message
stringlengths 2
39.6k
| message_type
stringclasses 2
values | message_id
int64 0
1
| conversation_id
int64 450
109k
| cluster
float64 2
2
| __index_level_0__
int64 900
217k
|
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Whoso in ignorance draws near to them and hears the Sirens' voice, he nevermore returns.
Homer, Odyssey
In the times of Jason and the Argonauts, it was well known that sirens use the sound of their songs to lure sailors into their demise. Yet only a few knew that every time sirens call a sailor by his name, his will weakens, making him more vulnerable.
For the purpose of this problem, both siren songs and names of the sailors will be represented as strings of lowercase English letters. The more times the sailor's name occurs as a contiguous substring of the song, the greater danger he is in.
Jason found out that sirens can sing one of the n+1 songs, which have the following structure: let s_i (0 ≤ i ≤ n) be the i-th song and t be a string of length n, then for every i < n: s_{i+1} = s_i t_i s_i. In other words i+1-st song is the concatenation of i-th song, i-th letter (0-indexed) of t and the i-th song.
Fortunately, he also knows s_0 and t. Jason wonders how many times a sailor's name is mentioned in a particular song. Answer q queries: given the sailor's name (w) and the index of a song (i) output the number of occurrences of w in s_i as a substring. As this number can be quite large, output its remainder modulo 10^9+7.
Input
In the first line of input there are two integers n, q ( 1 ≤ n, q ≤ 10^5) meaning that there are n+1 songs and q queries. In the next two lines strings s_0 and t follow (1 ≤ |s_0| ≤ 100, |t| = n).
Next q lines describe the queries; each one contains an integer k ( 0 ≤ k ≤ n), the index of the song sung by the sirens, and a non-empty string w, which is the name of a sailor. All strings in this problem consist only of lowercase English letters, and the sum of lengths of sailors' names does not exceed 10^6.
Output
Output q lines, i-th of them should contain the remainder modulo 10^9+7 of the number of occurrences of w in s_k.
Examples
Input
3 3
aa
bcd
2 aba
3 ca
3 aa
Output
2
2
8
Input
4 5
aba
bbac
1 a
3 baca
3 ab
2 bab
4 aba
Output
4
0
14
6
28
Note
In the first example songs of the sirens are as follows:
* Song 0: aa
* Song 1: aabaa
* Song 2: aabaacaabaa
* Song 3: aabaacaabaadaabaacaabaa
Submitted Solution:
```
def substr(string,substring):
# Initialize count and start to 0
count = 0
start = 0
# Search through the string till
# we reach the end of it
while start < len(string):
# Check if a substring is present from
# 'start' position till the end
pos = string.find(substring, start)
if pos != -1:
# If a substring is present, move 'start' to
# the next position from start of the substring
start = pos + 1
# Increment the count
count += 1
else:
# If no further substring is present
break
# return the value of count
return count
n,m=map(int,input().split())
string1=input()
string2=list(map(str,input().strip()))
result=[string1]
stack=string1
for i in string2:
res=stack+i+stack
result.append(res)
stack=res
print(result)
for _ in range(m):
l,string=map(str,input().split())
ini_str=result[int(l)]
res = substr(ini_str,string)
print(res)
```
|
instruction
| 0
| 18,753
| 2
| 37,506
|
No
|
output
| 1
| 18,753
| 2
| 37,507
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Whoso in ignorance draws near to them and hears the Sirens' voice, he nevermore returns.
Homer, Odyssey
In the times of Jason and the Argonauts, it was well known that sirens use the sound of their songs to lure sailors into their demise. Yet only a few knew that every time sirens call a sailor by his name, his will weakens, making him more vulnerable.
For the purpose of this problem, both siren songs and names of the sailors will be represented as strings of lowercase English letters. The more times the sailor's name occurs as a contiguous substring of the song, the greater danger he is in.
Jason found out that sirens can sing one of the n+1 songs, which have the following structure: let s_i (0 ≤ i ≤ n) be the i-th song and t be a string of length n, then for every i < n: s_{i+1} = s_i t_i s_i. In other words i+1-st song is the concatenation of i-th song, i-th letter (0-indexed) of t and the i-th song.
Fortunately, he also knows s_0 and t. Jason wonders how many times a sailor's name is mentioned in a particular song. Answer q queries: given the sailor's name (w) and the index of a song (i) output the number of occurrences of w in s_i as a substring. As this number can be quite large, output its remainder modulo 10^9+7.
Input
In the first line of input there are two integers n, q ( 1 ≤ n, q ≤ 10^5) meaning that there are n+1 songs and q queries. In the next two lines strings s_0 and t follow (1 ≤ |s_0| ≤ 100, |t| = n).
Next q lines describe the queries; each one contains an integer k ( 0 ≤ k ≤ n), the index of the song sung by the sirens, and a non-empty string w, which is the name of a sailor. All strings in this problem consist only of lowercase English letters, and the sum of lengths of sailors' names does not exceed 10^6.
Output
Output q lines, i-th of them should contain the remainder modulo 10^9+7 of the number of occurrences of w in s_k.
Examples
Input
3 3
aa
bcd
2 aba
3 ca
3 aa
Output
2
2
8
Input
4 5
aba
bbac
1 a
3 baca
3 ab
2 bab
4 aba
Output
4
0
14
6
28
Note
In the first example songs of the sirens are as follows:
* Song 0: aa
* Song 1: aabaa
* Song 2: aabaacaabaa
* Song 3: aabaacaabaadaabaacaabaa
Submitted Solution:
```
z=input()
n,Q=z.split()
n,Q=int(n),int(Q)
s=[]
s.append(input())
t=input()
#print(s,t)
i=0
for q in range(Q):
z=input()
count=0
k,w=z.split()
k=int(k)
while i<k:
s.append(s[i]+t[i]+s[i])
i+=1
x=s[i]
for y in range(len(x)):
if y+len(w)>len(x):
break
if x[y:y+len(w)]==w:
count+=1
print(count)
```
|
instruction
| 0
| 18,754
| 2
| 37,508
|
No
|
output
| 1
| 18,754
| 2
| 37,509
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Whoso in ignorance draws near to them and hears the Sirens' voice, he nevermore returns.
Homer, Odyssey
In the times of Jason and the Argonauts, it was well known that sirens use the sound of their songs to lure sailors into their demise. Yet only a few knew that every time sirens call a sailor by his name, his will weakens, making him more vulnerable.
For the purpose of this problem, both siren songs and names of the sailors will be represented as strings of lowercase English letters. The more times the sailor's name occurs as a contiguous substring of the song, the greater danger he is in.
Jason found out that sirens can sing one of the n+1 songs, which have the following structure: let s_i (0 ≤ i ≤ n) be the i-th song and t be a string of length n, then for every i < n: s_{i+1} = s_i t_i s_i. In other words i+1-st song is the concatenation of i-th song, i-th letter (0-indexed) of t and the i-th song.
Fortunately, he also knows s_0 and t. Jason wonders how many times a sailor's name is mentioned in a particular song. Answer q queries: given the sailor's name (w) and the index of a song (i) output the number of occurrences of w in s_i as a substring. As this number can be quite large, output its remainder modulo 10^9+7.
Input
In the first line of input there are two integers n, q ( 1 ≤ n, q ≤ 10^5) meaning that there are n+1 songs and q queries. In the next two lines strings s_0 and t follow (1 ≤ |s_0| ≤ 100, |t| = n).
Next q lines describe the queries; each one contains an integer k ( 0 ≤ k ≤ n), the index of the song sung by the sirens, and a non-empty string w, which is the name of a sailor. All strings in this problem consist only of lowercase English letters, and the sum of lengths of sailors' names does not exceed 10^6.
Output
Output q lines, i-th of them should contain the remainder modulo 10^9+7 of the number of occurrences of w in s_k.
Examples
Input
3 3
aa
bcd
2 aba
3 ca
3 aa
Output
2
2
8
Input
4 5
aba
bbac
1 a
3 baca
3 ab
2 bab
4 aba
Output
4
0
14
6
28
Note
In the first example songs of the sirens are as follows:
* Song 0: aa
* Song 1: aabaa
* Song 2: aabaacaabaa
* Song 3: aabaacaabaadaabaacaabaa
Submitted Solution:
```
# Python3 program to count occurrences of
# pattern in a text.
def KMPSearch(pat, txt):
M = len(pat)
N = len(txt)
# Create lps[] that will hold the longest
# prefix suffix values for pattern
lps = [None] * M
j = 0 # index for pat[]
# Preprocess the pattern (calculate lps[]
# array)
computeLPSArray(pat, M, lps)
i = 0 # index for txt[]
res = 0
next_i = 0
while (i < N):
if pat[j] == txt[i]:
j = j + 1
i = i + 1
if j == M:
# When we find pattern first time,
# we iterate again to check if there
# exists more pattern
j = lps[j - 1]
res = res + 1
# We start i to check for more than once
# appearance of pattern, we will reset i
# to previous start+1
if lps[j] != 0:
next_i = next_i + 1
i = next_i
j = 0
# Mismatch after j matches
elif ((i < N) and (pat[j] != txt[i])):
# Do not match lps[0..lps[j-1]]
# characters, they will match anyway
if (j != 0):
j = lps[j - 1]
else:
i = i + 1
return res
def computeLPSArray(pat, M, lps):
# Length of the previous longest
# prefix suffix
len = 0
i = 1
lps[0] = 0 # lps[0] is always 0
# The loop calculates lps[i] for
# i = 1 to M-1
while (i < M):
if pat[i] == pat[len]:
len = len + 1
lps[i] = len
i = i + 1
else: # (pat[i] != pat[len])
# This is tricky. Consider the example.
# AAACAAAA and i = 7. The idea is similar
# to search step.
if len != 0:
len = lps[len - 1]
# Also, note that we do not increment
# i here
else: # if (len == 0)
lps[i] = len
i = i + 1
n, q = map(int, input().split())
s = []
s.append(input())
t = input()
for i in range(1, n + 1):
s.append(s[i - 1] + t[i - 1] + s[i - 1])
for j in range(q):
k, w = map(str, input().split())
k = int(k)
print(KMPSearch(w, s[k]) % (pow(10, 9) + 7))
```
|
instruction
| 0
| 18,755
| 2
| 37,510
|
No
|
output
| 1
| 18,755
| 2
| 37,511
|
Provide a correct Python 3 solution for this coding contest problem.
You are playing a game called Guru Guru Gururin. In this game, you can move with the vehicle called Gururin. There are two commands you can give to Gururin: 'R' and 'L'. When 'R' is sent, Gururin rotates clockwise by 90 degrees. Otherwise, when 'L' is sent, Gururin rotates counterclockwise by 90 degrees.
During the game, you noticed that Gururin obtains magical power by performing special commands. In short, Gururin obtains magical power every time when it performs one round in the clockwise direction from north to north. In more detail, the conditions under which magical power can be obtained is as follows.
* At the beginning of the special commands, Gururin faces north.
* At the end of special commands, Gururin faces north.
* Except for the beginning and the end of special commands, Gururin does not face north.
* During the special commands, Gururin faces north, east, south, and west one or more times, respectively, after the command of 'R'.
At the beginning of the game, Gururin faces north. For example, if the sequence of commands Gururin received in order is 'RRRR' or 'RRLRRLRR', Gururin can obtain magical power. Otherwise, if the sequence of commands is 'LLLL' or 'RLLR', Gururin cannot obtain magical power.
Your task is to calculate how many times Gururin obtained magical power throughout the game. In other words, given the sequence of the commands Gururin received, calculate how many special commands exists in it.
Input
The input consists of a single test case in the format below.
$S$
The first line consists of a string $S$, which represents the sequence of commands Gururin received. $S$ consists of 'L' and 'R'. The length of $S$ is between $1$ and $10^3$ inclusive.
Output
Print the number of times Gururin obtained magical power throughout the game in one line.
Examples
Input
RRRRLLLLRRRR
Output
2
Input
RLLRLLLLRRRLLLRRR
Output
0
Input
LR
Output
0
Input
RRLRRLRRRRLRRLRRRRLRRLRRRRLRRLRR
Output
4
|
instruction
| 0
| 20,230
| 2
| 40,460
|
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
S = readline().strip()
ans = 0
d = 0; cur = 0
for c in S:
if c == "R":
if d == cur:
cur += 1
d = (d + 1) % 4
if d == 0 and cur == 4:
ans += 1
cur = 0
else:
d = (d - 1) % 4
if d == 0:
cur = 0
write("%d\n" % ans)
solve()
```
|
output
| 1
| 20,230
| 2
| 40,461
|
Provide a correct Python 3 solution for this coding contest problem.
You are playing a game called Guru Guru Gururin. In this game, you can move with the vehicle called Gururin. There are two commands you can give to Gururin: 'R' and 'L'. When 'R' is sent, Gururin rotates clockwise by 90 degrees. Otherwise, when 'L' is sent, Gururin rotates counterclockwise by 90 degrees.
During the game, you noticed that Gururin obtains magical power by performing special commands. In short, Gururin obtains magical power every time when it performs one round in the clockwise direction from north to north. In more detail, the conditions under which magical power can be obtained is as follows.
* At the beginning of the special commands, Gururin faces north.
* At the end of special commands, Gururin faces north.
* Except for the beginning and the end of special commands, Gururin does not face north.
* During the special commands, Gururin faces north, east, south, and west one or more times, respectively, after the command of 'R'.
At the beginning of the game, Gururin faces north. For example, if the sequence of commands Gururin received in order is 'RRRR' or 'RRLRRLRR', Gururin can obtain magical power. Otherwise, if the sequence of commands is 'LLLL' or 'RLLR', Gururin cannot obtain magical power.
Your task is to calculate how many times Gururin obtained magical power throughout the game. In other words, given the sequence of the commands Gururin received, calculate how many special commands exists in it.
Input
The input consists of a single test case in the format below.
$S$
The first line consists of a string $S$, which represents the sequence of commands Gururin received. $S$ consists of 'L' and 'R'. The length of $S$ is between $1$ and $10^3$ inclusive.
Output
Print the number of times Gururin obtained magical power throughout the game in one line.
Examples
Input
RRRRLLLLRRRR
Output
2
Input
RLLRLLLLRRRLLLRRR
Output
0
Input
LR
Output
0
Input
RRLRRLRRRRLRRLRRRRLRRLRRRRLRRLRR
Output
4
|
instruction
| 0
| 20,231
| 2
| 40,462
|
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
import itertools
sys.setrecursionlimit(10**5)
stdin = sys.stdin
bisect_left = bisect.bisect_left
bisect_right = bisect.bisect_right
def LI(): return list(map(int, stdin.readline().split()))
def LF(): return list(map(float, stdin.readline().split()))
def LI_(): return list(map(lambda x: int(x)-1, stdin.readline().split()))
def II(): return int(stdin.readline())
def IF(): return float(stdin.readline())
def LS(): return list(map(list, stdin.readline().split()))
def S(): return list(stdin.readline().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')
#A
def A():
s = S()
i = 0
now = 0
ans = 0
while i < len(s):
if s[i] == "R":
if now == 0:
now += 1
i += 1
while i < len(s):
if s[i] == "R":
now += 1
else:
now -= 1
now = now % 4
i += 1
if now == 0:
if s[i-1] == "R":
ans += 1
break
else:
now += 1
now = now % 4
i += 1
else:
now -= 1
now = now % 4
i += 1
print(ans)
return
#B
def B():
return
#C
def C():
return
#D
def D():
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#Solve
if __name__ == '__main__':
A()
```
|
output
| 1
| 20,231
| 2
| 40,463
|
Provide a correct Python 3 solution for this coding contest problem.
You are playing a game called Guru Guru Gururin. In this game, you can move with the vehicle called Gururin. There are two commands you can give to Gururin: 'R' and 'L'. When 'R' is sent, Gururin rotates clockwise by 90 degrees. Otherwise, when 'L' is sent, Gururin rotates counterclockwise by 90 degrees.
During the game, you noticed that Gururin obtains magical power by performing special commands. In short, Gururin obtains magical power every time when it performs one round in the clockwise direction from north to north. In more detail, the conditions under which magical power can be obtained is as follows.
* At the beginning of the special commands, Gururin faces north.
* At the end of special commands, Gururin faces north.
* Except for the beginning and the end of special commands, Gururin does not face north.
* During the special commands, Gururin faces north, east, south, and west one or more times, respectively, after the command of 'R'.
At the beginning of the game, Gururin faces north. For example, if the sequence of commands Gururin received in order is 'RRRR' or 'RRLRRLRR', Gururin can obtain magical power. Otherwise, if the sequence of commands is 'LLLL' or 'RLLR', Gururin cannot obtain magical power.
Your task is to calculate how many times Gururin obtained magical power throughout the game. In other words, given the sequence of the commands Gururin received, calculate how many special commands exists in it.
Input
The input consists of a single test case in the format below.
$S$
The first line consists of a string $S$, which represents the sequence of commands Gururin received. $S$ consists of 'L' and 'R'. The length of $S$ is between $1$ and $10^3$ inclusive.
Output
Print the number of times Gururin obtained magical power throughout the game in one line.
Examples
Input
RRRRLLLLRRRR
Output
2
Input
RLLRLLLLRRRLLLRRR
Output
0
Input
LR
Output
0
Input
RRLRRLRRRRLRRLRRRRLRRLRRRRLRRLRR
Output
4
|
instruction
| 0
| 20,232
| 2
| 40,464
|
"Correct Solution:
```
S=input()
count=0
direction=0
for i in range(len(S)):
if S[i]=='R':
direction+=1
else:
direction-=1
if direction==4:
count+=1
direction=0
elif direction==-4:
direction=0
print(count)
```
|
output
| 1
| 20,232
| 2
| 40,465
|
Provide a correct Python 3 solution for this coding contest problem.
You are playing a game called Guru Guru Gururin. In this game, you can move with the vehicle called Gururin. There are two commands you can give to Gururin: 'R' and 'L'. When 'R' is sent, Gururin rotates clockwise by 90 degrees. Otherwise, when 'L' is sent, Gururin rotates counterclockwise by 90 degrees.
During the game, you noticed that Gururin obtains magical power by performing special commands. In short, Gururin obtains magical power every time when it performs one round in the clockwise direction from north to north. In more detail, the conditions under which magical power can be obtained is as follows.
* At the beginning of the special commands, Gururin faces north.
* At the end of special commands, Gururin faces north.
* Except for the beginning and the end of special commands, Gururin does not face north.
* During the special commands, Gururin faces north, east, south, and west one or more times, respectively, after the command of 'R'.
At the beginning of the game, Gururin faces north. For example, if the sequence of commands Gururin received in order is 'RRRR' or 'RRLRRLRR', Gururin can obtain magical power. Otherwise, if the sequence of commands is 'LLLL' or 'RLLR', Gururin cannot obtain magical power.
Your task is to calculate how many times Gururin obtained magical power throughout the game. In other words, given the sequence of the commands Gururin received, calculate how many special commands exists in it.
Input
The input consists of a single test case in the format below.
$S$
The first line consists of a string $S$, which represents the sequence of commands Gururin received. $S$ consists of 'L' and 'R'. The length of $S$ is between $1$ and $10^3$ inclusive.
Output
Print the number of times Gururin obtained magical power throughout the game in one line.
Examples
Input
RRRRLLLLRRRR
Output
2
Input
RLLRLLLLRRRLLLRRR
Output
0
Input
LR
Output
0
Input
RRLRRLRRRRLRRLRRRRLRRLRRRRLRRLRR
Output
4
|
instruction
| 0
| 20,233
| 2
| 40,466
|
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
def f(n):
r = 0
t = 0
for c in n:
if c == 'R':
t += 1
if t == 4:
r += 1
t = 0
else:
t -= 1
if t == -4:
t = 0
return r
while 1:
n = S()
if n == 0:
break
rr.append(f(n))
break
return '\n'.join(map(str, rr))
print(main())
```
|
output
| 1
| 20,233
| 2
| 40,467
|
Provide a correct Python 3 solution for this coding contest problem.
You are playing a game called Guru Guru Gururin. In this game, you can move with the vehicle called Gururin. There are two commands you can give to Gururin: 'R' and 'L'. When 'R' is sent, Gururin rotates clockwise by 90 degrees. Otherwise, when 'L' is sent, Gururin rotates counterclockwise by 90 degrees.
During the game, you noticed that Gururin obtains magical power by performing special commands. In short, Gururin obtains magical power every time when it performs one round in the clockwise direction from north to north. In more detail, the conditions under which magical power can be obtained is as follows.
* At the beginning of the special commands, Gururin faces north.
* At the end of special commands, Gururin faces north.
* Except for the beginning and the end of special commands, Gururin does not face north.
* During the special commands, Gururin faces north, east, south, and west one or more times, respectively, after the command of 'R'.
At the beginning of the game, Gururin faces north. For example, if the sequence of commands Gururin received in order is 'RRRR' or 'RRLRRLRR', Gururin can obtain magical power. Otherwise, if the sequence of commands is 'LLLL' or 'RLLR', Gururin cannot obtain magical power.
Your task is to calculate how many times Gururin obtained magical power throughout the game. In other words, given the sequence of the commands Gururin received, calculate how many special commands exists in it.
Input
The input consists of a single test case in the format below.
$S$
The first line consists of a string $S$, which represents the sequence of commands Gururin received. $S$ consists of 'L' and 'R'. The length of $S$ is between $1$ and $10^3$ inclusive.
Output
Print the number of times Gururin obtained magical power throughout the game in one line.
Examples
Input
RRRRLLLLRRRR
Output
2
Input
RLLRLLLLRRRLLLRRR
Output
0
Input
LR
Output
0
Input
RRLRRLRRRRLRRLRRRRLRRLRRRRLRRLRR
Output
4
|
instruction
| 0
| 20,234
| 2
| 40,468
|
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LS()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
#A
def A():
s = S()
d = 0
ans = 0
k = 0
for i in s:
if i == "R":
if d == 0:
k = 1
d += 1
d %= 4
if d == 0 and k:ans += 1
else:
if d == 0:
k = 0
d -= 1
d %= 4
print(ans)
return
#B
def B():
return
#C
def C():
return
#D
def D():
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#I
def I_():
return
#J
def J():
return
#Solve
if __name__ == "__main__":
A()
```
|
output
| 1
| 20,234
| 2
| 40,469
|
Provide tags and a correct Python 3 solution for this coding contest problem.
[INSPION FullBand Master - INSPION](https://www.youtube.com/watch?v=kwsciXm_7sA)
[INSPION - IOLITE-SUNSTONE](https://www.youtube.com/watch?v=kwsciXm_7sA)
On another floor of the A.R.C. Markland-N, the young man Simon "Xenon" Jackson, takes a break after finishing his project early (as always). Having a lot of free time, he decides to put on his legendary hacker "X" instinct and fight against the gangs of the cyber world.
His target is a network of n small gangs. This network contains exactly n - 1 direct links, each of them connecting two gangs together. The links are placed in such a way that every pair of gangs is connected through a sequence of direct links.
By mining data, Xenon figured out that the gangs used a form of cross-encryption to avoid being busted: every link was assigned an integer from 0 to n - 2 such that all assigned integers are distinct and every integer was assigned to some link. If an intruder tries to access the encrypted data, they will have to surpass S password layers, with S being defined by the following formula:
$$$S = ∑_{1 ≤ u < v ≤ n} mex(u, v)$$$
Here, mex(u, v) denotes the smallest non-negative integer that does not appear on any link on the unique simple path from gang u to gang v.
Xenon doesn't know the way the integers are assigned, but it's not a problem. He decides to let his AI's instances try all the passwords on his behalf, but before that, he needs to know the maximum possible value of S, so that the AIs can be deployed efficiently.
Now, Xenon is out to write the AI scripts, and he is expected to finish them in two hours. Can you find the maximum possible S before he returns?
Input
The first line contains an integer n (2 ≤ n ≤ 3000), the number of gangs in the network.
Each of the next n - 1 lines contains integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i), indicating there's a direct link between gangs u_i and v_i.
It's guaranteed that links are placed in such a way that each pair of gangs will be connected by exactly one simple path.
Output
Print the maximum possible value of S — the number of password layers in the gangs' network.
Examples
Input
3
1 2
2 3
Output
3
Input
5
1 2
1 3
1 4
3 5
Output
10
Note
In the first example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, mex(1, 2) = 0, mex(1, 3) = 2 and mex(2, 3) = 1. Therefore, S = 0 + 2 + 1 = 3.
In the second example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, all non-zero mex value are listed below:
* mex(1, 3) = 1
* mex(1, 5) = 2
* mex(2, 3) = 1
* mex(2, 5) = 2
* mex(3, 4) = 1
* mex(4, 5) = 3
Therefore, S = 1 + 2 + 1 + 2 + 1 + 3 = 10.
|
instruction
| 0
| 20,377
| 2
| 40,754
|
Tags: combinatorics, dfs and similar, dp, greedy, trees
Correct Solution:
```
import sys
# Read input and build the graph
inp = [int(x) for x in sys.stdin.buffer.read().split()]; ii = 0
n = inp[ii]; ii += 1
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
u = inp[ii] - 1; ii += 1
v = inp[ii] - 1; ii += 1
coupl[u].append(v)
coupl[v].append(u)
# Relabel to speed up n^2 operations later on
bfs = [0]
found = [0]*n
found[0] = 1
for node in bfs:
for nei in coupl[node]:
if not found[nei]:
found[nei] = 1
bfs.append(nei)
new_label = [0]*n
for i in range(n):
new_label[bfs[i]] = i
coupl = [coupl[i] for i in bfs]
for c in coupl:
c[:] = [new_label[x] for x in c]
##### DP using multisource bfs
DP = [0] * (n * n)
size = [1] * (n * n)
P = [-1] * (n * n)
# Create the bfs ordering
bfs = [root * n + root for root in range(n)]
for ind in bfs:
P[ind] = ind
for ind in bfs:
node, root = divmod(ind, n)
for nei in coupl[node]:
ind2 = nei * n + root
if P[ind2] == -1:
bfs.append(ind2)
P[ind2] = ind
del bfs[:n]
# Do the DP
for ind in reversed(bfs):
node, root = divmod(ind, n)
pind = P[ind]
parent = pind//n
# Update size of (root, parent)
size[pind] += size[ind]
# Update DP value of (root, parent)
DP[root * n + parent] = DP[pind] = max(DP[pind], DP[ind] + size[ind] * size[root * n + node])
print(max(DP[root * n + root] for root in range(n)))
```
|
output
| 1
| 20,377
| 2
| 40,755
|
Provide tags and a correct Python 3 solution for this coding contest problem.
[INSPION FullBand Master - INSPION](https://www.youtube.com/watch?v=kwsciXm_7sA)
[INSPION - IOLITE-SUNSTONE](https://www.youtube.com/watch?v=kwsciXm_7sA)
On another floor of the A.R.C. Markland-N, the young man Simon "Xenon" Jackson, takes a break after finishing his project early (as always). Having a lot of free time, he decides to put on his legendary hacker "X" instinct and fight against the gangs of the cyber world.
His target is a network of n small gangs. This network contains exactly n - 1 direct links, each of them connecting two gangs together. The links are placed in such a way that every pair of gangs is connected through a sequence of direct links.
By mining data, Xenon figured out that the gangs used a form of cross-encryption to avoid being busted: every link was assigned an integer from 0 to n - 2 such that all assigned integers are distinct and every integer was assigned to some link. If an intruder tries to access the encrypted data, they will have to surpass S password layers, with S being defined by the following formula:
$$$S = ∑_{1 ≤ u < v ≤ n} mex(u, v)$$$
Here, mex(u, v) denotes the smallest non-negative integer that does not appear on any link on the unique simple path from gang u to gang v.
Xenon doesn't know the way the integers are assigned, but it's not a problem. He decides to let his AI's instances try all the passwords on his behalf, but before that, he needs to know the maximum possible value of S, so that the AIs can be deployed efficiently.
Now, Xenon is out to write the AI scripts, and he is expected to finish them in two hours. Can you find the maximum possible S before he returns?
Input
The first line contains an integer n (2 ≤ n ≤ 3000), the number of gangs in the network.
Each of the next n - 1 lines contains integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i), indicating there's a direct link between gangs u_i and v_i.
It's guaranteed that links are placed in such a way that each pair of gangs will be connected by exactly one simple path.
Output
Print the maximum possible value of S — the number of password layers in the gangs' network.
Examples
Input
3
1 2
2 3
Output
3
Input
5
1 2
1 3
1 4
3 5
Output
10
Note
In the first example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, mex(1, 2) = 0, mex(1, 3) = 2 and mex(2, 3) = 1. Therefore, S = 0 + 2 + 1 = 3.
In the second example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, all non-zero mex value are listed below:
* mex(1, 3) = 1
* mex(1, 5) = 2
* mex(2, 3) = 1
* mex(2, 5) = 2
* mex(3, 4) = 1
* mex(4, 5) = 3
Therefore, S = 1 + 2 + 1 + 2 + 1 + 3 = 10.
|
instruction
| 0
| 20,378
| 2
| 40,756
|
Tags: combinatorics, dfs and similar, dp, greedy, trees
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import deque
N=int(input())
E=[[] for i in range(N+1)]
for i in range(N-1):
x,y=map(int,input().split())
E[x].append(y)
E[y].append(x)
Q=deque()
USE=[0]*(N+1)
Q.append(1)
H=[0]*(N+1)
H[1]=1
USE[1]=1
P=[-1]*(N+1)
QI=deque()
while Q:
x=Q.pop()
for to in E[x]:
if USE[to]==0:
USE[to]=1
H[to]=H[x]+1
P[to]=x
Q.append(to)
QI.append((x,to,to,x))
P[to]=x
EH=[(h,ind+1) for ind,h in enumerate(H[1:])]
EH.sort(reverse=True)
COME=[1]*(N+1)
USE=[0]*(N+1)
for h,ind in EH:
USE[ind]=1
for to in E[ind]:
if USE[to]==0:
COME[to]+=COME[ind]
GO=[N-COME[i] for i in range(N+1)]
DP=[[0]*(N+1) for i in range(N+1)]
USE=[[0]*(N+1) for i in range(N+1)]
while QI:
l,frl,r,frr=QI.pop()
if P[l]==frl:
A1=COME[l]
else:
A1=GO[frl]
if P[r]==frr:
A2=COME[r]
else:
A2=GO[frr]
DP[l][r]=DP[r][l]=A1*A2+max(DP[frl][r],DP[l][frr])
for to in E[l]:
if to==frl or USE[to][r]==1:
continue
USE[to][r]=USE[r][to]=1
QI.appendleft((to,l,r,frr))
#for to in E[r]:
# if to==frr or USE[l][to]==1:
# continue
# USE[to][l]=USE[l][to]=1
# QI.append((l,frl,to,r))
ANS=0
for d in DP:
ANS=max(ANS,max(d))
print(ANS)
```
|
output
| 1
| 20,378
| 2
| 40,757
|
Provide tags and a correct Python 3 solution for this coding contest problem.
[INSPION FullBand Master - INSPION](https://www.youtube.com/watch?v=kwsciXm_7sA)
[INSPION - IOLITE-SUNSTONE](https://www.youtube.com/watch?v=kwsciXm_7sA)
On another floor of the A.R.C. Markland-N, the young man Simon "Xenon" Jackson, takes a break after finishing his project early (as always). Having a lot of free time, he decides to put on his legendary hacker "X" instinct and fight against the gangs of the cyber world.
His target is a network of n small gangs. This network contains exactly n - 1 direct links, each of them connecting two gangs together. The links are placed in such a way that every pair of gangs is connected through a sequence of direct links.
By mining data, Xenon figured out that the gangs used a form of cross-encryption to avoid being busted: every link was assigned an integer from 0 to n - 2 such that all assigned integers are distinct and every integer was assigned to some link. If an intruder tries to access the encrypted data, they will have to surpass S password layers, with S being defined by the following formula:
$$$S = ∑_{1 ≤ u < v ≤ n} mex(u, v)$$$
Here, mex(u, v) denotes the smallest non-negative integer that does not appear on any link on the unique simple path from gang u to gang v.
Xenon doesn't know the way the integers are assigned, but it's not a problem. He decides to let his AI's instances try all the passwords on his behalf, but before that, he needs to know the maximum possible value of S, so that the AIs can be deployed efficiently.
Now, Xenon is out to write the AI scripts, and he is expected to finish them in two hours. Can you find the maximum possible S before he returns?
Input
The first line contains an integer n (2 ≤ n ≤ 3000), the number of gangs in the network.
Each of the next n - 1 lines contains integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i), indicating there's a direct link between gangs u_i and v_i.
It's guaranteed that links are placed in such a way that each pair of gangs will be connected by exactly one simple path.
Output
Print the maximum possible value of S — the number of password layers in the gangs' network.
Examples
Input
3
1 2
2 3
Output
3
Input
5
1 2
1 3
1 4
3 5
Output
10
Note
In the first example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, mex(1, 2) = 0, mex(1, 3) = 2 and mex(2, 3) = 1. Therefore, S = 0 + 2 + 1 = 3.
In the second example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, all non-zero mex value are listed below:
* mex(1, 3) = 1
* mex(1, 5) = 2
* mex(2, 3) = 1
* mex(2, 5) = 2
* mex(3, 4) = 1
* mex(4, 5) = 3
Therefore, S = 1 + 2 + 1 + 2 + 1 + 3 = 10.
|
instruction
| 0
| 20,379
| 2
| 40,758
|
Tags: combinatorics, dfs and similar, dp, greedy, trees
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import deque
N=int(input())
E=[[] for i in range(N+1)]
for i in range(N-1):
x,y=map(int,input().split())
E[x].append(y)
E[y].append(x)
Q=deque()
USE=[0]*(N+1)
Q.append(1)
H=[0]*(N+1)
H[1]=1
USE[1]=1
P=[-1]*(N+1)
QI=deque()
while Q:
x=Q.pop()
for to in E[x]:
if USE[to]==0:
USE[to]=1
H[to]=H[x]+1
P[to]=x
Q.append(to)
QI.append((x,to,to,x))
P[to]=x
EH=[(h,ind+1) for ind,h in enumerate(H[1:])]
EH.sort(reverse=True)
COME=[1]*(N+1)
USE=[0]*(N+1)
for h,ind in EH:
USE[ind]=1
for to in E[ind]:
if USE[to]==0:
COME[to]+=COME[ind]
GO=[N-COME[i] for i in range(N+1)]
DP=[[0]*(N+1) for i in range(N+1)]
USE=[[0]*(N+1) for i in range(N+1)]
while QI:
l,frl,r,frr=QI.popleft()
if P[l]==frl:
A1=COME[l]
else:
A1=GO[frl]
if P[r]==frr:
A2=COME[r]
else:
A2=GO[frr]
DP[l][r]=DP[r][l]=A1*A2+max(DP[frl][r],DP[l][frr])
for to in E[l]:
if to==frl or USE[to][r]==1:
continue
USE[to][r]=USE[r][to]=1
QI.append((to,l,r,frr))
#for to in E[r]:
# if to==frr or USE[l][to]==1:
# continue
# USE[to][l]=USE[l][to]=1
# QI.append((l,frl,to,r))
ANS=0
for d in DP:
ANS=max(ANS,max(d))
print(ANS)
```
|
output
| 1
| 20,379
| 2
| 40,759
|
Provide tags and a correct Python 3 solution for this coding contest problem.
[INSPION FullBand Master - INSPION](https://www.youtube.com/watch?v=kwsciXm_7sA)
[INSPION - IOLITE-SUNSTONE](https://www.youtube.com/watch?v=kwsciXm_7sA)
On another floor of the A.R.C. Markland-N, the young man Simon "Xenon" Jackson, takes a break after finishing his project early (as always). Having a lot of free time, he decides to put on his legendary hacker "X" instinct and fight against the gangs of the cyber world.
His target is a network of n small gangs. This network contains exactly n - 1 direct links, each of them connecting two gangs together. The links are placed in such a way that every pair of gangs is connected through a sequence of direct links.
By mining data, Xenon figured out that the gangs used a form of cross-encryption to avoid being busted: every link was assigned an integer from 0 to n - 2 such that all assigned integers are distinct and every integer was assigned to some link. If an intruder tries to access the encrypted data, they will have to surpass S password layers, with S being defined by the following formula:
$$$S = ∑_{1 ≤ u < v ≤ n} mex(u, v)$$$
Here, mex(u, v) denotes the smallest non-negative integer that does not appear on any link on the unique simple path from gang u to gang v.
Xenon doesn't know the way the integers are assigned, but it's not a problem. He decides to let his AI's instances try all the passwords on his behalf, but before that, he needs to know the maximum possible value of S, so that the AIs can be deployed efficiently.
Now, Xenon is out to write the AI scripts, and he is expected to finish them in two hours. Can you find the maximum possible S before he returns?
Input
The first line contains an integer n (2 ≤ n ≤ 3000), the number of gangs in the network.
Each of the next n - 1 lines contains integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i), indicating there's a direct link between gangs u_i and v_i.
It's guaranteed that links are placed in such a way that each pair of gangs will be connected by exactly one simple path.
Output
Print the maximum possible value of S — the number of password layers in the gangs' network.
Examples
Input
3
1 2
2 3
Output
3
Input
5
1 2
1 3
1 4
3 5
Output
10
Note
In the first example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, mex(1, 2) = 0, mex(1, 3) = 2 and mex(2, 3) = 1. Therefore, S = 0 + 2 + 1 = 3.
In the second example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, all non-zero mex value are listed below:
* mex(1, 3) = 1
* mex(1, 5) = 2
* mex(2, 3) = 1
* mex(2, 5) = 2
* mex(3, 4) = 1
* mex(4, 5) = 3
Therefore, S = 1 + 2 + 1 + 2 + 1 + 3 = 10.
|
instruction
| 0
| 20,380
| 2
| 40,760
|
Tags: combinatorics, dfs and similar, dp, greedy, trees
Correct Solution:
```
import sys
# Read input and build the graph
inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
n = inp[ii]; ii += 1
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
u = inp[ii] - 1; ii += 1
v = inp[ii] - 1; ii += 1
coupl[u].append(v)
coupl[v].append(u)
# Relable to speed up n^2 operations later on
bfs = [0]
found = [0]*n
found[0] = 1
for node in bfs:
for nei in coupl[node]:
if not found[nei]:
found[nei] = 1
bfs.append(nei)
naming = [0]*n
for i in range(n):
naming[bfs[i]] = i
coupl = [coupl[i] for i in bfs]
for c in coupl:
c[:] = [naming[x] for x in c]
# Precalculations for the DP
subsizes = []
Ps = []
for root in range(n):
P = [-1]*n
P[root] = root
bfs = [root]
for node in bfs:
for nei in coupl[node]:
if P[nei] == -1:
P[nei] = node
bfs.append(nei)
Ps += P
subsize = [1]*n
for node in reversed(bfs[1:]):
subsize[P[node]] += subsize[node]
other = [0]*n
for nei in coupl[root]:
other[nei] = subsize[root] - subsize[nei]
for node in bfs:
if P[node] != root:
other[node] = other[P[node]]
subsize[node] *= other[node]
subsizes += subsize
# DP using a multisource bfs
DP = [0]*(n * n)
bfs = [n * root + root for root in range(n)]
for ind in bfs:
root, node = divmod(ind, n)
for nei in coupl[node]:
ind2 = ind - node + nei
if Ps[ind2] == node:
DP[n * nei + root] = DP[ind2] = max(DP[ind2], DP[ind] + subsizes[ind2])
bfs.append(ind2)
print(max(DP))
```
|
output
| 1
| 20,380
| 2
| 40,761
|
Provide tags and a correct Python 3 solution for this coding contest problem.
[INSPION FullBand Master - INSPION](https://www.youtube.com/watch?v=kwsciXm_7sA)
[INSPION - IOLITE-SUNSTONE](https://www.youtube.com/watch?v=kwsciXm_7sA)
On another floor of the A.R.C. Markland-N, the young man Simon "Xenon" Jackson, takes a break after finishing his project early (as always). Having a lot of free time, he decides to put on his legendary hacker "X" instinct and fight against the gangs of the cyber world.
His target is a network of n small gangs. This network contains exactly n - 1 direct links, each of them connecting two gangs together. The links are placed in such a way that every pair of gangs is connected through a sequence of direct links.
By mining data, Xenon figured out that the gangs used a form of cross-encryption to avoid being busted: every link was assigned an integer from 0 to n - 2 such that all assigned integers are distinct and every integer was assigned to some link. If an intruder tries to access the encrypted data, they will have to surpass S password layers, with S being defined by the following formula:
$$$S = ∑_{1 ≤ u < v ≤ n} mex(u, v)$$$
Here, mex(u, v) denotes the smallest non-negative integer that does not appear on any link on the unique simple path from gang u to gang v.
Xenon doesn't know the way the integers are assigned, but it's not a problem. He decides to let his AI's instances try all the passwords on his behalf, but before that, he needs to know the maximum possible value of S, so that the AIs can be deployed efficiently.
Now, Xenon is out to write the AI scripts, and he is expected to finish them in two hours. Can you find the maximum possible S before he returns?
Input
The first line contains an integer n (2 ≤ n ≤ 3000), the number of gangs in the network.
Each of the next n - 1 lines contains integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i), indicating there's a direct link between gangs u_i and v_i.
It's guaranteed that links are placed in such a way that each pair of gangs will be connected by exactly one simple path.
Output
Print the maximum possible value of S — the number of password layers in the gangs' network.
Examples
Input
3
1 2
2 3
Output
3
Input
5
1 2
1 3
1 4
3 5
Output
10
Note
In the first example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, mex(1, 2) = 0, mex(1, 3) = 2 and mex(2, 3) = 1. Therefore, S = 0 + 2 + 1 = 3.
In the second example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, all non-zero mex value are listed below:
* mex(1, 3) = 1
* mex(1, 5) = 2
* mex(2, 3) = 1
* mex(2, 5) = 2
* mex(3, 4) = 1
* mex(4, 5) = 3
Therefore, S = 1 + 2 + 1 + 2 + 1 + 3 = 10.
|
instruction
| 0
| 20,381
| 2
| 40,762
|
Tags: combinatorics, dfs and similar, dp, greedy, trees
Correct Solution:
```
import sys
# Read input and build the graph
inp = [int(x) for x in sys.stdin.buffer.read().split()]; ii = 0
n = inp[ii]; ii += 1
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
u = inp[ii] - 1; ii += 1
v = inp[ii] - 1; ii += 1
coupl[u].append(v)
coupl[v].append(u)
# Relabel to speed up n^2 operations later on
bfs = [0]
found = [0]*n
found[0] = 1
for node in bfs:
for nei in coupl[node]:
if not found[nei]:
found[nei] = 1
bfs.append(nei)
new_label = [0]*n
for i in range(n):
new_label[bfs[i]] = i
coupl = [coupl[i] for i in bfs]
for c in coupl:
c[:] = [new_label[x] for x in c]
##### DP using multisource bfs
DP = [0] * (n * n)
size = [1] * (n * n)
P = [-1] * (n * n)
# Create the bfs ordering
bfs = [root * n + root for root in range(n)]
for ind in bfs:
P[ind] = ind
for ind in bfs:
node, root = divmod(ind, n)
for nei in coupl[node]:
ind2 = nei * n + root
if P[ind2] == -1:
bfs.append(ind2)
P[ind2] = ind
del bfs[:n]
# Do the DP
for ind in reversed(bfs):
node, root = divmod(ind, n)
ind2 = root * n + node
pind = P[ind]
parent = pind//n
# Update size of (root, parent)
size[pind] += size[ind]
# Update DP value of (root, parent)
DP[pind] = max(DP[pind], max(DP[ind], DP[ind2]) + size[ind] * size[ind2])
print(max(DP[root * n + root] for root in range(n)))
```
|
output
| 1
| 20,381
| 2
| 40,763
|
Provide tags and a correct Python 3 solution for this coding contest problem.
[INSPION FullBand Master - INSPION](https://www.youtube.com/watch?v=kwsciXm_7sA)
[INSPION - IOLITE-SUNSTONE](https://www.youtube.com/watch?v=kwsciXm_7sA)
On another floor of the A.R.C. Markland-N, the young man Simon "Xenon" Jackson, takes a break after finishing his project early (as always). Having a lot of free time, he decides to put on his legendary hacker "X" instinct and fight against the gangs of the cyber world.
His target is a network of n small gangs. This network contains exactly n - 1 direct links, each of them connecting two gangs together. The links are placed in such a way that every pair of gangs is connected through a sequence of direct links.
By mining data, Xenon figured out that the gangs used a form of cross-encryption to avoid being busted: every link was assigned an integer from 0 to n - 2 such that all assigned integers are distinct and every integer was assigned to some link. If an intruder tries to access the encrypted data, they will have to surpass S password layers, with S being defined by the following formula:
$$$S = ∑_{1 ≤ u < v ≤ n} mex(u, v)$$$
Here, mex(u, v) denotes the smallest non-negative integer that does not appear on any link on the unique simple path from gang u to gang v.
Xenon doesn't know the way the integers are assigned, but it's not a problem. He decides to let his AI's instances try all the passwords on his behalf, but before that, he needs to know the maximum possible value of S, so that the AIs can be deployed efficiently.
Now, Xenon is out to write the AI scripts, and he is expected to finish them in two hours. Can you find the maximum possible S before he returns?
Input
The first line contains an integer n (2 ≤ n ≤ 3000), the number of gangs in the network.
Each of the next n - 1 lines contains integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i), indicating there's a direct link between gangs u_i and v_i.
It's guaranteed that links are placed in such a way that each pair of gangs will be connected by exactly one simple path.
Output
Print the maximum possible value of S — the number of password layers in the gangs' network.
Examples
Input
3
1 2
2 3
Output
3
Input
5
1 2
1 3
1 4
3 5
Output
10
Note
In the first example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, mex(1, 2) = 0, mex(1, 3) = 2 and mex(2, 3) = 1. Therefore, S = 0 + 2 + 1 = 3.
In the second example, one can achieve the maximum S with the following assignment:
<image>
With this assignment, all non-zero mex value are listed below:
* mex(1, 3) = 1
* mex(1, 5) = 2
* mex(2, 3) = 1
* mex(2, 5) = 2
* mex(3, 4) = 1
* mex(4, 5) = 3
Therefore, S = 1 + 2 + 1 + 2 + 1 + 3 = 10.
|
instruction
| 0
| 20,382
| 2
| 40,764
|
Tags: combinatorics, dfs and similar, dp, greedy, trees
Correct Solution:
```
''' Testing Python performance @Pajenegod's solution '''
INF = 10 ** 10
def main():
#print = out.append
''' Cook your dish here! '''
# Read input and build the graph
n = get_int()
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
u, v = get_list()
coupl[u-1].append(v-1)
coupl[v-1].append(u-1)
# Relabel to speed up n^2 operations later on
bfs = [0]
found = [0] * n
found[0] = 1
for node in bfs:
for nei in coupl[node]:
if not found[nei]:
found[nei] = 1
bfs.append(nei)
new_label = [0] * n
for i in range(n):
new_label[bfs[i]] = i
coupl = [coupl[i] for i in bfs]
for c in coupl:
c[:] = [new_label[x] for x in c]
##### DP using multisource bfs
DP = [0] * (n * n)
size = [1] * (n * n)
P = [-1] * (n * n)
# Create the bfs ordering
bfs = [root * n + root for root in range(n)]
for ind in bfs:
P[ind] = ind
for ind in bfs:
node, root = divmod(ind, n)
for nei in coupl[node]:
ind2 = nei * n + root
if P[ind2] == -1:
bfs.append(ind2)
P[ind2] = ind
del bfs[:n]
# Do the DP
for ind in reversed(bfs):
node, root = divmod(ind, n)
ind2 = root * n + node
pind = P[ind]
parent = pind // n
# Update size of (root, parent)
size[pind] += size[ind]
# Update DP value of (root, parent)
DP[pind] = max(DP[pind], max(DP[ind], DP[ind2]) + size[ind] * size[ind2])
print(max(DP[root * n + root] for root in range(n)))
''' Pythonista fLite 1.1 '''
import sys
#from collections import defaultdict, Counter, deque
# from bisect import bisect_left, bisect_right
# from functools import reduce
# import math
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
out = []
get_int = lambda: int(input())
get_list = lambda: list(map(int, input().split()))
main()
#[main() for _ in range(int(input()))]
print(*out, sep='\n')
```
|
output
| 1
| 20,382
| 2
| 40,765
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
|
instruction
| 0
| 20,689
| 2
| 41,378
|
Tags: implementation
Correct Solution:
```
n = int(input())
dominos = set()
for r in range(n):
if r != 0:
input()
domino = list(map(int, list(input().strip())))
domino += reversed(list(map(int, list(input().strip()))))
numbers = []
for i in range(4):
numbers.append(domino[0] * 1000 + domino[1] * 100 + domino[2] * 10 +
domino[3])
domino = domino[1:] + domino[0:1]
#print(numbers)
dominos.add(min(numbers))
#print(dominos)
print(len(dominos))
```
|
output
| 1
| 20,689
| 2
| 41,379
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
|
instruction
| 0
| 20,690
| 2
| 41,380
|
Tags: implementation
Correct Solution:
```
p = {}
for i in range(int(input()) - 1):
t = input() + input()[:: -1]
p[min(t, t[1: ] + t[0], t[2: ] + t[: 2], t[3] + t[: 3])] = 0
input()
t = input() + input()[:: -1]
p[min(t, t[1: ] + t[0], t[2: ] + t[: 2], t[3] + t[: 3])] = 0
print(len(p))
```
|
output
| 1
| 20,690
| 2
| 41,381
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
|
instruction
| 0
| 20,691
| 2
| 41,382
|
Tags: implementation
Correct Solution:
```
def sp(n):
return n//10,n%10
def rsp(n):
return n%10,n//10
def dsp(fn,sn):
return sp(int(fn))+rsp(int(sn))
def left(b,k):
return b[k:]+b[:k]
def gbv(b):
return [left(b,x) for x in range(4)]
def att(ttc,vc,b):
if b not in ttc:
for c in gbv(b):
if c in ttc:
continue
ttc[c]=len(vc)
vc.append(1)
else:
vc[ttc[b]]+=1
n=int(input())
v=[]
for c in range(n-1):
v.append(dsp(input(),input()))
input()
v.append(dsp(input(),input()))
vc=[]
ttc=dict()
for c in v:
att(ttc,vc,c)
print(len(vc))
```
|
output
| 1
| 20,691
| 2
| 41,383
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
|
instruction
| 0
| 20,692
| 2
| 41,384
|
Tags: implementation
Correct Solution:
```
'''
12
34
'1234' -> '3142'
'''
def rotate(line: str) -> str:
# return f'{}{}{}{}'
return '{}{}{}{}'.format(line[2], line[0], line[3], line[1])
def equals(it1: str, it2: str) -> bool:
for i in range(4):
if it1 == it2:
return True
it1 = rotate(it1)
return False
def contains(cont: list, element: str) -> bool:
ln = len(cont)
for item in cont:
if equals(element, item):
return True
return False
n, cont, temp = int(input()), [], ''
for i in range(1, n * 3):
if i % 3 == 0:
input()
elif i % 3 == 1:
temp += input()
else:
temp += input()
if not contains(cont, temp):
cont.append(temp)
temp = ''
print(len(cont))
'''
4
51
26
**
54
35
**
25
61
**
45
53
'''
```
|
output
| 1
| 20,692
| 2
| 41,385
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
|
instruction
| 0
| 20,693
| 2
| 41,386
|
Tags: implementation
Correct Solution:
```
##################
# 1st October 2019
##################
###########################################################################
class Amulet:
# Method to determine if given amulet is similar.
@staticmethod
def areSimilar(amuOne,amuTwo):
listOne = amuOne.listAllRotations()
listTwo = amuTwo.listAllRotations()
for i in listOne:
for j in listTwo:
if i == j: return True
return False
# Method to rotate Amulet 90 degrees clockwise.
def rotate(self):
m = self.matrix
interim = [[m[1][0],m[0][0]],[m[1][1],m[0][1]]]
self.matrix = interim
# Method to list all rotations.
def listAllRotations(self):
rotations = []
for i in range(3):
rotations.append(self.matrix);self.rotate()
return rotations
# Constructor.
def __init__(self):
self.matrix = []
for i in range(2):
self.matrix.append(list(input()))
###########################################################################
# Getting Amulets from codeforces.
count,amulets = int(input()),[]
for i in range(count):
amulets.append(Amulet())
# Ignoring delimeters.
if i<count-1:input()
# For Storing equivalent classes.
eqClasses = []
# Algorithm.
for i in range(len(amulets)):
index,found = 0,False
# Determine if amulet belongs to existing classes.
while not found and index<len(eqClasses):
if Amulet.areSimilar(eqClasses[index][0],amulets[i]):
found = True
else: index = index +1
# If found adding to class otherwise creating new class.
if found:
eqClasses[index].append(amulets[i])
else: eqClasses.append([amulets[i]])
# Solution.
print(len(eqClasses))
###########################################################################
#######################################
# Programming Credits atifcppprogrammer
#######################################
```
|
output
| 1
| 20,693
| 2
| 41,387
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
|
instruction
| 0
| 20,694
| 2
| 41,388
|
Tags: implementation
Correct Solution:
```
d={}
I=input
n=int(I())
for i in range(n):
s=I()+I()
for _ in range(4):
s=s[2]+s[0]+s[3]+s[1]
if s in d:d[s]+=1;break
else:d[s]=1
if i+1<n:I()
print(len(d))
```
|
output
| 1
| 20,694
| 2
| 41,389
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
|
instruction
| 0
| 20,695
| 2
| 41,390
|
Tags: implementation
Correct Solution:
```
i,ans=set(),0
n=int(input())
for _ in range(n):
a,b=input(),input()
if _<n-1:input()
cur=(a[0],a[1],b[1],b[0])
for j in range(4):
if cur in i:
break
cur=list(cur)
cur=tuple([cur.pop()]+cur)
else:
i.add((cur))
ans+=1
print(ans)
```
|
output
| 1
| 20,695
| 2
| 41,391
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
|
instruction
| 0
| 20,696
| 2
| 41,392
|
Tags: implementation
Correct Solution:
```
nro_amu = int(input())
amu = []
for i in range(nro_amu):
fila1 = list(input())
fila2 = list(input())
fila1.extend(fila2)
amu.append(fila1)
if i+1 != nro_amu:
_ = input()
pilas = []
nro_pilas = 0
for a in amu:
existe = False
for i in range(nro_pilas):
for j in range(4):
if pilas[i][j] == a:
existe = True
if not existe:
pilas.append([a,[a[2],a[0],a[3],a[1]],[a[3],a[2],a[1],a[0]],[a[1],a[3],a[0],a[2]]])
nro_pilas = nro_pilas+1
print(nro_pilas)
```
|
output
| 1
| 20,696
| 2
| 41,393
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
Submitted Solution:
```
def rotate(item: str) -> str:
# return item[2]+item[0]+item[3]+item[1]
return f'{item[2]}{item[0]}{item[3]}{item[1]}'
def equals(it1: str, it2: str) -> bool:
for i in range(4):
if it1 == it2:
return True
it2 = rotate(it2)
return False
def contains(cont: list, element: str) -> bool:
for item in cont:
if equals(item, element):
return True
return False
n = int(input())
cont, helper = [], '',
for i in range(1, n * 3):
if i % 3 == 1:
helper = input()
elif i % 3 == 2:
helper += input()
# cont.append(helper) ????
if not contains(cont, helper):
cont.append(helper)
else:
input()
print(len(cont))
```
|
instruction
| 0
| 20,697
| 2
| 41,394
|
Yes
|
output
| 1
| 20,697
| 2
| 41,395
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
Submitted Solution:
```
n = int(input())
d = {}
for i in range(n):
s = ''
found = False
s = input() + input()
if i != n - 1:
input()
t = s[2] + s[0] + s[3] + s[1]
p = s[3] + s[2] + s[1] + s[0]
q = s[1] + s[3] + s[0] + s[2]
for key in d.keys():
for var in d[key].keys():
if s == var:
found = True
d[key][var] += 1
if not found:
d[s] = {}
d[s][s] = 1
d[s][t] = 0
d[s][p] = 0
d[s][q] = 0
print(len(d))
```
|
instruction
| 0
| 20,698
| 2
| 41,396
|
Yes
|
output
| 1
| 20,698
| 2
| 41,397
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
Submitted Solution:
```
d=dict();o=int(input())
for t in range(o):
x,y=input().replace('',' ').split()
i,j=input().replace('',' ').split()
if t!=o-1:input()
if not((x,y,i,j) in d or (y,j,x,i) in d or (j,i,y,x) in d or (i,x,j,y) in d):d[(x,y,i,j)]=1
print(len(d))
```
|
instruction
| 0
| 20,699
| 2
| 41,398
|
Yes
|
output
| 1
| 20,699
| 2
| 41,399
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
Submitted Solution:
```
import copy
def rotations(amulet):
rots_ans = [amulet]
a = [2, 0, 3, 1]
b = [3, 2, 1, 0]
c = [1, 3, 0, 2]
rots_ans.append(rotation(amulet, a))
rots_ans.append(rotation(amulet, b))
rots_ans.append(rotation(amulet, c))
return copy.deepcopy(rots_ans)
def rotation(amulet, x):
rot_ans = []
for i in x:
rot_ans.append(amulet[i])
return copy.deepcopy(rot_ans)
n = int(input())
input_list = []
for i in range((n * 3) - 1):
line = input()
if not line == '**':
input_list = input_list + [line[0]] + [line[1]]
amulets_list = []
for j in range(n):
amulets_list.append([input_list[0], input_list[1], input_list[2], input_list[3]])
for k in range(4):
del input_list[0]
answer = 0
amulets_types = []
for i in amulets_list:
if not i in amulets_types:
answer += 1
amulets_types = amulets_types + rotations(i)
print(answer)
```
|
instruction
| 0
| 20,700
| 2
| 41,400
|
Yes
|
output
| 1
| 20,700
| 2
| 41,401
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
Submitted Solution:
```
def comp(a,b):
a2 = a[1]+a[0]+a[3]+a[2]
a3 = a[0]+a[2]+a[1]+a[3]
a4 = a[2]+a[0]+a[3]+a[1]
a5 = a[1]+a[3]+a[0]+a[2]
a6 = a[3]+a[2]+a[1]+a[0]
#print(a,a2,a3,a4)
return (a == b or a2 == b or a3 == b or a4 == b or a5 == b or a6 == b)
n = int(input())
cont_parada = 0
re = ""
ans = 0
arr = []
for j in range(n):
arr.append(input()+input())
if cont_parada != n-1:
pula = input()
cont_parada+=1
mark = [0]*n
for i in range(n):
if mark[i]:
continue
for j in range(i+1,n):
if mark[j]:
continue
if comp(arr[i],arr[j]):
mark[j] = 1
ans+=1
print(ans)
```
|
instruction
| 0
| 20,701
| 2
| 41,402
|
No
|
output
| 1
| 20,701
| 2
| 41,403
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
Submitted Solution:
```
def comp(a,b):
a2 = a[1]+a[0]+a[3]+a[2]
a3 = a[0]+a[2]+a[1]+a[3]
a4 = a[2]+a[0]+a[3]+a[1]
a5 = a[1]+a[3]+a[0]+a[2]
a6 = a[3]+a[2]+a[1]+a[0]
a7 = a[2]+a[3]+a[1]+a[0]
a8 = a[3]+a[1]+a[2]+a[0]
#print(a,a2,a3,a4)
#print(a,a2,a3,a4,a5,a6)
return (a == b or a2 == b or a3 == b or a4 == b or a5 == b or a6 == b or a7 == b or a8 == b)
n = int(input())
cont_parada = 0
re = ""
ans = 0
arr = []
for j in range(n):
arr.append(input()+input())
if cont_parada != n-1:
pula = input()
cont_parada+=1
mark = [0]*n
for i in range(n):
if mark[i]:
continue
for j in range(i+1,n):
if mark[j]:
continue
if comp(arr[i],arr[j]):
mark[j] = 1
ans+=1
print(ans)
```
|
instruction
| 0
| 20,702
| 2
| 41,404
|
No
|
output
| 1
| 20,702
| 2
| 41,405
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
Submitted Solution:
```
def comp(a,b):
a2 = a[1]+a[0]+a[3]+a[2]
a3 = a[0]+a[2]+a[1]+a[3]
a4 = a[2]+a[0]+a[3]+a[1]
a5 = a[1]+a[3]+a[0]+a[2]
#print(a,a2,a3,a4)
return (a == b or a2 == b or a3 == b or a4 == b or a5 == b)
n = int(input())
cont_parada = 0
re = ""
ans = 0
arr = []
for j in range(n):
arr.append(input()+input())
if cont_parada != n-1:
pula = input()
cont_parada+=1
mark = [0]*n
for i in range(n):
if mark[i]:
continue
for j in range(i+1,n):
if mark[j]:
continue
if comp(arr[i],arr[j]):
mark[j] = 1
ans+=1
print(ans)
```
|
instruction
| 0
| 20,703
| 2
| 41,406
|
No
|
output
| 1
| 20,703
| 2
| 41,407
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Cheaterius is a famous in all the Berland astrologist, magician and wizard, and he also is a liar and a cheater. One of his latest inventions is Cheaterius' amulets! They bring luck and wealth, but are rather expensive. Cheaterius makes them himself. The technology of their making is kept secret. But we know that throughout long nights Cheaterius glues together domino pairs with super glue to get squares 2 × 2 which are the Cheaterius' magic amulets!
<image> That's what one of Cheaterius's amulets looks like
After a hard night Cheaterius made n amulets. Everyone of them represents a square 2 × 2, every quarter contains 1 to 6 dots. Now he wants sort them into piles, every pile must contain similar amulets. Two amulets are called similar if they can be rotated by 90, 180 or 270 degrees so that the following condition is met: the numbers of dots in the corresponding quarters should be the same. It is forbidden to turn over the amulets.
Write a program that by the given amulets will find the number of piles on Cheaterius' desk.
Input
The first line contains an integer n (1 ≤ n ≤ 1000), where n is the number of amulets. Then the amulet's descriptions are contained. Every description occupies two lines and contains two numbers (from 1 to 6) in each line. Between every pair of amulets the line "**" is located.
Output
Print the required number of piles.
Examples
Input
4
31
23
**
31
23
**
13
32
**
32
13
Output
1
Input
4
51
26
**
54
35
**
25
61
**
45
53
Output
2
Submitted Solution:
```
n=int(input())
m=[]
for i in range(n):
a=input()
if a[0]=='0' or a[0]=='1' or a[0]=='2' or a[0]=='3' or a[0]=='4' or a[0]=='5' or a[0]=='6' or a[0]=='7' or a[0]=='8' or a[0]=='9':
a=int(a)
m.append(a)
kol=0
for i in range(len(m)):
if type(m[i])==int and m[i]<18:
kol+=1
if type(m[i])==str and (m[i]=='ABSINTH' or m[i]=='BEER' or m[i]=='BRANDY' or m[i]=='CHAMPAGNE' or m[i]=='GIN' or m[i]=='RUM' or m[i]=='SAKE' or m[i]=='TEQUILA' or m[i]=='VODKA' or m[i]=='WHISKEY' or m[i]=='WINE' ):
kol+=1
print(kol)
```
|
instruction
| 0
| 20,704
| 2
| 41,408
|
No
|
output
| 1
| 20,704
| 2
| 41,409
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
|
instruction
| 0
| 20,875
| 2
| 41,750
|
Tags: combinatorics, constructive algorithms, math, number theory
Correct Solution:
```
m, n, k = map(int, input().split())
mod = 10**9+7
def powermod(x,y):
res = 1
while(y>0):
if y&1:
res=res*x%mod
x=x*x%mod
y=y>>1
return res
if (m%2 != n%2 and k == -1):
print (0)
else:
print (powermod(2, (m-1)*(n-1)))
```
|
output
| 1
| 20,875
| 2
| 41,751
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
|
instruction
| 0
| 20,877
| 2
| 41,754
|
Tags: combinatorics, constructive algorithms, math, number theory
Correct Solution:
```
n, m, k = map(int, input().split())
if (k == -1 and (n % 2 != m % 2)):
print(0)
else:
ans = pow(2, (n - 1) * (m- 1), 1000000007)
print(ans)
```
|
output
| 1
| 20,877
| 2
| 41,755
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
|
instruction
| 0
| 20,878
| 2
| 41,756
|
Tags: combinatorics, constructive algorithms, math, number theory
Correct Solution:
```
from math import *
n, m, k = map(int,input().split(' '))
ans = (2**((n-1)*(m-1)%1000000006))%(1000000007)
if (k==-1):
if (n+m)%2==1:
ans=0;
print(ans)
```
|
output
| 1
| 20,878
| 2
| 41,757
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
|
instruction
| 0
| 20,879
| 2
| 41,758
|
Tags: combinatorics, constructive algorithms, math, number theory
Correct Solution:
```
import math
n,m,k = map(int,input().split())
if(k == 1 or (n + m) % 2 == 0):
print(pow(2,(n - 1) * (m - 1),1000000007))
else:
print("0")
```
|
output
| 1
| 20,879
| 2
| 41,759
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
|
instruction
| 0
| 20,880
| 2
| 41,760
|
Tags: combinatorics, constructive algorithms, math, number theory
Correct Solution:
```
n,m,k = [int(i) for i in input().split()]
if (n + m) % 2 == 1 and k == -1:
print(0)
exit(0)
result = 1
power = (n-1)*(m-1)
print(pow(2,power,1000000007))
```
|
output
| 1
| 20,880
| 2
| 41,761
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
|
instruction
| 0
| 20,881
| 2
| 41,762
|
Tags: combinatorics, constructive algorithms, math, number theory
Correct Solution:
```
mod=1000000007
def pow(n):
if n==0:
return 1
if n==1:
return 2
if n%2==1:
return (pow(n//2)**2*2)%mod
else:
return (pow(n//2)**2) % mod
n,m,k=map(int,input().split())
if (n%2+m%2)%2==1 and k==-1:
print(0)
exit()
if n==1 or m==1:
print(1)
else:
print(pow((n-1)*(m-1)))
```
|
output
| 1
| 20,881
| 2
| 41,763
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
|
instruction
| 0
| 20,882
| 2
| 41,764
|
Tags: combinatorics, constructive algorithms, math, number theory
Correct Solution:
```
def fast_pow(a, b):
if a == 0:
return 1
val1 = (fast_pow(a // 2, b))
if a % 2:
return (val1 * val1 % 1000000007) * b % 1000000007
return val1 * val1 % 1000000007
n, m, k = map(int, input().split())
if (n + m) % 2 and k == -1:
print(0)
exit()
print(fast_pow((n - 1) * (m - 1), 2))
```
|
output
| 1
| 20,882
| 2
| 41,765
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
Submitted Solution:
```
n,m,k=list(map(int,input().split()))
if ((n+m)%2==1 and k==-1):
print(0)
else:
print(pow(2,(n-1)*(m-1),10**9+7))
```
|
instruction
| 0
| 20,883
| 2
| 41,766
|
Yes
|
output
| 1
| 20,883
| 2
| 41,767
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
Submitted Solution:
```
MOD = 1000000007
def calc(n, m, k):
if k == -1 and n%2 != m%2:
return 0
n = (n-1)
m = (m-1)
n = (n * m)
r = 1
s = 2
while n > 0:
if n % 2 == 1:
r = (r*s) % MOD
n //= 2
s = (s*s) % MOD
return r
if __name__ == '__main__':
n, m, k = input().split(' ')
print(calc(int(n), int(m), int(k)))
```
|
instruction
| 0
| 20,884
| 2
| 41,768
|
Yes
|
output
| 1
| 20,884
| 2
| 41,769
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
Submitted Solution:
```
def fastpow(base, exp, mod):
res = 1;
while exp > 0:
if (exp % 2) == 1:
res = (res * base) % mod
base = (base * base) % mod
exp //= 2
return res % mod
a, b, c = input().split()
a = int(a)
b = int(b)
c = int(c)
ans = fastpow(2, (a-1), 1000000007)
ans = fastpow(ans, (b-1), 1000000007)
if (a+b)%2==1 :
if c == -1 :
ans = 0
print(ans)
```
|
instruction
| 0
| 20,885
| 2
| 41,770
|
Yes
|
output
| 1
| 20,885
| 2
| 41,771
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
Submitted Solution:
```
mod = 1000000007
n,m,k=map(int,input().split())
if(k == -1 and (n+m)%2 == 1):
print(0)
else:
print(pow(pow(2, m - 1, mod), n - 1, mod))
```
|
instruction
| 0
| 20,886
| 2
| 41,772
|
Yes
|
output
| 1
| 20,886
| 2
| 41,773
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
Submitted Solution:
```
n, m, k = map(int, input().split())
if (k == -1) and (n % 2 != m % 2) :
print(0)
raise SystemExit(0)
ans = (n - 1) * (m - 1)
ans %= (1e9 + 7)
ans = int(ans)
res = 1 << ans
print(int(res))
```
|
instruction
| 0
| 20,887
| 2
| 41,774
|
No
|
output
| 1
| 20,887
| 2
| 41,775
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
Submitted Solution:
```
def main():
def binpow(a, n):
res = 1
while(n):
if n % 2:
res = (res * a) % mod
a = (a * a) % mod
n //= 2
return res
read = lambda: tuple(map(int, input().split()))
mod = 1000000007
def mul(a, b):
v = a * b
return v - mod if v > mod else v
m, n, k = read()
if m > n:
m, n = n, m
a = (m + n) - 1
m -= 1
d = m * (m + 1)
c = a * m - d
if 2 * m > a:
c += 1
#print(c, m, a)
if m == 0 and n % 2 == 0 and k == -1:
print(0)
else:
if c <= 0 or m == 0:
print(1)
else:
print(binpow(2, c))
main()
```
|
instruction
| 0
| 20,888
| 2
| 41,776
|
No
|
output
| 1
| 20,888
| 2
| 41,777
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
Submitted Solution:
```
a = input().split()
b = (int(a[0])-1)*(int(a[1]) - 1)
d = 2
result = 1
if( (int(a[0]) % 2 == int(a[1]) % 2 ) or (int(a[0]) % 2 != int(a[1]) % 2 and int(a[2]) == 1)):
if(b!= 0 ):
while (b) :
if(b % 2 != 0):
b-=1
result *= d
result = result % 1000000007
else:
b /= 2
d *= d
d %= 1000000007
print(result)
else:
print(0)
```
|
instruction
| 0
| 20,889
| 2
| 41,778
|
No
|
output
| 1
| 20,889
| 2
| 41,779
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.
Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.
Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).
Output
Print a single number denoting the answer modulo 1000000007.
Examples
Input
1 1 -1
Output
1
Input
1 3 1
Output
1
Input
3 3 -1
Output
16
Note
In the first example the only way is to put -1 into the only block.
In the second example the only way is to put 1 into every block.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
"""
created by shhuan at 2017/11/19 21:52
"""
MOD = 1000000007
N, M, K = map(int, input().split())
def my_pow_2(x):
if x == 0:
return 1
h = my_pow_2(x//2)
if x & 1:
return (h*h*2) % MOD
else:
return (h*h) % MOD
if K == -1 and N%2 != M%2:
print(-1)
else:
print(my_pow_2((N-1)*(M-1)))
```
|
instruction
| 0
| 20,890
| 2
| 41,780
|
No
|
output
| 1
| 20,890
| 2
| 41,781
|
Provide a correct Python 3 solution for this coding contest problem.
Once upon a time when people still believed in magic, there was a great wizard Aranyaka Gondlir. After twenty years of hard training in a deep forest, he had finally mastered ultimate magic, and decided to leave the forest for his home.
Arriving at his home village, Aranyaka was very surprised at the extraordinary desolation. A gloom had settled over the village. Even the whisper of the wind could scare villagers. It was a mere shadow of what it had been.
What had happened? Soon he recognized a sure sign of an evil monster that is immortal. Even the great wizard could not kill it, and so he resolved to seal it with magic. Aranyaka could cast a spell to create a monster trap: once he had drawn a line on the ground with his magic rod, the line would function as a barrier wall that any monster could not get over. Since he could only draw straight lines, he had to draw several lines to complete a monster trap, i.e., magic barrier walls enclosing the monster. If there was a gap between barrier walls, the monster could easily run away through the gap.
For instance, a complete monster trap without any gaps is built by the barrier walls in the left figure, where “M” indicates the position of the monster. In contrast, the barrier walls in the right figure have a loophole, even though it is almost complete.
<image>
Your mission is to write a program to tell whether or not the wizard has successfully sealed the monster.
Input
The input consists of multiple data sets, each in the following format.
n
x1 y1 x'1 y'1
x2 y2 x'2 y'2
...
xn yn x'n y'n
The first line of a data set contains a positive integer n, which is the number of the line segments drawn by the wizard. Each of the following n input lines contains four integers x, y, x', and y', which represent the x- and y-coordinates of two points (x, y) and (x', y' ) connected by a line segment. You may assume that all line segments have non-zero lengths. You may also assume that n is less than or equal to 100 and that all coordinates are between -50 and 50, inclusive.
For your convenience, the coordinate system is arranged so that the monster is always on the origin (0, 0). The wizard never draws lines crossing (0, 0).
You may assume that any two line segments have at most one intersection point and that no three line segments share the same intersection point. You may also assume that the distance between any two intersection points is greater than 10-5.
An input line containing a zero indicates the end of the input.
Output
For each data set, print “yes” or “no” in a line. If a monster trap is completed, print “yes”. Otherwise, i.e., if there is a loophole, print “no”.
Example
Input
8
-7 9 6 9
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
8
-7 9 5 7
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
0
Output
yes
no
|
instruction
| 0
| 21,101
| 2
| 42,202
|
"Correct Solution:
```
def string_to_complex(s):
a, b, c, d = map(int, s.split())
return (a + b * 1j, c + d * 1j)
def dot(c1, c2):
return c1.real * c2.real + c1.imag * c2.imag
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def cross_point(p1, p2, p3, p4):
crs1 = cross(p2 - p1, p3 - p1)
crs2 = cross(p2 - p1, p4 - p1)
if crs1 == 0 and crs2 == 0:
if p1 == p3 or p1 == p4:
return p1
elif p2 == p3 or p2 == p4:
return p2
else:
return None
crs3 = cross(p4 - p3, p1 - p3)
crs4 = cross(p4 - p3, p2 - p3)
if crs1 * crs2 <= 0 and crs3 * crs4 <= 0:
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = abs(cross(base, hypo1)) / abs(base)
d2 = abs(cross(base, hypo2)) / abs(base)
return p1 + d1 / (d1 + d2) * (p2 - p1)
else:
return None
def contain(polygon):
flag = False
for a, b in zip(polygon[0:], polygon[1:] + [polygon[0]]):
if a.imag > b.imag:
a, b = b, a
if a.imag <= 0 and b.imag > 0 and cross(a, b) > 0:
flag = not flag
return flag
def solve():
def dfs_contain(goal, edges, cross_points):
cur_edge = edges[-1]
for next_edge, next_cp in adj_edge[cur_edge]:
if next_edge == goal:
if contain(cross_points + [next_cp]):
break
elif next_edge not in edges and unchecked[next_edge] and DP[next_edge]:
edges.append(next_edge)
cross_points.append(next_cp)
if dfs_contain(goal, edges, cross_points):
break
e = edges.pop()
DP[e] = False
cross_points.pop()
else:
return False
return True
from sys import stdin
lines = stdin.readlines()
from itertools import combinations
while True:
n = int(lines[0])
if n == 0:
break
edges = enumerate(map(string_to_complex, lines[1:1+n]))
adj_edge = [[] for i in range(n)]
for e1, e2 in combinations(edges, 2):
n1, t1 = e1
n2, t2 = e2
cp = cross_point(*t1, *t2)
if cp:
adj_edge[n1].append((n2, cp))
adj_edge[n2].append((n1, cp))
flag = True
while flag:
for i, ae in enumerate(adj_edge):
if len(ae) == 1:
ne, cp = ae.pop()
adj_edge[ne].remove((i, cp))
break
else:
flag = False
unchecked = [True] * n
for e in range(n):
unchecked[e] = False
DP = [True] * n
if dfs_contain(e, [e], []):
print('yes')
break
else:
print('no')
del lines[:1+n]
solve()
```
|
output
| 1
| 21,101
| 2
| 42,203
|
Provide a correct Python 3 solution for this coding contest problem.
Once upon a time when people still believed in magic, there was a great wizard Aranyaka Gondlir. After twenty years of hard training in a deep forest, he had finally mastered ultimate magic, and decided to leave the forest for his home.
Arriving at his home village, Aranyaka was very surprised at the extraordinary desolation. A gloom had settled over the village. Even the whisper of the wind could scare villagers. It was a mere shadow of what it had been.
What had happened? Soon he recognized a sure sign of an evil monster that is immortal. Even the great wizard could not kill it, and so he resolved to seal it with magic. Aranyaka could cast a spell to create a monster trap: once he had drawn a line on the ground with his magic rod, the line would function as a barrier wall that any monster could not get over. Since he could only draw straight lines, he had to draw several lines to complete a monster trap, i.e., magic barrier walls enclosing the monster. If there was a gap between barrier walls, the monster could easily run away through the gap.
For instance, a complete monster trap without any gaps is built by the barrier walls in the left figure, where “M” indicates the position of the monster. In contrast, the barrier walls in the right figure have a loophole, even though it is almost complete.
<image>
Your mission is to write a program to tell whether or not the wizard has successfully sealed the monster.
Input
The input consists of multiple data sets, each in the following format.
n
x1 y1 x'1 y'1
x2 y2 x'2 y'2
...
xn yn x'n y'n
The first line of a data set contains a positive integer n, which is the number of the line segments drawn by the wizard. Each of the following n input lines contains four integers x, y, x', and y', which represent the x- and y-coordinates of two points (x, y) and (x', y' ) connected by a line segment. You may assume that all line segments have non-zero lengths. You may also assume that n is less than or equal to 100 and that all coordinates are between -50 and 50, inclusive.
For your convenience, the coordinate system is arranged so that the monster is always on the origin (0, 0). The wizard never draws lines crossing (0, 0).
You may assume that any two line segments have at most one intersection point and that no three line segments share the same intersection point. You may also assume that the distance between any two intersection points is greater than 10-5.
An input line containing a zero indicates the end of the input.
Output
For each data set, print “yes” or “no” in a line. If a monster trap is completed, print “yes”. Otherwise, i.e., if there is a loophole, print “no”.
Example
Input
8
-7 9 6 9
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
8
-7 9 5 7
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
0
Output
yes
no
|
instruction
| 0
| 21,102
| 2
| 42,204
|
"Correct Solution:
```
def string_to_complex(s):
a, b, c, d = map(int, s.split())
return (a + b * 1j, c + d * 1j)
def dot(c1, c2):
return c1.real * c2.real + c1.imag * c2.imag
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def cross_point(p1, p2, p3, p4):
crs1 = cross(p2 - p1, p3 - p1)
crs2 = cross(p2 - p1, p4 - p1)
if crs1 == 0 and crs2 == 0:
if p1 == p3 or p1 == p4:
return p1
elif p2 == p3 or p2 == p4:
return p2
else:
return None
crs3 = cross(p4 - p3, p1 - p3)
crs4 = cross(p4 - p3, p2 - p3)
if crs1 * crs2 <= 0 and crs3 * crs4 <= 0:
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = abs(cross(base, hypo1)) / abs(base)
d2 = abs(cross(base, hypo2)) / abs(base)
return p1 + d1 / (d1 + d2) * (p2 - p1)
else:
return None
def contain(polygon):
flag = False
for a, b in zip(polygon[0:], polygon[1:] + [polygon[0]]):
if a.imag > b.imag:
a, b = b, a
if a.imag <= 0 and b.imag > 0 and cross(a, b) > 0:
flag = not flag
return flag
def solve():
def dfs_contain(goal, edges, cross_points):
cur_edge = edges[-1]
for next_edge, next_cp in adj_edge[cur_edge]:
if next_edge == goal:
if contain(cross_points + [next_cp]):
break
elif next_edge not in edges and DP[next_edge]:
edges.append(next_edge)
cross_points.append(next_cp)
if dfs_contain(goal, edges, cross_points):
break
e = edges.pop()
DP[e] = False
cross_points.pop()
else:
return False
return True
from sys import stdin
lines = stdin.readlines()
from itertools import combinations
while True:
n = int(lines[0])
if n == 0:
break
edges = enumerate(map(string_to_complex, lines[1:1+n]))
adj_edge = [[] for i in range(n)]
for e1, e2 in combinations(edges, 2):
n1, t1 = e1
n2, t2 = e2
cp = cross_point(*t1, *t2)
if cp:
adj_edge[n1].append((n2, cp))
adj_edge[n2].append((n1, cp))
flag = True
while flag:
for i, ae in enumerate(adj_edge):
if len(ae) == 1:
ne, cp = ae.pop()
adj_edge[ne].remove((i, cp))
break
else:
flag = False
for e in range(n):
DP = [False] * e + [True] * (n - e)
if dfs_contain(e, [e], []):
print('yes')
break
else:
print('no')
del lines[:1+n]
solve()
```
|
output
| 1
| 21,102
| 2
| 42,205
|
Provide a correct Python 3 solution for this coding contest problem.
Once upon a time when people still believed in magic, there was a great wizard Aranyaka Gondlir. After twenty years of hard training in a deep forest, he had finally mastered ultimate magic, and decided to leave the forest for his home.
Arriving at his home village, Aranyaka was very surprised at the extraordinary desolation. A gloom had settled over the village. Even the whisper of the wind could scare villagers. It was a mere shadow of what it had been.
What had happened? Soon he recognized a sure sign of an evil monster that is immortal. Even the great wizard could not kill it, and so he resolved to seal it with magic. Aranyaka could cast a spell to create a monster trap: once he had drawn a line on the ground with his magic rod, the line would function as a barrier wall that any monster could not get over. Since he could only draw straight lines, he had to draw several lines to complete a monster trap, i.e., magic barrier walls enclosing the monster. If there was a gap between barrier walls, the monster could easily run away through the gap.
For instance, a complete monster trap without any gaps is built by the barrier walls in the left figure, where “M” indicates the position of the monster. In contrast, the barrier walls in the right figure have a loophole, even though it is almost complete.
<image>
Your mission is to write a program to tell whether or not the wizard has successfully sealed the monster.
Input
The input consists of multiple data sets, each in the following format.
n
x1 y1 x'1 y'1
x2 y2 x'2 y'2
...
xn yn x'n y'n
The first line of a data set contains a positive integer n, which is the number of the line segments drawn by the wizard. Each of the following n input lines contains four integers x, y, x', and y', which represent the x- and y-coordinates of two points (x, y) and (x', y' ) connected by a line segment. You may assume that all line segments have non-zero lengths. You may also assume that n is less than or equal to 100 and that all coordinates are between -50 and 50, inclusive.
For your convenience, the coordinate system is arranged so that the monster is always on the origin (0, 0). The wizard never draws lines crossing (0, 0).
You may assume that any two line segments have at most one intersection point and that no three line segments share the same intersection point. You may also assume that the distance between any two intersection points is greater than 10-5.
An input line containing a zero indicates the end of the input.
Output
For each data set, print “yes” or “no” in a line. If a monster trap is completed, print “yes”. Otherwise, i.e., if there is a loophole, print “no”.
Example
Input
8
-7 9 6 9
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
8
-7 9 5 7
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
0
Output
yes
no
|
instruction
| 0
| 21,103
| 2
| 42,206
|
"Correct Solution:
```
def string_to_complex(s):
a, b, c, d = map(int, s.split())
return (a + b * 1j, c + d * 1j)
def dot(c1, c2):
return c1.real * c2.real + c1.imag * c2.imag
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def cross_point(p1, p2, p3, p4):
crs1 = cross(p2 - p1, p3 - p1)
crs2 = cross(p2 - p1, p4 - p1)
if crs1 == 0 and crs2 == 0:
if p1 == p3 or p1 == p4:
return p1
elif p2 == p3 or p2 == p4:
return p2
else:
return None
crs3 = cross(p4 - p3, p1 - p3)
crs4 = cross(p4 - p3, p2 - p3)
if crs1 * crs2 <= 0 and crs3 * crs4 <= 0:
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = abs(cross(base, hypo1)) / abs(base)
d2 = abs(cross(base, hypo2)) / abs(base)
return p1 + d1 / (d1 + d2) * (p2 - p1)
else:
return None
def contain(polygon):
flag = False
for a, b in zip(polygon[0:], polygon[1:] + [polygon[0]]):
if a.imag > b.imag:
a, b = b, a
if a.imag <= 0 and b.imag > 0 and cross(a, b) > 0:
flag = not flag
return flag
from cmath import phase, pi
def complex_compare_key(c):
return (c.real, c.imag)
def complex_compare_key2(c):
return c.real
def pos_phase(base, c1, c2):
p = phase((c2 - base) / (c1 - base))
if p < 0:
p += 2 * pi
return p
def solve():
from sys import stdin
lines = stdin.readlines()
from itertools import combinations
while True:
n = int(lines[0])
if n == 0:
break
edges = enumerate(map(string_to_complex, lines[1:1+n]))
adj_edge_cp = [[] for i in range(n)]
adj_cross_point = {}
for e1, e2 in combinations(edges, 2):
n1, t1 = e1
n2, t2 = e2
cp = cross_point(*t1, *t2)
if cp:
adj_edge_cp[n1].append(cp)
adj_edge_cp[n2].append(cp)
adj_cross_point[cp] = (n1, n2)
for cp_list in adj_edge_cp:
cp_list.sort(key=complex_compare_key)
# creating adjacency list of intersection points
for cp, t in adj_cross_point.items():
e1, e2 = t
e1_cp = adj_edge_cp[e1]
e2_cp = adj_edge_cp[e2]
cp_l = []
if len(e1_cp) == 1:
pass
else:
if cp == e1_cp[0]:
cp_l.append(e1_cp[1])
elif cp == e1_cp[-1]:
cp_l.append(e1_cp[-2])
else:
cp_idx = e1_cp.index(cp)
cp_l.append(e1_cp[cp_idx - 1])
cp_l.append(e1_cp[cp_idx + 1])
if len(e2_cp) == 1:
pass
else:
if cp == e2_cp[0]:
cp_l.append(e2_cp[1])
elif cp == e2_cp[-1]:
cp_l.append(e2_cp[-2])
else:
cp_idx = e2_cp.index(cp)
cp_l.append(e2_cp[cp_idx - 1])
cp_l.append(e2_cp[cp_idx + 1])
adj_cross_point[cp] = cp_l
# branch cut operation
flag = True
while flag:
for cp, cp_list in adj_cross_point.items():
if len(cp_list) == 1:
next_cp = cp_list.pop()
adj_cross_point[next_cp].remove(cp)
break
else:
flag = False
del_cp_list = []
for cp, cp_list in adj_cross_point.items():
if not cp_list:
del_cp_list.append(cp)
for cp in del_cp_list:
del adj_cross_point[cp]
# polygon trace with left(right)-hand method
cp_list = list(adj_cross_point)
cp_list.sort(key=complex_compare_key2)
visited = dict(zip(cp_list, [False] * len(cp_list)))
for start in cp_list:
if start.real >= 0:
print("no")
break
if visited[start]:
continue
visited[start] = True
path = [start - 1, start]
flag = True
while flag:
pre_cp = path[-2]
cp = path[-1]
ang = 2 * pi
for p in adj_cross_point[cp]:
if p == pre_cp:
continue
new_ang = pos_phase(cp, pre_cp, p)
if new_ang < ang:
next_cp = p
ang = new_ang
visited[next_cp] = True
if next_cp == start:
if contain(path[1:]):
print("yes")
flag = False
else:
break
else:
path.append(next_cp)
else:
break
else:
print("no")
del lines[:1+n]
solve()
```
|
output
| 1
| 21,103
| 2
| 42,207
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time when people still believed in magic, there was a great wizard Aranyaka Gondlir. After twenty years of hard training in a deep forest, he had finally mastered ultimate magic, and decided to leave the forest for his home.
Arriving at his home village, Aranyaka was very surprised at the extraordinary desolation. A gloom had settled over the village. Even the whisper of the wind could scare villagers. It was a mere shadow of what it had been.
What had happened? Soon he recognized a sure sign of an evil monster that is immortal. Even the great wizard could not kill it, and so he resolved to seal it with magic. Aranyaka could cast a spell to create a monster trap: once he had drawn a line on the ground with his magic rod, the line would function as a barrier wall that any monster could not get over. Since he could only draw straight lines, he had to draw several lines to complete a monster trap, i.e., magic barrier walls enclosing the monster. If there was a gap between barrier walls, the monster could easily run away through the gap.
For instance, a complete monster trap without any gaps is built by the barrier walls in the left figure, where “M” indicates the position of the monster. In contrast, the barrier walls in the right figure have a loophole, even though it is almost complete.
<image>
Your mission is to write a program to tell whether or not the wizard has successfully sealed the monster.
Input
The input consists of multiple data sets, each in the following format.
n
x1 y1 x'1 y'1
x2 y2 x'2 y'2
...
xn yn x'n y'n
The first line of a data set contains a positive integer n, which is the number of the line segments drawn by the wizard. Each of the following n input lines contains four integers x, y, x', and y', which represent the x- and y-coordinates of two points (x, y) and (x', y' ) connected by a line segment. You may assume that all line segments have non-zero lengths. You may also assume that n is less than or equal to 100 and that all coordinates are between -50 and 50, inclusive.
For your convenience, the coordinate system is arranged so that the monster is always on the origin (0, 0). The wizard never draws lines crossing (0, 0).
You may assume that any two line segments have at most one intersection point and that no three line segments share the same intersection point. You may also assume that the distance between any two intersection points is greater than 10-5.
An input line containing a zero indicates the end of the input.
Output
For each data set, print “yes” or “no” in a line. If a monster trap is completed, print “yes”. Otherwise, i.e., if there is a loophole, print “no”.
Example
Input
8
-7 9 6 9
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
8
-7 9 5 7
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
0
Output
yes
no
Submitted Solution:
```
def string_to_complex(s):
a, b, c, d = map(int, s.split())
return (a + b * 1j, c + d * 1j)
def dot(c1, c2):
return c1.real * c2.real + c1.imag * c2.imag
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def cross_point(p1, p2, p3, p4):
crs1 = cross(p2 - p1, p3 - p1)
crs2 = cross(p2 - p1, p4 - p1)
if crs1 == 0 and crs2 == 0:
if p1 == p3 or p1 == p4:
return p1
elif p2 == p3 or p2 == p4:
return p2
else:
return None
crs3 = cross(p4 - p3, p1 - p3)
crs4 = cross(p4 - p3, p2 - p3)
if crs1 * crs2 <= 0 and crs3 * crs4 <= 0:
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = abs(cross(base, hypo1)) / abs(base)
d2 = abs(cross(base, hypo2)) / abs(base)
return p1 + d1 / (d1 + d2) * (p2 - p1)
else:
return None
def contain(polygon):
flag = False
for a, b in zip(polygon[0:], polygon[1:] + [polygon[0]]):
if a.imag > b.imag:
a, b = b, a
if a.imag <= 0 and b.imag > 0 and cross(a, b) > 0:
flag = not flag
return flag
def solve():
def dfs_contain(goal, edges, cross_points):
cur_edge = edges[-1]
for next_edge, next_cp in adj_edge[cur_edge]:
if next_edge == goal:
if contain(cross_points + [next_cp]):
break
elif next_edge not in edges and unchecked[next_edge] and DP[next_edge]:
edges.append(next_edge)
cross_points.append(next_cp)
if dfs_contain(goal, edges, cross_points):
break
e = edges.pop()
DP[e] = False
cross_points.pop()
else:
return False
return True
from sys import stdin
file_input = stdin.readlines()
from itertools import combinations
while True:
n = int(lines[0])
if n == 0:
break
edges = enumerate(map(string_to_complex, lines[1:1+n]))
adj_edge = [[] for i in range(n)]
for e1, e2 in combinations(edges, 2):
n1, t1 = e1
n2, t2 = e2
cp = cross_point(*t1, *t2)
if cp:
adj_edge[n1].append((n2, cp))
adj_edge[n2].append((n1, cp))
flag = True
while flag:
for i, ae in enumerate(adj_edge):
if len(ae) == 1:
ne, cp = ae.pop()
adj_edge[ne].remove((i, cp))
break
else:
flag = False
unchecked = [True] * n
for e in range(n):
unchecked[e] = False
DP = [True] * n
if dfs_contain(e, [e], []):
print('yes')
break
else:
print('no')
del lines[:1+n]
solve()
```
|
instruction
| 0
| 21,104
| 2
| 42,208
|
No
|
output
| 1
| 21,104
| 2
| 42,209
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time when people still believed in magic, there was a great wizard Aranyaka Gondlir. After twenty years of hard training in a deep forest, he had finally mastered ultimate magic, and decided to leave the forest for his home.
Arriving at his home village, Aranyaka was very surprised at the extraordinary desolation. A gloom had settled over the village. Even the whisper of the wind could scare villagers. It was a mere shadow of what it had been.
What had happened? Soon he recognized a sure sign of an evil monster that is immortal. Even the great wizard could not kill it, and so he resolved to seal it with magic. Aranyaka could cast a spell to create a monster trap: once he had drawn a line on the ground with his magic rod, the line would function as a barrier wall that any monster could not get over. Since he could only draw straight lines, he had to draw several lines to complete a monster trap, i.e., magic barrier walls enclosing the monster. If there was a gap between barrier walls, the monster could easily run away through the gap.
For instance, a complete monster trap without any gaps is built by the barrier walls in the left figure, where “M” indicates the position of the monster. In contrast, the barrier walls in the right figure have a loophole, even though it is almost complete.
<image>
Your mission is to write a program to tell whether or not the wizard has successfully sealed the monster.
Input
The input consists of multiple data sets, each in the following format.
n
x1 y1 x'1 y'1
x2 y2 x'2 y'2
...
xn yn x'n y'n
The first line of a data set contains a positive integer n, which is the number of the line segments drawn by the wizard. Each of the following n input lines contains four integers x, y, x', and y', which represent the x- and y-coordinates of two points (x, y) and (x', y' ) connected by a line segment. You may assume that all line segments have non-zero lengths. You may also assume that n is less than or equal to 100 and that all coordinates are between -50 and 50, inclusive.
For your convenience, the coordinate system is arranged so that the monster is always on the origin (0, 0). The wizard never draws lines crossing (0, 0).
You may assume that any two line segments have at most one intersection point and that no three line segments share the same intersection point. You may also assume that the distance between any two intersection points is greater than 10-5.
An input line containing a zero indicates the end of the input.
Output
For each data set, print “yes” or “no” in a line. If a monster trap is completed, print “yes”. Otherwise, i.e., if there is a loophole, print “no”.
Example
Input
8
-7 9 6 9
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
8
-7 9 5 7
-5 5 6 5
-10 -5 10 -5
-6 9 -9 -6
6 9 9 -6
-1 -2 -3 10
1 -2 3 10
-2 -3 2 -3
0
Output
yes
no
Submitted Solution:
```
def string_to_complex(s):
a, b, c, d = map(int, s.split())
return (a + b * 1j, c + d * 1j)
def dot(c1, c2):
return c1.real * c2.real + c1.imag * c2.imag
def cross(c1, c2):
return c1.real * c2.imag - c1.imag * c2.real
def cross_point(p1, p2, p3, p4):
crs1 = cross(p2 - p1, p3 - p1)
crs2 = cross(p2 - p1, p4 - p1)
if crs1 == 0 and crs2 == 0:
if p1 == p3 or p1 == p4:
return p1
elif p2 == p3 or p2 == p4:
return p2
else:
return None
crs3 = cross(p4 - p3, p1 - p3)
crs4 = cross(p4 - p3, p2 - p3)
if crs1 * crs2 <= 0 and crs3 * crs4 <= 0:
base = p4 - p3
hypo1 = p1 - p3
hypo2 = p2 - p3
d1 = abs(cross(base, hypo1)) / abs(base)
d2 = abs(cross(base, hypo2)) / abs(base)
return p1 + d1 / (d1 + d2) * (p2 - p1)
else:
return None
def contain(polygon):
flag = False
for a, b in zip(polygon[0:], polygon[1:] + [polygon[0]]):
if a.imag > b.imag:
a, b = b, a
if a.imag <= 0 and b.imag > 0 and cross(a, b) > 0:
flag = not flag
return flag
def solve():
def dfs_contain(goal, edges, cross_points):
cur_edge = edges[-1]
for next_edge, next_cp in adj_edge[cur_edge]:
if next_edge == goal:
if contain(cross_points + [next_cp]):
break
elif next_edge not in edges and unchecked[next_edge] and DP[next_edge]:
edges.append(next_edge)
cross_points.append(next_cp)
if dfs_contain(goal, edges, cross_points):
break
e = edges.pop()
DP[e] = False
cross_points.pop()
else:
return False
return True
from sys import stdin
file_input = stdin.radlines()
from itertools import combinations
while True:
n = int(lines[0])
if n == 0:
break
edges = enumerate(map(string_to_complex, lines[1:1+n]))
adj_edge = [[] for i in range(n)]
for e1, e2 in combinations(edges, 2):
n1, t1 = e1
n2, t2 = e2
cp = cross_point(*t1, *t2)
if cp:
adj_edge[n1].append((n2, cp))
adj_edge[n2].append((n1, cp))
flag = True
while flag:
for i, ae in enumerate(adj_edge):
if len(ae) == 1:
ne, cp = ae.pop()
adj_edge[ne].remove((i, cp))
break
else:
flag = False
unchecked = [True] * n
for e in range(n):
unchecked[e] = False
DP = [True] * n
if dfs_contain(e, [e], []):
print('yes')
break
else:
print('no')
del lines[:1+n]
solve()
```
|
instruction
| 0
| 21,105
| 2
| 42,210
|
No
|
output
| 1
| 21,105
| 2
| 42,211
|
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