output
74.6757620881
Submitted Solution:
```
import sys, math
def dist(a,b):
return pow((a[0] - b[0])**2 + (a[1] - b[1])**2 + (a[2] - b[2])**2,0.5)
def vxv(v1,v2):
x = -v1[2] * v2[1] + v1[1] * v2[2]
y = -v1[0] * v2[2] + v1[2] * v2[0]
z = -v1[1] * v2[0] + v1[0] * v2[1]
return (x,y,z)
readline = sys.stdin.readline
s1,s2,s3,v1,v2,v3,R = map(int,readline().split())
n = int(readline())
pm = []
p = []
for _ in range(n):
o1,o2,o3,r,m = map(int,readline().split())
p.append((o1,o2,o3,r))
for _ in range(m):
pm1,pm2,pm3 = map(int,readline().split())
pm.append((pm1 + o1, pm2 + o2, pm3 + o3))
nv = (v1,v2,v3)
m = (s1,s2,s3)
distance = []
for x in pm:
tp = (x[0] - m[0], m[1] - x[1], x[2] - m[2])
tp1 = vxv(tp,nv)
d = dist(tp1,(0,0,0))/dist(nv,(0,0,0))
if d <= R:
dd = pow(dist(x,m)**2-d**2,0.5)
dnv = dist(nv, (0, 0, 0))
nnv = (dd * nv[0] / dnv + m[0], dd * nv[1] / dnv + m[1], dd * nv[2] / dnv + m[2])
if dist(nnv, x) < dist(x, m):
distance.append(dd - pow(R**2 - d**2,0.5))
for i in range(len(p)):
pi =(p[i][0],p[i][1],p[i][2])
tp = (pi[0] - m[0], pi[1] - m[1], pi[2] - m[2])
tp1 = vxv(tp,nv)
d = dist(tp1,(0,0,0))/dist(nv,(0,0,0))
if d - p[i][3] <= R:
dd = pow(dist(pi,m)**2-d**2,0.5)
dnv = dist(nv,(0,0,0))
nnv = (dd * nv[0]/dnv + m[0], dd * nv[1]/dnv + m[1],dd * nv[2]/dnv + m[2])
if dist(nnv,p[i]) < dist(p[i],m):
dr = pow((R + p[i][3])**2 - d**2,0.5)
distance.append(dd - dr)
if len(distance):
distance.sort()
print(distance[0]/dist(nv,(0,0,0)))
else:
print(-1)
```
|
instruction
| 0
| 3,052
| 3
| 6,104
|
No
|
output
| 1
| 3,052
| 3
| 6,105
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def main():
n, I = map(int, input().split())
A = list(map(int, input().split()))
table = [0]*21
table[0] = 1
for i in range(1, 21):
table[i] = table[i-1]*2
import bisect
def is_ok(K):
k = bisect.bisect_left(table, K)
if n*k <= 8*I:
return True
else:
return False
ok = 1
ng = len(set(A))+1
while ok+1 <ng:
c = (ok+ng)//2
if is_ok(c):
ok = c
else:
ng = c
K = ok
#print(K)
from collections import Counter
C = Counter(A)
if len(C) <= K:
print(0)
exit()
C = list(C.items())
C.sort(key=lambda x: x[0])
#print(C)
S = []
for k, v in C:
S.append(v)
#print(S)
T = list(reversed(S))
from itertools import accumulate
CS = [0]+S
CT = [0]+T
CS = list(accumulate(CS))
CT = list(accumulate(CT))
#print(CS)
#print(CT)
p = len(C)-K
ans = 5*10**5
for x in range(0, p+1):
y = p-x
ans = min(ans, CS[x]+CT[y])
print(ans)
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 3,416
| 3
| 6,832
|
Yes
|
output
| 1
| 3,416
| 3
| 6,833
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
import sys
import math
# n=int(sys.stdin.readline())
n,y=list(map(int,sys.stdin.readline().strip().split()))
a=list(map(int,sys.stdin.readline().strip().split()))
a.sort()
a.append(0)
# print(a)
y=y*8
arr=[]
i=0
k=2**(y//n)
# print(k)
while(i<n):
# x=[a[i],1,math.ceil(math.log(a[i],2))+1]
# x=[a[i],1]
x=1
# print(i)
while(a[i]==a[i+1] and i<n):
# x[1]+=1
x+=1
i+=1
# print(i)
arr.append(x)
i+=1
# print(a,arr)
arr=[0]+arr
# print(arr)
for i in range(1,len(arr)):
arr[i]=arr[i]+arr[i-1]
# print(arr)
op=n
# print(k)
if(len(arr)<k):
print(0)
exit(0)
for i in range(1,len(arr)):
# xxx=arr[i]-arr[min(i-1,i-k)]
if(i<k):
xxx=arr[i]-arr[i-1]
else:
xxx=arr[i]-arr[i-k]
# print(xxx)
op=min(op,n-xxx)
print(op)
```
|
instruction
| 0
| 3,417
| 3
| 6,834
|
Yes
|
output
| 1
| 3,417
| 3
| 6,835
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
R=lambda:map(int,input().split())
n,I=R()
a=[-1]+sorted(R())
b=[i for i in range(n)if a[i]<a[i+1]]
print(n-max(y-x for x,y in zip(b,b[1<<8*I//n:]+[n])))
```
|
instruction
| 0
| 3,418
| 3
| 6,836
|
Yes
|
output
| 1
| 3,418
| 3
| 6,837
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
import math
n, I = map(int, input().split())
a = list(map(int,input().split()))
a.sort()
vals = 2**((I*8)//n)
K = len(a)
if K>vals:
d = {}
for i in a:
d[i] = d.get(i, 0) + 1
d = list(d.items())
s = sum([val for key, val in d[:vals]])
maxs = s
for i in range(vals, len(d)):
s = s - d[i-vals][1] + d[i][1]
maxs = max(maxs, s)
print(n - maxs)
else:
print(0)
```
|
instruction
| 0
| 3,419
| 3
| 6,838
|
Yes
|
output
| 1
| 3,419
| 3
| 6,839
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from math import log
from collections import deque
n, i = map(int, input().split())
l = list(map(int, input().split()))
quota = 8 * i / n
k = 1
for i in range(100):
if log(k * 2, 2) <= quota:
k *= 2
else:
break
l = sorted(l)
a = l[0]
b = l[0] + k
d = deque()
d.append(a)
i = 1
ans = 1
cnt = 1
while (i < n):
#print(d, cnt, ans)
x = l[i]
if x >= b:
while (d != deque() and d[0] <= l[i] - k):
d.popleft()
cnt -= 1
d.append(x)
cnt += 1
else:
d.append(x)
cnt += 1
ans = max(cnt, ans)
i += 1
print(n - ans)
```
|
instruction
| 0
| 3,420
| 3
| 6,840
|
No
|
output
| 1
| 3,420
| 3
| 6,841
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
n,I=map(int,input().split())
print(n,I)
a=list(map(int,input().split()))
I*=8
k=I//n
if k>=20:
res=0
else:
K=2**k
print(K,I)
cc={}
for ai in a:
cc[ai]=cc.get(ai,0)+1
pres=[]
dd=sorted(cc)
for i in dd:
pres.append(cc[i])
for i in range(1,len(pres)):
pres[i]+=pres[i-1]
if K>=len(pres):
res=0
else:
mm=0
for i in range(K,len(pres)):
if mm<pres[i]-pres[i-K]:
mm=pres[i]-pres[i-K]
res=n-mm
print(res)
```
|
instruction
| 0
| 3,421
| 3
| 6,842
|
No
|
output
| 1
| 3,421
| 3
| 6,843
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
n, l = map(int, input().split())
a = [*map(int, input().split())]
a.sort()
from math import log, ceil, exp
t = ceil(log(len(set(a)))) * n
if t <= l * 8:
print('0')
exit()
r = 0
for i in range(n, -1, -1):
if ceil(log(i)) * n <= l * 8:
r = i
break
from collections import Counter
c = Counter(a).most_common()[::-1]
ans = 0
for i in c[:len(set(a)) - r]:
ans += i[1]
print(ans)
```
|
instruction
| 0
| 3,422
| 3
| 6,844
|
No
|
output
| 1
| 3,422
| 3
| 6,845
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
#576_C
import math
l = [int(i) for i in input().split(" ")]
n = l[0]
bt = l[1]
ln = sorted([int(i) for i in input().split(" ")])
K = 1
dists = [1]
for i in range(1, len(ln)):
if ln[i] != ln[i - 1]:
K += 1
dists.append(1)
else:
dists[-1] += 1
rdists = dists[:]
ldists = dists[::-1]
minVal = math.floor(2 ** ((bt * 8) / n))
dels = 0
inds = 0
minVal = K - minVal
while dels < minVal:
if ldists[-1] < rdists[-1]:
inds += ldists[-1]
ldists.pop()
else:
inds += rdists[-1]
rdists.pop()
dels += 1
print(inds)
```
|
instruction
| 0
| 3,423
| 3
| 6,846
|
No
|
output
| 1
| 3,423
| 3
| 6,847
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are an intergalactic surgeon and you have an alien patient. For the purposes of this problem, we can and we will model this patient's body using a 2 × (2k + 1) rectangular grid. The alien has 4k + 1 distinct organs, numbered 1 to 4k + 1.
In healthy such aliens, the organs are arranged in a particular way. For example, here is how the organs of a healthy such alien would be positioned, when viewed from the top, for k = 4:
<image>
Here, the E represents empty space.
In general, the first row contains organs 1 to 2k + 1 (in that order from left to right), and the second row contains organs 2k + 2 to 4k + 1 (in that order from left to right) and then empty space right after.
Your patient's organs are complete, and inside their body, but they somehow got shuffled around! Your job, as an intergalactic surgeon, is to put everything back in its correct position. All organs of the alien must be in its body during the entire procedure. This means that at any point during the procedure, there is exactly one cell (in the grid) that is empty. In addition, you can only move organs around by doing one of the following things:
* You can switch the positions of the empty space E with any organ to its immediate left or to its immediate right (if they exist). In reality, you do this by sliding the organ in question to the empty space;
* You can switch the positions of the empty space E with any organ to its immediate top or its immediate bottom (if they exist) only if the empty space is on the leftmost column, rightmost column or in the centermost column. Again, you do this by sliding the organ in question to the empty space.
Your job is to figure out a sequence of moves you must do during the surgical procedure in order to place back all 4k + 1 internal organs of your patient in the correct cells. If it is impossible to do so, you must say so.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 4) denoting the number of test cases. The next lines contain descriptions of the test cases.
Each test case consists of three lines. The first line contains a single integer k (1 ≤ k ≤ 15) which determines the size of the grid. Then two lines follow. Each of them contains 2k + 1 space-separated integers or the letter E. They describe the first and second rows of organs, respectively. It is guaranteed that all 4k + 1 organs are present and there is exactly one E.
Output
For each test case, first, print a single line containing either:
* SURGERY COMPLETE if it is possible to place back all internal organs in the correct locations;
* SURGERY FAILED if it is impossible.
If it is impossible, then this is the only line of output for the test case. However, if it is possible, output a few more lines describing the sequence of moves to place the organs in the correct locations.
The sequence of moves will be a (possibly empty) string of letters u, d, l or r, representing sliding the organ that's directly above, below, to the left or to the right of the empty space, respectively, into the empty space. Print the sequence of moves in the following line, as such a string.
For convenience, you may use shortcuts to reduce the size of your output. You may use uppercase letters as shortcuts for sequences of moves. For example, you could choose T to represent the string lddrr. These shortcuts may also include other shortcuts on their own! For example, you could choose E to represent TruT, etc.
You may use any number of uppercase letters (including none) as shortcuts. The only requirements are the following:
* The total length of all strings in your output for a single case is at most 10^4;
* There must be no cycles involving the shortcuts that are reachable from the main sequence;
* The resulting sequence of moves is finite, after expanding all shortcuts. Note that the final sequence of moves (after expanding) may be much longer than 10^4; the only requirement is that it's finite.
As an example, if T = lddrr, E = TruT and R = rrr, then TurTlER expands to:
* TurTlER
* lddrrurTlER
* lddrrurlddrrlER
* lddrrurlddrrlTruTR
* lddrrurlddrrllddrrruTR
* lddrrurlddrrllddrrrulddrrR
* lddrrurlddrrllddrrrulddrrrrr
To use shortcuts, print each one of them in a single line as the uppercase letter, then space, and then the string that this shortcut represents. They may be printed in any order. At the end of all of those, print a single line containing DONE.
Note: You still need to print DONE even if you don't plan on using shortcuts.
Your sequence does not need to be the shortest. Any valid sequence of moves (satisfying the requirements above) will be accepted.
Example
Input
2
3
1 2 3 5 6 E 7
8 9 10 4 11 12 13
11
34 45 6 22 16 43 38 44 5 4 41 14 7 29 28 19 9 18 42 8 17 33 1
E 15 40 36 31 24 10 2 21 11 32 23 30 27 35 25 13 12 39 37 26 20 3
Output
SURGERY COMPLETE
IR
R SrS
S rr
I lldll
DONE
SURGERY FAILED
Note
There are three shortcuts defined in the first sample output:
* R = SrS
* S = rr
* I = lldll
The sequence of moves is IR and it expands to:
* IR
* lldllR
* lldllSrS
* lldllrrrS
* lldllrrrrr
|
instruction
| 0
| 3,461
| 3
| 6,922
|
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
def flatten(grid):
k = len(grid[0]) // 2
seek = list(range(2*k + 2)) + list(range(2*k + 2, 4*k + 2))[::-1]
return [seek[v] for v in grid[0] + grid[1][::-1] if v]
def solve(grid):
grid = list(map(list, grid))
k = len(grid[0]) // 2
flat = flatten(grid)
m = {
'L': 'l'*2*k + 'u' + 'r'*2*k + 'd',
'R': 'u' + 'l'*2*k + 'd' + 'r'*2*k,
'C': 'l'*k + 'u' + 'r'*k + 'd',
'D': 'CC' + 'R'*(2*k + 1) + 'CC' + 'R'*(2*k + 2),
'F': 'R'*(k - 1) + 'DD' + 'R'*(2*k + 1) + 'D' + 'L'*2*k + 'DD' + 'L'*k,
'G': 'FF',
}
[(i, j)] = [(i, j) for i in range(2) for j in range(2*k + 1) if grid[i][j] == 0]
st = 'r'*(2*k - j) + 'd'*(1 - i)
for v in range(2, 4*k + 2):
ct = flat.index(v)
if ct >= 2:
st += 'L'*(ct - 2) + 'GR'*(ct - 2) + 'G'
flat = flat[ct - 1: ct + 1] + flat[:ct - 1] + flat[ct + 1:]
if ct >= 1:
st += 'G'
flat = flat[1:3] + flat[:1] + flat[3:]
st += 'L'
flat = flat[1:] + flat[:1]
if flat[0] == 1: return st, m
def main():
def get_line():
return [0 if x == 'E' else int(x) for x in input().split()]
for cas in range(int(input())):
k = int(input())
grid = [get_line() for i in range(2)]
assert all(len(row) == 2*k + 1 for row in grid)
res = solve(grid)
if res is None:
print('SURGERY FAILED')
else:
print('SURGERY COMPLETE')
st, m = res
print(st)
for shortcut in m.items(): print(*shortcut)
print('DONE')
main()
```
|
output
| 1
| 3,461
| 3
| 6,923
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are an intergalactic surgeon and you have an alien patient. For the purposes of this problem, we can and we will model this patient's body using a 2 × (2k + 1) rectangular grid. The alien has 4k + 1 distinct organs, numbered 1 to 4k + 1.
In healthy such aliens, the organs are arranged in a particular way. For example, here is how the organs of a healthy such alien would be positioned, when viewed from the top, for k = 4:
<image>
Here, the E represents empty space.
In general, the first row contains organs 1 to 2k + 1 (in that order from left to right), and the second row contains organs 2k + 2 to 4k + 1 (in that order from left to right) and then empty space right after.
Your patient's organs are complete, and inside their body, but they somehow got shuffled around! Your job, as an intergalactic surgeon, is to put everything back in its correct position. All organs of the alien must be in its body during the entire procedure. This means that at any point during the procedure, there is exactly one cell (in the grid) that is empty. In addition, you can only move organs around by doing one of the following things:
* You can switch the positions of the empty space E with any organ to its immediate left or to its immediate right (if they exist). In reality, you do this by sliding the organ in question to the empty space;
* You can switch the positions of the empty space E with any organ to its immediate top or its immediate bottom (if they exist) only if the empty space is on the leftmost column, rightmost column or in the centermost column. Again, you do this by sliding the organ in question to the empty space.
Your job is to figure out a sequence of moves you must do during the surgical procedure in order to place back all 4k + 1 internal organs of your patient in the correct cells. If it is impossible to do so, you must say so.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 4) denoting the number of test cases. The next lines contain descriptions of the test cases.
Each test case consists of three lines. The first line contains a single integer k (1 ≤ k ≤ 15) which determines the size of the grid. Then two lines follow. Each of them contains 2k + 1 space-separated integers or the letter E. They describe the first and second rows of organs, respectively. It is guaranteed that all 4k + 1 organs are present and there is exactly one E.
Output
For each test case, first, print a single line containing either:
* SURGERY COMPLETE if it is possible to place back all internal organs in the correct locations;
* SURGERY FAILED if it is impossible.
If it is impossible, then this is the only line of output for the test case. However, if it is possible, output a few more lines describing the sequence of moves to place the organs in the correct locations.
The sequence of moves will be a (possibly empty) string of letters u, d, l or r, representing sliding the organ that's directly above, below, to the left or to the right of the empty space, respectively, into the empty space. Print the sequence of moves in the following line, as such a string.
For convenience, you may use shortcuts to reduce the size of your output. You may use uppercase letters as shortcuts for sequences of moves. For example, you could choose T to represent the string lddrr. These shortcuts may also include other shortcuts on their own! For example, you could choose E to represent TruT, etc.
You may use any number of uppercase letters (including none) as shortcuts. The only requirements are the following:
* The total length of all strings in your output for a single case is at most 10^4;
* There must be no cycles involving the shortcuts that are reachable from the main sequence;
* The resulting sequence of moves is finite, after expanding all shortcuts. Note that the final sequence of moves (after expanding) may be much longer than 10^4; the only requirement is that it's finite.
As an example, if T = lddrr, E = TruT and R = rrr, then TurTlER expands to:
* TurTlER
* lddrrurTlER
* lddrrurlddrrlER
* lddrrurlddrrlTruTR
* lddrrurlddrrllddrrruTR
* lddrrurlddrrllddrrrulddrrR
* lddrrurlddrrllddrrrulddrrrrr
To use shortcuts, print each one of them in a single line as the uppercase letter, then space, and then the string that this shortcut represents. They may be printed in any order. At the end of all of those, print a single line containing DONE.
Note: You still need to print DONE even if you don't plan on using shortcuts.
Your sequence does not need to be the shortest. Any valid sequence of moves (satisfying the requirements above) will be accepted.
Example
Input
2
3
1 2 3 5 6 E 7
8 9 10 4 11 12 13
11
34 45 6 22 16 43 38 44 5 4 41 14 7 29 28 19 9 18 42 8 17 33 1
E 15 40 36 31 24 10 2 21 11 32 23 30 27 35 25 13 12 39 37 26 20 3
Output
SURGERY COMPLETE
IR
R SrS
S rr
I lldll
DONE
SURGERY FAILED
Note
There are three shortcuts defined in the first sample output:
* R = SrS
* S = rr
* I = lldll
The sequence of moves is IR and it expands to:
* IR
* lldllR
* lldllSrS
* lldllrrrS
* lldllrrrrr
|
instruction
| 0
| 3,462
| 3
| 6,924
|
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
def solve(k, grid):
seek = *range(2*k + 2), *range(4*k + 1, 2*k + 1, -1)
flat = [seek[v] for v in grid[0] + grid[1][::-1] if v]
m = {
'L': 'l'*2*k + 'u' + 'r'*2*k + 'd',
'R': 'u' + 'l'*2*k + 'd' + 'r'*2*k,
'C': 'l'*k + 'u' + 'r'*k + 'd',
'D': 'CC' + 'R'*(2*k + 1) + 'CC' + 'R'*(2*k + 2),
'F': 'R'*(k - 1) + 'DD' + 'R'*(2*k + 1) + 'D' + 'L'*2*k + 'DD' + 'L'*k,
'G': 'FF',
}
[(i, j)] = [(i, j) for i in range(2) for j in range(2*k + 1) if grid[i][j] == 0]
st = 'r'*(2*k - j) + 'd'*(1 - i)
for v in range(2, 4*k + 2):
ct = flat.index(v)
if ct >= 2:
st += 'L'*(ct - 2) + 'GR'*(ct - 2) + 'G'
flat = flat[ct - 1: ct + 1] + flat[:ct - 1] + flat[ct + 1:]
if ct >= 1:
st += 'G'
flat = flat[1:3] + flat[:1] + flat[3:]
st += 'L'
flat = flat[1:] + flat[:1]
if flat[0] == 1: return st, m
def main():
def get_line():
return [0 if x == 'E' else int(x) for x in input().split()]
for cas in range(int(input())):
res = solve(int(input()), [get_line() for i in range(2)])
if res is None:
print('SURGERY FAILED')
else:
print('SURGERY COMPLETE')
st, m = res
print(st)
for shortcut in m.items(): print(*shortcut)
print('DONE')
main()
```
|
output
| 1
| 3,462
| 3
| 6,925
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are an intergalactic surgeon and you have an alien patient. For the purposes of this problem, we can and we will model this patient's body using a 2 × (2k + 1) rectangular grid. The alien has 4k + 1 distinct organs, numbered 1 to 4k + 1.
In healthy such aliens, the organs are arranged in a particular way. For example, here is how the organs of a healthy such alien would be positioned, when viewed from the top, for k = 4:
<image>
Here, the E represents empty space.
In general, the first row contains organs 1 to 2k + 1 (in that order from left to right), and the second row contains organs 2k + 2 to 4k + 1 (in that order from left to right) and then empty space right after.
Your patient's organs are complete, and inside their body, but they somehow got shuffled around! Your job, as an intergalactic surgeon, is to put everything back in its correct position. All organs of the alien must be in its body during the entire procedure. This means that at any point during the procedure, there is exactly one cell (in the grid) that is empty. In addition, you can only move organs around by doing one of the following things:
* You can switch the positions of the empty space E with any organ to its immediate left or to its immediate right (if they exist). In reality, you do this by sliding the organ in question to the empty space;
* You can switch the positions of the empty space E with any organ to its immediate top or its immediate bottom (if they exist) only if the empty space is on the leftmost column, rightmost column or in the centermost column. Again, you do this by sliding the organ in question to the empty space.
Your job is to figure out a sequence of moves you must do during the surgical procedure in order to place back all 4k + 1 internal organs of your patient in the correct cells. If it is impossible to do so, you must say so.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 4) denoting the number of test cases. The next lines contain descriptions of the test cases.
Each test case consists of three lines. The first line contains a single integer k (1 ≤ k ≤ 15) which determines the size of the grid. Then two lines follow. Each of them contains 2k + 1 space-separated integers or the letter E. They describe the first and second rows of organs, respectively. It is guaranteed that all 4k + 1 organs are present and there is exactly one E.
Output
For each test case, first, print a single line containing either:
* SURGERY COMPLETE if it is possible to place back all internal organs in the correct locations;
* SURGERY FAILED if it is impossible.
If it is impossible, then this is the only line of output for the test case. However, if it is possible, output a few more lines describing the sequence of moves to place the organs in the correct locations.
The sequence of moves will be a (possibly empty) string of letters u, d, l or r, representing sliding the organ that's directly above, below, to the left or to the right of the empty space, respectively, into the empty space. Print the sequence of moves in the following line, as such a string.
For convenience, you may use shortcuts to reduce the size of your output. You may use uppercase letters as shortcuts for sequences of moves. For example, you could choose T to represent the string lddrr. These shortcuts may also include other shortcuts on their own! For example, you could choose E to represent TruT, etc.
You may use any number of uppercase letters (including none) as shortcuts. The only requirements are the following:
* The total length of all strings in your output for a single case is at most 10^4;
* There must be no cycles involving the shortcuts that are reachable from the main sequence;
* The resulting sequence of moves is finite, after expanding all shortcuts. Note that the final sequence of moves (after expanding) may be much longer than 10^4; the only requirement is that it's finite.
As an example, if T = lddrr, E = TruT and R = rrr, then TurTlER expands to:
* TurTlER
* lddrrurTlER
* lddrrurlddrrlER
* lddrrurlddrrlTruTR
* lddrrurlddrrllddrrruTR
* lddrrurlddrrllddrrrulddrrR
* lddrrurlddrrllddrrrulddrrrrr
To use shortcuts, print each one of them in a single line as the uppercase letter, then space, and then the string that this shortcut represents. They may be printed in any order. At the end of all of those, print a single line containing DONE.
Note: You still need to print DONE even if you don't plan on using shortcuts.
Your sequence does not need to be the shortest. Any valid sequence of moves (satisfying the requirements above) will be accepted.
Example
Input
2
3
1 2 3 5 6 E 7
8 9 10 4 11 12 13
11
34 45 6 22 16 43 38 44 5 4 41 14 7 29 28 19 9 18 42 8 17 33 1
E 15 40 36 31 24 10 2 21 11 32 23 30 27 35 25 13 12 39 37 26 20 3
Output
SURGERY COMPLETE
IR
R SrS
S rr
I lldll
DONE
SURGERY FAILED
Note
There are three shortcuts defined in the first sample output:
* R = SrS
* S = rr
* I = lldll
The sequence of moves is IR and it expands to:
* IR
* lldllR
* lldllSrS
* lldllrrrS
* lldllrrrrr
|
instruction
| 0
| 3,463
| 3
| 6,926
|
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
def solve(kk,grid):
sk = *range(kk * 2 + 2),*range(4 * kk + 1,2 * kk + 1,-1)
flat = [sk[v] for v in grid[0] + grid[1][::-1] if v]
dir = {
'L': 'l' * 2 * kk + 'u' + 'r' * 2 * kk + 'd',
'R': 'u' + 'l' * 2 * kk + 'd' + 'r' * 2 * kk,
'C': 'l' * kk + 'u' + 'r' * kk + 'd',
'D': 'CC' + 'R' * (2 * kk + 1) + 'CC' + 'R' * (2 * kk + 2),
'F': 'R' * (kk - 1) + 'DD' + 'R' * (2 * kk + 1) + 'D' + 'L' * 2 * kk + 'DD' +
'L' * kk,
'G': 'FF'
}
[(i,j)] = [(i,j) for i in range(2) for j in range(2 * kk + 1) if grid[i][j] == 0]
st = 'r' * (2 * kk - j) + 'd' * (1 - i)
for i in range(2,4 * kk + 2):
tt = flat.index(i)
if tt >= 2:
st += 'L' * (tt - 2) + 'GR' * (tt - 2) + 'G'
flat = flat[tt - 1 : tt + 1] + flat[:tt - 1] + flat[tt + 1:]
if tt >= 1:
st += 'G'
flat = flat[1 : 3] + flat[:1] + flat[3:]
st += 'L'
flat = flat[1:] + flat[:1]
if flat[0] == 1: return st,dir
def getline():
return [0 if x == 'E' else int(x) for x in input().split()]
for case in range(int(input())):
ret = solve(int(input()),[getline() for i in range(2)])
if ret is None:
print('SURGERY FAILED')
else:
print('SURGERY COMPLETE')
st,vv = ret
print(st)
for shortcut in vv.items() : print(*shortcut)
print('DONE')
```
|
output
| 1
| 3,463
| 3
| 6,927
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are an intergalactic surgeon and you have an alien patient. For the purposes of this problem, we can and we will model this patient's body using a 2 × (2k + 1) rectangular grid. The alien has 4k + 1 distinct organs, numbered 1 to 4k + 1.
In healthy such aliens, the organs are arranged in a particular way. For example, here is how the organs of a healthy such alien would be positioned, when viewed from the top, for k = 4:
<image>
Here, the E represents empty space.
In general, the first row contains organs 1 to 2k + 1 (in that order from left to right), and the second row contains organs 2k + 2 to 4k + 1 (in that order from left to right) and then empty space right after.
Your patient's organs are complete, and inside their body, but they somehow got shuffled around! Your job, as an intergalactic surgeon, is to put everything back in its correct position. All organs of the alien must be in its body during the entire procedure. This means that at any point during the procedure, there is exactly one cell (in the grid) that is empty. In addition, you can only move organs around by doing one of the following things:
* You can switch the positions of the empty space E with any organ to its immediate left or to its immediate right (if they exist). In reality, you do this by sliding the organ in question to the empty space;
* You can switch the positions of the empty space E with any organ to its immediate top or its immediate bottom (if they exist) only if the empty space is on the leftmost column, rightmost column or in the centermost column. Again, you do this by sliding the organ in question to the empty space.
Your job is to figure out a sequence of moves you must do during the surgical procedure in order to place back all 4k + 1 internal organs of your patient in the correct cells. If it is impossible to do so, you must say so.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 4) denoting the number of test cases. The next lines contain descriptions of the test cases.
Each test case consists of three lines. The first line contains a single integer k (1 ≤ k ≤ 15) which determines the size of the grid. Then two lines follow. Each of them contains 2k + 1 space-separated integers or the letter E. They describe the first and second rows of organs, respectively. It is guaranteed that all 4k + 1 organs are present and there is exactly one E.
Output
For each test case, first, print a single line containing either:
* SURGERY COMPLETE if it is possible to place back all internal organs in the correct locations;
* SURGERY FAILED if it is impossible.
If it is impossible, then this is the only line of output for the test case. However, if it is possible, output a few more lines describing the sequence of moves to place the organs in the correct locations.
The sequence of moves will be a (possibly empty) string of letters u, d, l or r, representing sliding the organ that's directly above, below, to the left or to the right of the empty space, respectively, into the empty space. Print the sequence of moves in the following line, as such a string.
For convenience, you may use shortcuts to reduce the size of your output. You may use uppercase letters as shortcuts for sequences of moves. For example, you could choose T to represent the string lddrr. These shortcuts may also include other shortcuts on their own! For example, you could choose E to represent TruT, etc.
You may use any number of uppercase letters (including none) as shortcuts. The only requirements are the following:
* The total length of all strings in your output for a single case is at most 10^4;
* There must be no cycles involving the shortcuts that are reachable from the main sequence;
* The resulting sequence of moves is finite, after expanding all shortcuts. Note that the final sequence of moves (after expanding) may be much longer than 10^4; the only requirement is that it's finite.
As an example, if T = lddrr, E = TruT and R = rrr, then TurTlER expands to:
* TurTlER
* lddrrurTlER
* lddrrurlddrrlER
* lddrrurlddrrlTruTR
* lddrrurlddrrllddrrruTR
* lddrrurlddrrllddrrrulddrrR
* lddrrurlddrrllddrrrulddrrrrr
To use shortcuts, print each one of them in a single line as the uppercase letter, then space, and then the string that this shortcut represents. They may be printed in any order. At the end of all of those, print a single line containing DONE.
Note: You still need to print DONE even if you don't plan on using shortcuts.
Your sequence does not need to be the shortest. Any valid sequence of moves (satisfying the requirements above) will be accepted.
Example
Input
2
3
1 2 3 5 6 E 7
8 9 10 4 11 12 13
11
34 45 6 22 16 43 38 44 5 4 41 14 7 29 28 19 9 18 42 8 17 33 1
E 15 40 36 31 24 10 2 21 11 32 23 30 27 35 25 13 12 39 37 26 20 3
Output
SURGERY COMPLETE
IR
R SrS
S rr
I lldll
DONE
SURGERY FAILED
Note
There are three shortcuts defined in the first sample output:
* R = SrS
* S = rr
* I = lldll
The sequence of moves is IR and it expands to:
* IR
* lldllR
* lldllSrS
* lldllrrrS
* lldllrrrrr
|
instruction
| 0
| 3,464
| 3
| 6,928
|
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
import sys
OP_CW1 = 'A'
OP_CW4 = 'B'
OP_CW16 = 'C'
OP_CCW = 'E'
OP_L = 'U'
OP_INV_L = 'V'
OP_R = 'W'
OP_INV_R = 'X'
OP_S = 'S'
OP_MOVE_LL1 = 'O'
OP_MOVE_LL4 = 'P'
OP_MOVE_LL16 = 'Q'
OP_MOVE_L_THROW = 'R'
def GetK(board): return len(board) // 4
def Slot(board, slotval=-1): return board.index(slotval)
def KS(board, slotval=-1): return GetK(board), Slot(board, slotval)
def Width(K): return K * 2 + 1
def PrintPermumation(head, board, slotval=None):
K = GetK(board)
W = Width(K)
if slotval is None: slotval = K
slot = Slot(board, slotval)
print(head, '===', file=sys.stderr)
tostr = lambda x: 'EE' if x == slotval else '%2d' % x
for row in [board[:W], board[W:]]:
print(' '.join(tostr(x) for x in row), file=sys.stderr)
def ApplyOps(board, seq, ops=None, slotval=-1):
K, slot = KS(board, slotval)
W = Width(K)
for c in seq:
if c == 'l':
assert slot % W > 0, slot
s1 = slot - 1
elif c == 'r':
assert slot % W + 1 < W, slot
s1 = slot + 1
elif c == 'd':
assert slot % W in [0, K, K * 2], slot
assert slot // W == 0
s1 = slot + W
elif c == 'u':
assert slot % W in [0, K, K * 2], slot
assert slot // W == 1
s1 = slot - W
else:
s1 = None
if s1 is not None:
board[s1], board[slot] = board[slot], board[s1]
else:
P = ops[c]
tmp = board[:]
for i in range(len(board)): board[i] = tmp[P[i]]
slot = Slot(board, slotval)
def LoopedView(board):
K, slot = KS(board)
W = Width(slot)
assert slot == K
def translate(x):
if x == -1: return x
if x <= W: return x - 1
# W+1 -> 2W-2
return 3 * W - 1 - x
r = []
for rng in [range(slot + 1, W), reversed(range(W, W * 2)), range(0, slot)]:
for i in rng:
r.append(translate(board[i]))
return r
def Mod(x, m):
x %= m
if x < 0: x += m
return x
def CanonicalizeView(view):
i = view.index(0)
view[:] = view[i:] + view[:i]
def ToPermutation(K, ops, seq):
board = list(range(K * 4 + 2))
ApplyOps(board, seq, ops, slotval=K)
return board
def InitOps(K, debug=False):
W = Width(K)
ops = {}
ops_codes = {}
def ToP(seq): return ToPermutation(K, ops, seq)
def Assign(rep, seq):
ops[rep] = ToP(seq)
ops_codes[rep] = seq
Assign(OP_CW1, 'l' * K + 'd' + 'r' * (K * 2) + 'u' + 'l' * K)
if debug: PrintPermumation('CW1', ops[OP_CW1])
Assign(OP_CW4, OP_CW1 * 4)
if debug: PrintPermumation('CW4', ops[OP_CW4])
Assign(OP_CW16, OP_CW4 * 4)
Assign(OP_CCW, OP_CW1 * (W * 2 - 2))
if debug: PrintPermumation('CCW', ops[OP_CCW])
Assign(OP_L, 'l' * K + 'd' + 'r' * K + 'u')
if debug: PrintPermumation('L', ops[OP_L])
Assign(OP_INV_L, OP_L * (K * 2))
if debug: PrintPermumation('INV_L', ops[OP_INV_L])
Assign(OP_R, 'r' * K + 'd' + 'l' * K + 'u')
if debug: PrintPermumation('R', ops[OP_R])
Assign(OP_INV_R, OP_R * (K * 2))
if debug: PrintPermumation('INV_R', ops[OP_INV_R])
Assign(OP_S, OP_INV_R + OP_INV_L + OP_R + OP_L)
if debug: PrintPermumation('S', ops[OP_S])
Assign(OP_MOVE_LL1, OP_CW1 + OP_S + OP_CW1)
Assign(OP_MOVE_LL4, OP_MOVE_LL1 * 4)
Assign(OP_MOVE_LL16, OP_MOVE_LL4 * 4)
Assign(OP_MOVE_L_THROW, OP_S * 2 + OP_CW1)
if debug:
PrintPermumation('MOVE_LL', ops[OP_MOVE_LL1])
PrintPermumation('MOVE_L_THROW', ops[OP_MOVE_L_THROW])
return ops, ops_codes
def Solve(board, debug=False):
ans = []
K, slot = KS(board)
W = Width(K)
if slot % W < K:
t = 'r' * (K - slot % W)
else:
t = 'l' * (slot % W - K)
if slot // W == 1:
t = t + 'u'
ans.append(t)
ApplyOps(board, t)
slot = Slot(board)
assert slot == K
ops, ops_codes = InitOps(K)
def ApplyMultipleOp(num, items):
for k, op in reversed(items):
while num >= k:
num -= k
ApplyOps(board, op, ops)
ans.append(op)
for k in range(1, W * 2 - 1):
view = LoopedView(board)
if debug: print('=' * 20, k, '\n', view, file=sys.stderr)
num_cw = Mod(+(K * 2 - view.index(k)), len(view))
CanonicalizeView(view)
num_move = view.index(k) - k
if debug: print(num_cw, num_move, file=sys.stderr)
ApplyMultipleOp(num_cw, [(1, OP_CW1), (4, OP_CW4), (16, OP_CW16)])
if debug: PrintPermumation('t', board, slotval=-1)
ApplyMultipleOp(num_move // 2, [(1, OP_MOVE_LL1), (4, OP_MOVE_LL4), (16, OP_MOVE_LL16)])
if debug: PrintPermumation('tt', board, slotval=-1)
if num_move % 2 > 0:
if k == W * 2 - 2: return False
ApplyOps(board, OP_MOVE_L_THROW, ops)
ans.append(OP_MOVE_L_THROW)
view = LoopedView(board)
num_cw = Mod(K * 3 + 1 - view.index(0), len(view))
ApplyMultipleOp(num_cw, [(1, OP_CW1), (4, OP_CW4), (16, OP_CW16)])
if debug: PrintPermumation('ttt', board, slotval=-1)
f = 'r' * K + 'd'
ApplyOps(board, f)
ans.append(f)
if debug: PrintPermumation('ttt', board, slotval=-1)
print('SURGERY COMPLETE')
print(''.join(ans))
for k, v in ops_codes.items():
print(k, v)
print('DONE')
return True
#ops = InitOps(11, debug=True)
#import random
#board = list(range(10 * 4 + 2))
#random.shuffle(board)
#print(board)
##board = [4, 35, 3, 15, 38, 1, 31, 7, 42, 26, 19, 27, 5, 11, 2, 33, 40, 16, 20, 12, 18, 24, 28, 6, 36, 22, 9, 29, 30, 43, 21, 10, 17, 0, 13, 8, 45, 25, 37, 32, 44, 14, 39, 34, 23, 41]
#if True:
# board[board.index(0)] = -1
# a = Solve(board, debug=True)
# print(a, file=sys.stderr)
for _ in range(int(sys.stdin.readline())):
sys.stdin.readline()
tr = lambda s: -1 if s == 'E' else int(s)
row1 = list(map(tr, sys.stdin.readline().split()))
row2 = list(map(tr, sys.stdin.readline().split()))
if not Solve(row1 + row2): print('SURGERY FAILED')
```
|
output
| 1
| 3,464
| 3
| 6,929
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are an intergalactic surgeon and you have an alien patient. For the purposes of this problem, we can and we will model this patient's body using a 2 × (2k + 1) rectangular grid. The alien has 4k + 1 distinct organs, numbered 1 to 4k + 1.
In healthy such aliens, the organs are arranged in a particular way. For example, here is how the organs of a healthy such alien would be positioned, when viewed from the top, for k = 4:
<image>
Here, the E represents empty space.
In general, the first row contains organs 1 to 2k + 1 (in that order from left to right), and the second row contains organs 2k + 2 to 4k + 1 (in that order from left to right) and then empty space right after.
Your patient's organs are complete, and inside their body, but they somehow got shuffled around! Your job, as an intergalactic surgeon, is to put everything back in its correct position. All organs of the alien must be in its body during the entire procedure. This means that at any point during the procedure, there is exactly one cell (in the grid) that is empty. In addition, you can only move organs around by doing one of the following things:
* You can switch the positions of the empty space E with any organ to its immediate left or to its immediate right (if they exist). In reality, you do this by sliding the organ in question to the empty space;
* You can switch the positions of the empty space E with any organ to its immediate top or its immediate bottom (if they exist) only if the empty space is on the leftmost column, rightmost column or in the centermost column. Again, you do this by sliding the organ in question to the empty space.
Your job is to figure out a sequence of moves you must do during the surgical procedure in order to place back all 4k + 1 internal organs of your patient in the correct cells. If it is impossible to do so, you must say so.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 4) denoting the number of test cases. The next lines contain descriptions of the test cases.
Each test case consists of three lines. The first line contains a single integer k (1 ≤ k ≤ 15) which determines the size of the grid. Then two lines follow. Each of them contains 2k + 1 space-separated integers or the letter E. They describe the first and second rows of organs, respectively. It is guaranteed that all 4k + 1 organs are present and there is exactly one E.
Output
For each test case, first, print a single line containing either:
* SURGERY COMPLETE if it is possible to place back all internal organs in the correct locations;
* SURGERY FAILED if it is impossible.
If it is impossible, then this is the only line of output for the test case. However, if it is possible, output a few more lines describing the sequence of moves to place the organs in the correct locations.
The sequence of moves will be a (possibly empty) string of letters u, d, l or r, representing sliding the organ that's directly above, below, to the left or to the right of the empty space, respectively, into the empty space. Print the sequence of moves in the following line, as such a string.
For convenience, you may use shortcuts to reduce the size of your output. You may use uppercase letters as shortcuts for sequences of moves. For example, you could choose T to represent the string lddrr. These shortcuts may also include other shortcuts on their own! For example, you could choose E to represent TruT, etc.
You may use any number of uppercase letters (including none) as shortcuts. The only requirements are the following:
* The total length of all strings in your output for a single case is at most 10^4;
* There must be no cycles involving the shortcuts that are reachable from the main sequence;
* The resulting sequence of moves is finite, after expanding all shortcuts. Note that the final sequence of moves (after expanding) may be much longer than 10^4; the only requirement is that it's finite.
As an example, if T = lddrr, E = TruT and R = rrr, then TurTlER expands to:
* TurTlER
* lddrrurTlER
* lddrrurlddrrlER
* lddrrurlddrrlTruTR
* lddrrurlddrrllddrrruTR
* lddrrurlddrrllddrrrulddrrR
* lddrrurlddrrllddrrrulddrrrrr
To use shortcuts, print each one of them in a single line as the uppercase letter, then space, and then the string that this shortcut represents. They may be printed in any order. At the end of all of those, print a single line containing DONE.
Note: You still need to print DONE even if you don't plan on using shortcuts.
Your sequence does not need to be the shortest. Any valid sequence of moves (satisfying the requirements above) will be accepted.
Example
Input
2
3
1 2 3 5 6 E 7
8 9 10 4 11 12 13
11
34 45 6 22 16 43 38 44 5 4 41 14 7 29 28 19 9 18 42 8 17 33 1
E 15 40 36 31 24 10 2 21 11 32 23 30 27 35 25 13 12 39 37 26 20 3
Output
SURGERY COMPLETE
IR
R SrS
S rr
I lldll
DONE
SURGERY FAILED
Note
There are three shortcuts defined in the first sample output:
* R = SrS
* S = rr
* I = lldll
The sequence of moves is IR and it expands to:
* IR
* lldllR
* lldllSrS
* lldllrrrS
* lldllrrrrr
Submitted Solution:
```
import sys
OP_CW1 = 'A'
OP_CW4 = 'B'
OP_CW16 = 'C'
OP_CCW = 'E'
OP_L = 'U'
OP_INV_L = 'V'
OP_R = 'W'
OP_INV_R = 'X'
OP_S = 'S'
OP_MOVE_LL1 = 'O'
OP_MOVE_LL4 = 'P'
OP_MOVE_LL16 = 'Q'
OP_MOVE_L_THROW = 'R'
def GetK(board): return len(board) // 4
def Slot(board, slotval=-1): return board.index(slotval)
def KS(board, slotval=-1): return GetK(board), Slot(board, slotval)
def Width(K): return K * 2 + 1
def PrintPermumation(head, board, slotval=None):
K = GetK(board)
W = Width(K)
if slotval is None: slotval = K
slot = Slot(board, slotval)
print(head, '===', file=sys.stderr)
tostr = lambda x: 'EE' if x == slotval else '%2d' % x
for row in [board[:W], board[W:]]:
print(' '.join(tostr(x) for x in row), file=sys.stderr)
def ApplyOps(board, seq, ops=None, slotval=-1):
K, slot = KS(board, slotval)
W = Width(K)
for c in seq:
if c == 'l':
assert slot % W > 0, slot
s1 = slot - 1
elif c == 'r':
assert slot % W + 1 < W, slot
s1 = slot + 1
elif c == 'd':
assert slot % W in [0, K, K * 2], slot
assert slot // W == 0
s1 = slot + W
elif c == 'u':
assert slot % W in [0, K, K * 2], slot
assert slot // W == 1
s1 = slot - W
else:
s1 = None
if s1 is not None:
board[s1], board[slot] = board[slot], board[s1]
else:
P = ops[c]
tmp = board[:]
for i in range(len(board)): board[i] = tmp[P[i]]
slot = Slot(board, slotval)
def LoopedView(board):
K, slot = KS(board)
W = Width(slot)
assert slot == K
def translate(x):
if x == -1: return x
if x <= W: return x - 1
# W+1 -> 2W-2
return 3 * W - 1 - x
r = []
for rng in [range(slot + 1, W), reversed(range(W, W * 2)), range(0, slot)]:
for i in rng:
r.append(translate(board[i]))
return r
def Mod(x, m):
x %= m
if x < 0: x += m
return x
def CanonicalizeView(view):
i = view.index(0)
view[:] = view[i:] + view[:i]
def ToPermutation(K, ops, seq):
board = list(range(K * 4 + 2))
ApplyOps(board, seq, ops, slotval=K)
return board
def InitOps(K, debug=False):
W = Width(K)
ops = {}
ops_codes = {}
def ToP(seq): return ToPermutation(K, ops, seq)
def Assign(rep, seq):
ops[rep] = ToP(seq)
ops_codes[rep] = seq
Assign(OP_CW1, 'l' * K + 'd' + 'r' * (K * 2) + 'u' + 'l' * K)
if debug: PrintPermumation('CW1', ops[OP_CW1])
Assign(OP_CW4, OP_CW1 * 4)
if debug: PrintPermumation('CW4', ops[OP_CW4])
Assign(OP_CW16, OP_CW4 * 4)
Assign(OP_CCW, OP_CW1 * (W * 2 - 2))
if debug: PrintPermumation('CCW', ops[OP_CCW])
Assign(OP_L, 'l' * K + 'd' + 'r' * K + 'u')
if debug: PrintPermumation('L', ops[OP_L])
Assign(OP_INV_L, OP_L * (K * 2))
if debug: PrintPermumation('INV_L', ops[OP_INV_L])
Assign(OP_R, 'r' * K + 'd' + 'l' * K + 'u')
if debug: PrintPermumation('R', ops[OP_R])
Assign(OP_INV_R, OP_R * (K * 2))
if debug: PrintPermumation('INV_R', ops[OP_INV_R])
Assign(OP_S, OP_INV_R + OP_INV_L + OP_R + OP_L)
if debug: PrintPermumation('S', ops[OP_S])
Assign(OP_MOVE_LL1, OP_CW1 + OP_S + OP_CW1)
Assign(OP_MOVE_LL4, OP_MOVE_LL1 * 4)
Assign(OP_MOVE_LL16, OP_MOVE_LL4 * 4)
Assign(OP_MOVE_L_THROW, OP_S * 2 + OP_CW1)
if debug:
PrintPermumation('MOVE_LL', ops[OP_MOVE_LL1])
PrintPermumation('MOVE_L_THROW', ops[OP_MOVE_L_THROW])
return ops, ops_codes
def Solve(board, debug=False):
ans = []
K, slot = KS(board)
W = Width(K)
if slot % W < K:
t = 'r' * (K - slot % W)
else:
t = 'l' * (slot % W - K)
if slot // W == 1:
t = t + 'u'
ans.append(t)
ApplyOps(board, t)
slot = Slot(board)
assert slot == K
ops, ops_codes = InitOps(K)
def ApplyMultipleOp(num, items):
for k, op in reversed(items):
while num >= k:
num -= k
ApplyOps(board, op, ops)
ans.append(op)
for k in range(1, W * 2 - 1):
view = LoopedView(board)
if debug: print('=' * 20, k, '\n', view, file=sys.stderr)
num_cw = Mod(+(K * 2 - view.index(k)), len(view))
CanonicalizeView(view)
num_move = view.index(k) - k
if debug: print(num_cw, num_move, file=sys.stderr)
ApplyMultipleOp(num_cw, [(1, OP_CW1), (4, OP_CW4), (16, OP_CW16)])
if debug: PrintPermumation('t', board, slotval=-1)
ApplyMultipleOp(num_move // 2, [(1, OP_MOVE_LL1), (4, OP_MOVE_LL4), (16, OP_MOVE_LL16)])
if debug: PrintPermumation('tt', board, slotval=-1)
if num_move % 2 > 0:
if k == W * 2 - 2: return False
ApplyOps(board, OP_MOVE_L_THROW, ops)
ans.append(OP_MOVE_L_THROW)
view = LoopedView(board)
num_cw = Mod(K * 3 + 1 - view.index(0), len(view))
ApplyMultipleOp(num_cw, [(1, OP_CW1), (4, OP_CW4), (16, OP_CW16)])
if debug: PrintPermumation('ttt', board, slotval=-1)
f = 'r' * K + 'd'
ApplyOps(board, f)
ans.append(f)
if debug: PrintPermumation('ttt', board, slotval=-1)
print('SURGERY COMPLETE')
print(''.join(ans))
for k, v in ops_codes.items():
print(k, v)
return True
#ops = InitOps(11, debug=True)
#import random
#board = list(range(10 * 4 + 2))
#random.shuffle(board)
#print(board)
##board = [4, 35, 3, 15, 38, 1, 31, 7, 42, 26, 19, 27, 5, 11, 2, 33, 40, 16, 20, 12, 18, 24, 28, 6, 36, 22, 9, 29, 30, 43, 21, 10, 17, 0, 13, 8, 45, 25, 37, 32, 44, 14, 39, 34, 23, 41]
#if True:
# board[board.index(0)] = -1
# a = Solve(board, debug=True)
# print(a, file=sys.stderr)
for _ in range(int(sys.stdin.readline())):
sys.stdin.readline()
tr = lambda s: -1 if s == 'E' else int(s)
row1 = list(map(tr, sys.stdin.readline().split()))
row2 = list(map(tr, sys.stdin.readline().split()))
if not Solve(row1 + row2): print('SURGERY FAILED')
```
|
instruction
| 0
| 3,465
| 3
| 6,930
|
No
|
output
| 1
| 3,465
| 3
| 6,931
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are an intergalactic surgeon and you have an alien patient. For the purposes of this problem, we can and we will model this patient's body using a 2 × (2k + 1) rectangular grid. The alien has 4k + 1 distinct organs, numbered 1 to 4k + 1.
In healthy such aliens, the organs are arranged in a particular way. For example, here is how the organs of a healthy such alien would be positioned, when viewed from the top, for k = 4:
<image>
Here, the E represents empty space.
In general, the first row contains organs 1 to 2k + 1 (in that order from left to right), and the second row contains organs 2k + 2 to 4k + 1 (in that order from left to right) and then empty space right after.
Your patient's organs are complete, and inside their body, but they somehow got shuffled around! Your job, as an intergalactic surgeon, is to put everything back in its correct position. All organs of the alien must be in its body during the entire procedure. This means that at any point during the procedure, there is exactly one cell (in the grid) that is empty. In addition, you can only move organs around by doing one of the following things:
* You can switch the positions of the empty space E with any organ to its immediate left or to its immediate right (if they exist). In reality, you do this by sliding the organ in question to the empty space;
* You can switch the positions of the empty space E with any organ to its immediate top or its immediate bottom (if they exist) only if the empty space is on the leftmost column, rightmost column or in the centermost column. Again, you do this by sliding the organ in question to the empty space.
Your job is to figure out a sequence of moves you must do during the surgical procedure in order to place back all 4k + 1 internal organs of your patient in the correct cells. If it is impossible to do so, you must say so.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 4) denoting the number of test cases. The next lines contain descriptions of the test cases.
Each test case consists of three lines. The first line contains a single integer k (1 ≤ k ≤ 15) which determines the size of the grid. Then two lines follow. Each of them contains 2k + 1 space-separated integers or the letter E. They describe the first and second rows of organs, respectively. It is guaranteed that all 4k + 1 organs are present and there is exactly one E.
Output
For each test case, first, print a single line containing either:
* SURGERY COMPLETE if it is possible to place back all internal organs in the correct locations;
* SURGERY FAILED if it is impossible.
If it is impossible, then this is the only line of output for the test case. However, if it is possible, output a few more lines describing the sequence of moves to place the organs in the correct locations.
The sequence of moves will be a (possibly empty) string of letters u, d, l or r, representing sliding the organ that's directly above, below, to the left or to the right of the empty space, respectively, into the empty space. Print the sequence of moves in the following line, as such a string.
For convenience, you may use shortcuts to reduce the size of your output. You may use uppercase letters as shortcuts for sequences of moves. For example, you could choose T to represent the string lddrr. These shortcuts may also include other shortcuts on their own! For example, you could choose E to represent TruT, etc.
You may use any number of uppercase letters (including none) as shortcuts. The only requirements are the following:
* The total length of all strings in your output for a single case is at most 10^4;
* There must be no cycles involving the shortcuts that are reachable from the main sequence;
* The resulting sequence of moves is finite, after expanding all shortcuts. Note that the final sequence of moves (after expanding) may be much longer than 10^4; the only requirement is that it's finite.
As an example, if T = lddrr, E = TruT and R = rrr, then TurTlER expands to:
* TurTlER
* lddrrurTlER
* lddrrurlddrrlER
* lddrrurlddrrlTruTR
* lddrrurlddrrllddrrruTR
* lddrrurlddrrllddrrrulddrrR
* lddrrurlddrrllddrrrulddrrrrr
To use shortcuts, print each one of them in a single line as the uppercase letter, then space, and then the string that this shortcut represents. They may be printed in any order. At the end of all of those, print a single line containing DONE.
Note: You still need to print DONE even if you don't plan on using shortcuts.
Your sequence does not need to be the shortest. Any valid sequence of moves (satisfying the requirements above) will be accepted.
Example
Input
2
3
1 2 3 5 6 E 7
8 9 10 4 11 12 13
11
34 45 6 22 16 43 38 44 5 4 41 14 7 29 28 19 9 18 42 8 17 33 1
E 15 40 36 31 24 10 2 21 11 32 23 30 27 35 25 13 12 39 37 26 20 3
Output
SURGERY COMPLETE
IR
R SrS
S rr
I lldll
DONE
SURGERY FAILED
Note
There are three shortcuts defined in the first sample output:
* R = SrS
* S = rr
* I = lldll
The sequence of moves is IR and it expands to:
* IR
* lldllR
* lldllSrS
* lldllrrrS
* lldllrrrrr
Submitted Solution:
```
def solve(kk,grid):
sk = *range(kk * 2 + 2),*range(4 * kk + 1,2 * kk + 1,-1)
flat = [sk[v] for v in grid[0] + grid[1][::-1] if v]
dir = {
'L': 'l' * 2 * kk + 'u' + 'r' * 2 * kk + 'd',
'R': 'u' + 'l' * 2 * kk + 'd' + 'r' * 2 * kk,
'C': 'CC' + 'R' * (2 * kk + 1) + 'CC' + 'R' * (2 * kk + 2),
'F': 'R' * (kk - 1) + 'DD' + 'R' * (2 * kk + 1) + 'D' + 'L' * 2 * kk + 'DD' +
'L' * kk,
'G': 'FF'
}
[(i,j)] = [(i,j) for i in range(2) for j in range(2 * kk + 1) if grid[i][j] == 0]
st = 'r' * (2 * kk - j) + 'd' * (1 - i)
for i in range(2,4 * kk + 2):
tt = flat.index(i)
if tt >= 2:
st += 'L' * (tt - 2) + 'GR' * (tt - 2) + 'G'
flat = flat[tt - 1 : tt + 1] + flat[:tt - 1] + flat[tt + 1:]
if tt >= 1:
st += 'G'
flat = flat[1 : 3] + flat[:1] + flat[3:]
st += 'L'
flat = flat[1:] + flat[:1]
if flat[0] == 1: return st,dir
def getline():
return [0 if x == 'E' else int(x) for x in input().split()]
for case in range(int(input())):
ret = solve(int(input()),[getline() for i in range(2)])
if ret is None:
print('SURGERY FAILED')
else:
print('SURGERY COMPLETE')
st,vv = ret
print(st)
for shortcut in vv.items() : print(*shortcut)
print('DONE')
```
|
instruction
| 0
| 3,466
| 3
| 6,932
|
No
|
output
| 1
| 3,466
| 3
| 6,933
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.
The i-th conveyor belt is in one of three states:
* If it is clockwise, snakes can only go from room i to (i+1) mod n.
* If it is anticlockwise, snakes can only go from room (i+1) mod n to i.
* If it is off, snakes can travel in either direction.
<image>
Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.
Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.
The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.
The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.
* If s_{i} = '>', the i-th conveyor belt goes clockwise.
* If s_{i} = '<', the i-th conveyor belt goes anticlockwise.
* If s_{i} = '-', the i-th conveyor belt is off.
It is guaranteed that the sum of n among all test cases does not exceed 300 000.
Output
For each test case, output the number of returnable rooms.
Example
Input
4
4
-><-
5
>>>>>
3
<--
2
<>
Output
3
5
3
0
Note
In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.
In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
Submitted Solution:
```
if __name__ == '__main__':
for _ in range (int(input())):
n = int(input())
s = input()
if len(set(s)) == 1:
print(len(s))
elif len(set(s)) == 2 :
if '<' in set(s) and '-' in set(s):
print(len(s))
elif '>' in set(s) and '-' in set(s):
print(len(s))
else:
print(0)
else:
ans = 0
for i in range (len(s)):
if i == 0 and s[len(s)-1]=='-':
ans+=1
elif s[i]=='-':
ans+=1
elif i != 0:
if s[i]!='-' and s[i-1]=='-':
ans+=1
print(ans)
```
|
instruction
| 0
| 3,573
| 3
| 7,146
|
Yes
|
output
| 1
| 3,573
| 3
| 7,147
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.
The i-th conveyor belt is in one of three states:
* If it is clockwise, snakes can only go from room i to (i+1) mod n.
* If it is anticlockwise, snakes can only go from room (i+1) mod n to i.
* If it is off, snakes can travel in either direction.
<image>
Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.
Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.
The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.
The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.
* If s_{i} = '>', the i-th conveyor belt goes clockwise.
* If s_{i} = '<', the i-th conveyor belt goes anticlockwise.
* If s_{i} = '-', the i-th conveyor belt is off.
It is guaranteed that the sum of n among all test cases does not exceed 300 000.
Output
For each test case, output the number of returnable rooms.
Example
Input
4
4
-><-
5
>>>>>
3
<--
2
<>
Output
3
5
3
0
Note
In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.
In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
Submitted Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
path=list(map(str,input().strip()))
All=0
co=0
if ('>' not in path) or ('<' not in path):
All=1
returnable=[0]*(n)
for i in range(n):
if(path[i]=='-'):
returnable[(i+1)%n]=1
returnable[i]=1
total=0
for i in range(n):
total+=returnable[i]
if(All==1):
print(n)
else:
print(total)
```
|
instruction
| 0
| 3,574
| 3
| 7,148
|
Yes
|
output
| 1
| 3,574
| 3
| 7,149
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.
The i-th conveyor belt is in one of three states:
* If it is clockwise, snakes can only go from room i to (i+1) mod n.
* If it is anticlockwise, snakes can only go from room (i+1) mod n to i.
* If it is off, snakes can travel in either direction.
<image>
Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.
Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.
The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.
The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.
* If s_{i} = '>', the i-th conveyor belt goes clockwise.
* If s_{i} = '<', the i-th conveyor belt goes anticlockwise.
* If s_{i} = '-', the i-th conveyor belt is off.
It is guaranteed that the sum of n among all test cases does not exceed 300 000.
Output
For each test case, output the number of returnable rooms.
Example
Input
4
4
-><-
5
>>>>>
3
<--
2
<>
Output
3
5
3
0
Note
In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.
In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
Submitted Solution:
```
import sys
for i in range(0,int(sys.stdin.readline())):
l=sys.stdin.readline()
s=list(sys.stdin.readline())
s=s[0:len(s)-1]
rr=0
pc=s[len(s)-1]
for c in s:
if c=="-":
if pc!="-":
rr+=1
rr+=1
pc=c
if (not ">" in s) or (not "<" in s):
rr=len(s)
sys.stdout.write(str(rr))
sys.stdout.write("\n")
```
|
instruction
| 0
| 3,575
| 3
| 7,150
|
Yes
|
output
| 1
| 3,575
| 3
| 7,151
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.
The i-th conveyor belt is in one of three states:
* If it is clockwise, snakes can only go from room i to (i+1) mod n.
* If it is anticlockwise, snakes can only go from room (i+1) mod n to i.
* If it is off, snakes can travel in either direction.
<image>
Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.
Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.
The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.
The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.
* If s_{i} = '>', the i-th conveyor belt goes clockwise.
* If s_{i} = '<', the i-th conveyor belt goes anticlockwise.
* If s_{i} = '-', the i-th conveyor belt is off.
It is guaranteed that the sum of n among all test cases does not exceed 300 000.
Output
For each test case, output the number of returnable rooms.
Example
Input
4
4
-><-
5
>>>>>
3
<--
2
<>
Output
3
5
3
0
Note
In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.
In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
Submitted Solution:
```
import sys
import random
# import numpy as np
import math
import copy
from heapq import heappush, heappop, heapify
from functools import cmp_to_key
from bisect import bisect_left, bisect_right
from collections import defaultdict, deque, Counter
# sys.setrecursionlimit(1000000)
# input aliases
input = sys.stdin.readline
getS = lambda: input().strip()
getN = lambda: int(input())
getList = lambda: list(map(int, input().split()))
getZList = lambda: [int(x) - 1 for x in input().split()]
INF = float("inf")
MOD = 10 ** 9 + 7
divide = lambda x: pow(x, MOD - 2, MOD)
def solve():
n = getN()
S = getS()
if "<" not in S:
print(n)
return
if ">" not in S:
print(n)
return
S = S + S[0]
ans = 0
for i in range(n):
if S[i] == "-" or S[i+1] == "-":
ans += 1
print(ans)
def main():
n = getN()
for _ in range(n):
solve()
return
if __name__ == "__main__":
main()
# solve()
```
|
instruction
| 0
| 3,576
| 3
| 7,152
|
Yes
|
output
| 1
| 3,576
| 3
| 7,153
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.
The i-th conveyor belt is in one of three states:
* If it is clockwise, snakes can only go from room i to (i+1) mod n.
* If it is anticlockwise, snakes can only go from room (i+1) mod n to i.
* If it is off, snakes can travel in either direction.
<image>
Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.
Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.
The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.
The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.
* If s_{i} = '>', the i-th conveyor belt goes clockwise.
* If s_{i} = '<', the i-th conveyor belt goes anticlockwise.
* If s_{i} = '-', the i-th conveyor belt is off.
It is guaranteed that the sum of n among all test cases does not exceed 300 000.
Output
For each test case, output the number of returnable rooms.
Example
Input
4
4
-><-
5
>>>>>
3
<--
2
<>
Output
3
5
3
0
Note
In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.
In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
Submitted Solution:
```
import sys
tests = int(sys.stdin.readline())
for i in range(tests):
rooms = int(sys.stdin.readline())
belts = sys.stdin.readline().strip()
# Test for cull circle
if "<" not in belts:
print(rooms)
continue
if ">" not in belts:
print(rooms)
continue
if "<" not in belts and ">" not in belts:
print(rooms)
continue
ret_rooms = 0
for room in range(rooms):
if belts[(room) % rooms] == ">" and belts[(room + 1) % rooms] == "<":
pass # trapped
elif (belts[(room - 1) % rooms] == "-") or (belts[(room + 1) % rooms] == "-"):
ret_rooms += 1
print(ret_rooms)
```
|
instruction
| 0
| 3,577
| 3
| 7,154
|
No
|
output
| 1
| 3,577
| 3
| 7,155
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.
The i-th conveyor belt is in one of three states:
* If it is clockwise, snakes can only go from room i to (i+1) mod n.
* If it is anticlockwise, snakes can only go from room (i+1) mod n to i.
* If it is off, snakes can travel in either direction.
<image>
Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.
Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.
The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.
The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.
* If s_{i} = '>', the i-th conveyor belt goes clockwise.
* If s_{i} = '<', the i-th conveyor belt goes anticlockwise.
* If s_{i} = '-', the i-th conveyor belt is off.
It is guaranteed that the sum of n among all test cases does not exceed 300 000.
Output
For each test case, output the number of returnable rooms.
Example
Input
4
4
-><-
5
>>>>>
3
<--
2
<>
Output
3
5
3
0
Note
In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.
In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
s = input()
t = [[] for i in range(n)]
if n <= 2:
print(0)
else:
for i in range(n):
if s[i] == '>':
t[i].append((i+1)%n)
elif s[i] == '<':
t[(i+1)%n].append(i)
else:
temp1 = (i+1)%n
t[temp1].append(i)
t[i].append(temp1)
c = 0
for i in t:
if not i:
c += 1
print(n-c)
```
|
instruction
| 0
| 3,578
| 3
| 7,156
|
No
|
output
| 1
| 3,578
| 3
| 7,157
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.
The i-th conveyor belt is in one of three states:
* If it is clockwise, snakes can only go from room i to (i+1) mod n.
* If it is anticlockwise, snakes can only go from room (i+1) mod n to i.
* If it is off, snakes can travel in either direction.
<image>
Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.
Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.
The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.
The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.
* If s_{i} = '>', the i-th conveyor belt goes clockwise.
* If s_{i} = '<', the i-th conveyor belt goes anticlockwise.
* If s_{i} = '-', the i-th conveyor belt is off.
It is guaranteed that the sum of n among all test cases does not exceed 300 000.
Output
For each test case, output the number of returnable rooms.
Example
Input
4
4
-><-
5
>>>>>
3
<--
2
<>
Output
3
5
3
0
Note
In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.
In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
Submitted Solution:
```
t =int(input())
for _ in range(t):
n = int(input())
s = input()
l = [0]*len(s)
for i in range(len(s)):
if i==len(s)-1:
if s[i]=='-':
l[i] = 1
l[0] = 1
if s[i]=='>':
l[i] = 1
if s[i]=='<':
l[0] = 1
elif i == 0:
if s[i]=='-':
l[i] = 1
l[len(s)-1] = 1
if s[i]=='>':
l[i] = 1
if s[i]=='<':
l[i] = 1
else:
if s[i]=='-':
l[i] = 1
l[i+1] = 1
if s[i]=='>':
l[i] = 1
if s[i]=='<':
l[i+1] = 1
cnt=0
for i in l:
if i==1:
cnt+=1
if len(s)==2 and cnt==2:
print(0)
else:
print(cnt)
```
|
instruction
| 0
| 3,579
| 3
| 7,158
|
No
|
output
| 1
| 3,579
| 3
| 7,159
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.
The i-th conveyor belt is in one of three states:
* If it is clockwise, snakes can only go from room i to (i+1) mod n.
* If it is anticlockwise, snakes can only go from room (i+1) mod n to i.
* If it is off, snakes can travel in either direction.
<image>
Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.
Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.
The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.
The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.
* If s_{i} = '>', the i-th conveyor belt goes clockwise.
* If s_{i} = '<', the i-th conveyor belt goes anticlockwise.
* If s_{i} = '-', the i-th conveyor belt is off.
It is guaranteed that the sum of n among all test cases does not exceed 300 000.
Output
For each test case, output the number of returnable rooms.
Example
Input
4
4
-><-
5
>>>>>
3
<--
2
<>
Output
3
5
3
0
Note
In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.
In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
Submitted Solution:
```
for i in range(int(input())):
n=int(input())
a=input()
i=0;j=1;cnt=0
while j<n:
if (a[i]=='<' and a[j]=='>') or (a[i]=='>' and a[j]=='<'):
cnt+=1
i+=1;j+=1
if (a[0]=='<' and a[-1]=='>') or (a[0]=='>' and a[-1]=='<'):
cnt+=1
print(n-cnt)
```
|
instruction
| 0
| 3,580
| 3
| 7,160
|
No
|
output
| 1
| 3,580
| 3
| 7,161
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Golorps are mysterious creatures who feed on variables. Golorp's name is a program in some programming language. Some scientists believe that this language is Befunge; golorps are tantalizingly silent.
Variables consumed by golorps can take values from 0 to 9, inclusive. For each golorp its daily diet is defined by its name. Some golorps are so picky that they can't be fed at all. Besides, all golorps are very health-conscious and try to eat as little as possible. Given a choice of several valid sequences of variable values, each golorp will choose lexicographically smallest one.
For the purposes of this problem you can assume that a golorp consists of jaws and a stomach. The number of variables necessary to feed a golorp is defined by the shape of its jaws. Variables can get to the stomach only via the jaws.
A hungry golorp is visiting you. You know its name; feed it or figure out that it's impossible.
Input
The input is a single string (between 13 and 1024 characters long) — the name of the visiting golorp. All names are similar and will resemble the ones given in the samples. The name is guaranteed to be valid.
Output
Output lexicographically smallest sequence of variable values fit for feeding this golorp. Values should be listed in the order in which they get into the jaws. If the golorp is impossible to feed, output "false".
Examples
Input
?(_-_/___*__):-___>__.
Output
0010
Input
?(__-_+_/_____):-__>__,_____<__.
Output
false
Input
?(______________________/____+_______*__-_____*______-___):-__<___,___<____,____<_____,_____<______,______<_______.
Output
0250341
Input
?(__+___+__-___):-___>__.
Output
0101
Submitted Solution:
```
import re
line=input()
line = line.rstrip('.')
a, b, c = line.partition("):-")
rels = c.split(',')
relations = set()
for rel in rels:
if "<" in rel:
x, _, y = rel.partition("<")
relations.add((len(x), len(y)))
else:
x, _, y = rel.partition(">")
relations.add((len(y), len(x)))
a = a.lstrip("?(")
args = re.split(r"(\+|-|\*|/)", a)
args = [len(args[i]) for i in range(0, len(args), 2)]
# print(args)
# print(relations)
edges = {k:[] for k in args}
for j, i in relations:
edges[i].append(j)
seen = {k:0 for k in args}
q = []
def topsort(k):
if seen[k] == 1:
print("false")
quit()
seen[k] = 1
for other in edges[k]:
topsort(other)
q.append(k)
seen[k] = 2
return True
for k in args:
if seen[k] == 0:
topsort(k)
vals = {}
for k in q:
others = edges[k]
maxval = -1
for other in others:
maxval = max(maxval, vals.get(other, -1))
vals[k] = maxval+1
for k in args:
print(vals[k], end='')
print()
```
|
instruction
| 0
| 3,733
| 3
| 7,466
|
No
|
output
| 1
| 3,733
| 3
| 7,467
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 4,319
| 3
| 8,638
|
Tags: dp
Correct Solution:
```
s = input()
n = int(input())
t = [j for j, q in enumerate(s) if q == 'T']
l, r = [0] * 101, [0] * 101
for i, (a, b) in enumerate(zip([-1] + t, t + [len(s)])):
v = b - a
u = v - 1
for k in range(i, 0, -1):
l[k] = max(l[k] + u, l[k - 1] + v)
r[k] = max(r[k] - u, r[k - 1] - v)
u, v = -u, -v
r[i + 1] = l[i] - 1
l[0] += u
r[0] -= u
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
```
|
output
| 1
| 4,319
| 3
| 8,639
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 4,320
| 3
| 8,640
|
Tags: dp
Correct Solution:
```
import sys
import math
INF = -100000000
memo = dict()
def func(line, r):
if line in memo and r in memo[line]:
return memo[line][r]
if len(line) == 1:
which = line[0] == 'T'
if r % 2 == 1:
which = not which
if which:
return [INF, INF, 0, 0]
else:
return [1, INF, INF, INF]
best = [INF, INF, INF, INF]
for i in range(r + 1):
a = func(line[:len(line) // 2], i)
b = func(line[len(line) // 2:], r - i)
for (j, k) in [(j, k) for j in range(4) for k in range(4)]:
D = j < 2
if a[j] == INF or b[k] == INF: continue
aa = -a[j] if j % 2 else a[j]
bb = -b[k] if k % 2 else b[k]
d1, d2 = 0, 1
if k < 2:
aa = aa + bb
if not D: d1, d2 = 2, 3
else:
aa = -aa + bb
if D: d1, d2 = 2, 3
if aa >= 0: best[d1] = max(best[d1], aa)
if aa <= 0: best[d2] = max(best[d2], -aa)
if not line in memo:
memo[line] = dict()
memo[line][r] = best
return best
line = input().strip()
k = int(input())
ans = max(func(line, k))
if ans == INF: print("0")
else: print(ans)
```
|
output
| 1
| 4,320
| 3
| 8,641
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 4,321
| 3
| 8,642
|
Tags: dp
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from math import inf
def main():
s = input().strip()
n = int(input())
dp = [[[-inf,-inf] for _ in range(n+1)]for _ in range(len(s))]
x = int(s[0]!='F')
for i in range(n+1):
dp[0][i][x] = (x^1)
x ^= 1
for i in range(1,len(s)):
for j in range(n+1):
x = int(s[i]!='F')
for k in range(j,-1,-1):
dp[i][j][0] = max(dp[i][j][0],dp[i-1][k][x]+(x^1))
dp[i][j][1] = max(dp[i][j][1],dp[i-1][k][x^1]-(x^1))
x ^= 1
dp1 = [[[inf,inf] for _ in range(n+1)]for _ in range(len(s))]
x = int(s[0]!='F')
for i in range(n+1):
dp1[0][i][x] = (x^1)
x ^= 1
for i in range(1,len(s)):
for j in range(n+1):
x = int(s[i]!='F')
for k in range(j,-1,-1):
dp1[i][j][0] = min(dp1[i][j][0],dp1[i-1][k][x]+(x^1))
dp1[i][j][1] = min(dp1[i][j][1],dp1[i-1][k][x^1]-(x^1))
x ^= 1
print(max(max(dp[-1][n]),abs(min(dp1[-1][n]))))
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
```
|
output
| 1
| 4,321
| 3
| 8,643
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 4,322
| 3
| 8,644
|
Tags: dp
Correct Solution:
```
from sys import stdin
s=stdin.readline().strip()
k=int(stdin.readline().strip())
n=len(s)
dp=[[[[ None for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
dp1=[[[[ None for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
def sol(i,j,t,c):
if i==n:
if j!=0:
return -10000000
else:
return 0
if dp[i][j][t][c]!=None:
return dp[i][j][t][c]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol(i+1,j,t,-1)
else:
ans=1+sol(i+1,j,t,-1)
if j>0:
ans=max(ans,sol(i,j-1, t,0))
else :
ans=sol(i+1,j,not t,-1)
if j>0:
ans=max(ans,sol(i,j-1, t,1))
dp[i][j][t][c]=ans
return dp[i][j][t][c]
def sol1(i,j,t,c):
if i==n:
if j!=0:
return 10000000
else:
return 0
if dp1[i][j][t][c]!=None:
return dp1[i][j][t][c]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol1(i+1,j,t,-1)
else:
ans=1+sol1(i+1,j,t,-1)
if j>0:
ans=min(ans,sol1(i,j-1, t,0))
else :
ans=sol1(i+1,j,not t,-1)
if j>0:
ans=min(ans,sol1(i,j-1, t,1))
dp1[i][j][t][c]=ans
return dp1[i][j][t][c]
print(max(sol(0,k,0,-1),-sol1(0,k,0,-1)))
```
|
output
| 1
| 4,322
| 3
| 8,645
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 4,323
| 3
| 8,646
|
Tags: dp
Correct Solution:
```
s = input()
n = int(input())
l, r = [-1e9] * 101, [-1e9] * 101
l[0] = r[0] = 0
for q in s:
for j in range(n, -1, -1):
x = max(r[j], l[j - 1] + 1) if q == 'T' else max(l[j] + 1, r[j - 1])
y = max(l[j], r[j - 1] + 1) if q == 'T' else max(r[j] - 1, l[j - 1])
l[j], r[j] = x, y
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
# Made By Mostafa_Khaled
```
|
output
| 1
| 4,323
| 3
| 8,647
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 4,324
| 3
| 8,648
|
Tags: dp
Correct Solution:
```
from sys import stdin
s=stdin.readline().strip()
k=int(stdin.readline().strip())
n=len(s)
dp=[[[[[ None for ty in range(2)]for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
def minn(a,b):
if a<b:
return a
return b
def maxx(a,b):
if a>b:
return a
return b
def sol(i,j,t,c,mt):
if i==n:
if j!=0:
if mt:
return -10001
else:
return 10001
else:
return 0
if dp[i][j][t][c][mt]!=None:
return dp[i][j][t][c][mt]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol(i+1,j,t,-1,mt)
else:
ans=1+sol(i+1,j,t,-1,mt)
if j>0:
if mt:
ans=maxx(ans,sol(i,j-1, t,0,mt))
else:
ans=minn(ans,sol(i,j-1, t,0,mt))
else :
ans=sol(i+1,j,not t,-1,mt)
if j>0:
if mt:
ans=maxx(ans,sol(i,j-1, t,1,mt))
else:
ans=minn(ans,sol(i,j-1, t,1,mt))
dp[i][j][t][c][mt]=ans
return dp[i][j][t][c][mt]
print(max(-sol(0,k,0,-1,0),sol(0,k,0,-1,1)))
```
|
output
| 1
| 4,324
| 3
| 8,649
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 4,325
| 3
| 8,650
|
Tags: dp
Correct Solution:
```
from sys import stdin
s=stdin.readline().strip()
k=int(stdin.readline().strip())
n=len(s)
dp=[[[[[ None for ty in range(2)]for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
def sol(i,j,t,c,mt):
if i==n:
if j!=0:
if mt:
return -10001
else:
return 10001
else:
return 0
if dp[i][j][t][c][mt]!=None:
return dp[i][j][t][c][mt]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol(i+1,j,t,-1,mt)
else:
ans=1+sol(i+1,j,t,-1,mt)
if j>0:
if mt:
ans=max(ans,sol(i,j-1, t,0,mt))
else:
ans=min(ans,sol(i,j-1, t,0,mt))
else :
ans=sol(i+1,j,not t,-1,mt)
if j>0:
if mt:
ans=max(ans,sol(i,j-1, t,1,mt))
else:
ans=min(ans,sol(i,j-1, t,1,mt))
dp[i][j][t][c][mt]=ans
return dp[i][j][t][c][mt]
print(max(-sol(0,k,0,-1,0),sol(0,k,0,-1,1)))
```
|
output
| 1
| 4,325
| 3
| 8,651
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
|
instruction
| 0
| 4,326
| 3
| 8,652
|
Tags: dp
Correct Solution:
```
s = input()
n = int(input())
t = [j for j, q in enumerate(s) if q == 'T']
l, r = [0] * 101, [0] * 101
for i, (a, b) in enumerate(zip([-1] + t, t + [len(s)])):
v = b - a
u = v - 1
if i:
l[i] = l[i - 1] + v
else:
u, v = -u, -v
r[i + 1] = l[i] - 1
for k in range(i - 1, 0, -1):
l[k] = max(l[k] - u, l[k - 1] - v)
r[k] = max(r[k] + u, r[k - 1] + v)
u, v = -u, -v
l[0] -= u
r[0] += u
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
```
|
output
| 1
| 4,326
| 3
| 8,653
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
s = input()
n = int(input())
l, r = [-1e9] * 101, [-1e9] * 101
l[0] = r[0] = 0
for q in s:
for j in range(n, -1, -1):
x = max(r[j], l[j - 1] + 1) if q == 'T' else max(l[j] + 1, r[j - 1])
y = max(l[j], r[j - 1] + 1) if q == 'T' else max(r[j] - 1, l[j - 1])
l[j], r[j] = x, y
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
```
|
instruction
| 0
| 4,327
| 3
| 8,654
|
Yes
|
output
| 1
| 4,327
| 3
| 8,655
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
from sys import stdin
s=stdin.readline().strip()
k=int(stdin.readline().strip())
n=len(s)
dp=[[[[[ None for ty in range(2)]for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
def min(a,b):
if a<b:
return a
return b
def max(a,b):
if a>b:
return a
return b
def sol(i,j,t,c,mt):
if i==n:
if j!=0:
if mt:
return -10001
else:
return 10001
else:
return 0
if dp[i][j][t][c][mt]!=None:
return dp[i][j][t][c][mt]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol(i+1,j,t,-1,mt)
else:
ans=1+sol(i+1,j,t,-1,mt)
if j>0:
if mt:
ans=max(ans,sol(i,j-1, t,0,mt))
else:
ans=min(ans,sol(i,j-1, t,0,mt))
else :
ans=sol(i+1,j,not t,-1,mt)
if j>0:
if mt:
ans=max(ans,sol(i,j-1, t,1,mt))
else:
ans=min(ans,sol(i,j-1, t,1,mt))
dp[i][j][t][c][mt]=ans
return dp[i][j][t][c][mt]
print(max(-sol(0,k,0,-1,0),sol(0,k,0,-1,1)))
```
|
instruction
| 0
| 4,328
| 3
| 8,656
|
Yes
|
output
| 1
| 4,328
| 3
| 8,657
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
from sys import stdin
import sys
sys.setrecursionlimit(10000000)
s=stdin.readline().strip()
n=int(stdin.readline().strip())
dp1=[[[[-110 for i in range(51)] for j in range(101)] for k in range(2)] for k1 in range(2)]
dp2=[[[[-110 for i in range(51)] for j in range(101)] for k in range(2)] for k1 in range(2)]
inf=1000
def sol(pos,el,cam,dir):
if pos>=len(s):
if el==0:
return 0
else:
return -inf
if el<0:
return -inf
if dp1[dir][cam][pos][el]!=-110:
return dp1[dir][cam][pos][el]
e=s[pos]
if s[pos]=="F" and cam==1:
e="T"
if s[pos]=="T" and cam==1:
e="F"
ans=-inf
if e=="F":
if dir==1:
ans=max(1+sol(pos+1,el,0,dir),sol(pos,el-1,(cam+1)%2,(dir)%2))
else:
ans=max(-1+sol(pos+1,el,0,dir),sol(pos,el-1,(cam+1)%2,(dir)%2))
else:
ans=max(sol(pos+1,el,0,(dir+1)%2),sol(pos,el-1,(cam+1)%2,dir))
dp1[dir][cam][pos][el]=ans
return ans
def sol2(pos,el,cam,dir):
if pos>=len(s):
if el==0:
return 0
else:
return inf
if el<0:
return inf
if dp2[dir][cam][pos][el]!=-110:
return dp2[dir][cam][pos][el]
e=s[pos]
if s[pos]=="F" and cam==1:
e="T"
if s[pos]=="T" and cam==1:
e="F"
ans=inf
if e=="F":
if dir==1:
ans=min(1+sol2(pos+1,el,0,dir),sol2(pos,el-1,(cam+1)%2,(dir)))
else:
ans=min(-1+sol2(pos+1,el,0,dir),sol2(pos,el-1,(cam+1)%2,(dir)))
else:
ans=min(sol2(pos+1,el,0,(dir+1)%2),sol2(pos,el-1,(cam+1)%2,dir))
dp2[dir][cam][pos][el]=ans
return ans
print(max(sol(0,n,0,1),-sol2(0,n,0,1)))
```
|
instruction
| 0
| 4,329
| 3
| 8,658
|
Yes
|
output
| 1
| 4,329
| 3
| 8,659
|
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