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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sarah has always been a lover of nature, and a couple of years ago she saved up enough money to travel the world and explore all the things built by nature over its lifetime on earth. During this time she visited some truly special places which were left untouched for centuries, from watching icebergs in freezing weather to scuba-diving in oceans and admiring the sea life, residing unseen. These experiences were enhanced with breathtaking views built by mountains over time and left there for visitors to see for years on end. Over time, all these expeditions took a toll on Sarah and culminated in her decision to settle down in the suburbs and live a quiet life.
However, as Sarah's love for nature never faded, she started growing flowers in her garden in an attempt to stay connected with nature. At the beginning she planted only blue orchids, but over time she started using different flower types to add variety to her collection of flowers. This collection of flowers can be represented as an array of N flowers and the i-th of them has a type associated with it, denoted as A_i. Each resident, passing by her collection and limited by the width of his view, can only see K contiguous flowers at each moment in time. To see the whole collection, the resident will look at the first K contiguous flowers A_1, A_2, ..., A_K, then shift his view by one flower and look at the next section of K contiguous flowers A_2, A_3, ..., A_{K+1} and so on until they scan the whole collection, ending with section A_{N-K+1}, ..., A_{N-1}, A_N.
Each resident determines the beautiness of a section of K flowers as the number of distinct flower types in that section. Furthermore, the beautiness of the whole collection is calculated by summing the beautiness values of each contiguous section. Formally, beautiness B_i of a section starting at the i-th position is calculated as B_i = distinct(A_i, A_{i+1}, ..., A_{i+K-1}), and beautiness of the collection B is calculated as B=B_1 + B_2 + ... + B_{N-K+1}.
In addition, as Sarah wants to keep her collection of flowers have a fresh feel, she can also pick two points L and R, dispose flowers between those two points and plant new flowers, all of them being the same type.
You will be given Q queries and each of those queries will be of the following two types:
1. You will be given three integers L, R, X describing that Sarah has planted flowers of type X between positions L and R inclusive. Formally collection is changed such that A[i]=X for all i in range [L.. R].
2. You will be given integer K, width of the resident's view and you have to determine the beautiness value B resident has associated with the collection
For each query of second type print the result β beautiness B of the collection.
Input
First line contains two integers N and Q \;(1 β€ N, Q β€ 10^5) β number of flowers and the number of queries, respectively.
The second line contains N integers A_1, A_2, ..., A_N\;(1 β€ A_i β€ 10^9) β where A_i represents type of the i-th flower.
Each of the next Q lines describe queries and start with integer Tβ\{1, 2\}.
* If T = 1, there will be three more integers in the line L, R, X\;(1 β€ L, R β€ N;\; 1 β€ X β€ 10^9) β L and R describing boundaries and X describing the flower type
* If T = 2, there will be one more integer in the line K\;(1 β€ K β€ N) β resident's width of view
Output
For each query of the second type print the beautiness B of the collection.
Example
Input
5 5
1 2 3 4 5
2 3
1 1 2 5
2 4
1 2 4 5
2 2
Output
9
6
4
Note
Let's look at the example.
Initially the collection is [1, 2, 3, 4, 5]. In the first query K = 3, we consider sections of three flowers with the first being [1, 2, 3]. Since beautiness of the section is the number of distinct flower types in that section, B_1 = 3. Second section is [2, 3, 4] and B_2 = 3. Third section is [3, 4, 5] and B_3 = 3, since the flower types are all distinct. The beautiness value resident has associated with the collection is B = B_1 + B_2 + B_3 = 3 + 3 + 3 = 9.
After the second query, the collection becomes [5, 5, 3, 4, 5].
For the third query K = 4, so we consider sections of four flowers with the first being [5, 5, 3, 4]. There are three distinct flower types [5, 3, 4] in this section, so B_1 = 3. Second section [5, 3, 4, 5] also has 3 distinct flower types, so B_2 = 3. The beautiness value resident has associated with the collection is B = B_1 + B_2 = 3 + 3 = 6
After the fourth query, the collection becomes [5, 5, 5, 5, 5].
For the fifth query K = 2 and in this case all the four sections are same with each of them being [5, 5]. Beautiness of [5, 5] is 1 since there is only one distinct element in this section [5]. Beautiness of the whole collection is B = B_1 + B_2 + B_3 + B_4 = 1 + 1 + 1 + 1 = 4
Submitted Solution:
```
n,q = map(int,input().split())
a = list(map(int,input().split()))
count = 0
for i in range(0,q):
querie = list(map(int,input().split()))
if (querie[0] == 1):
for i in range(querie[1] - 1,querie[2]):
a[i] = querie[3]
if (querie[0] == 2):
k = querie[1]
j = 0
count = 0
while(i <= len(a) - k):
l = []
for j in range(i,querie[1] + i):
l.append(a[j])
i = i + 1
count = count + len(set(l))
print(count)
```
|
instruction
| 0
| 5,234
| 3
| 10,468
|
No
|
output
| 1
| 5,234
| 3
| 10,469
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sarah has always been a lover of nature, and a couple of years ago she saved up enough money to travel the world and explore all the things built by nature over its lifetime on earth. During this time she visited some truly special places which were left untouched for centuries, from watching icebergs in freezing weather to scuba-diving in oceans and admiring the sea life, residing unseen. These experiences were enhanced with breathtaking views built by mountains over time and left there for visitors to see for years on end. Over time, all these expeditions took a toll on Sarah and culminated in her decision to settle down in the suburbs and live a quiet life.
However, as Sarah's love for nature never faded, she started growing flowers in her garden in an attempt to stay connected with nature. At the beginning she planted only blue orchids, but over time she started using different flower types to add variety to her collection of flowers. This collection of flowers can be represented as an array of N flowers and the i-th of them has a type associated with it, denoted as A_i. Each resident, passing by her collection and limited by the width of his view, can only see K contiguous flowers at each moment in time. To see the whole collection, the resident will look at the first K contiguous flowers A_1, A_2, ..., A_K, then shift his view by one flower and look at the next section of K contiguous flowers A_2, A_3, ..., A_{K+1} and so on until they scan the whole collection, ending with section A_{N-K+1}, ..., A_{N-1}, A_N.
Each resident determines the beautiness of a section of K flowers as the number of distinct flower types in that section. Furthermore, the beautiness of the whole collection is calculated by summing the beautiness values of each contiguous section. Formally, beautiness B_i of a section starting at the i-th position is calculated as B_i = distinct(A_i, A_{i+1}, ..., A_{i+K-1}), and beautiness of the collection B is calculated as B=B_1 + B_2 + ... + B_{N-K+1}.
In addition, as Sarah wants to keep her collection of flowers have a fresh feel, she can also pick two points L and R, dispose flowers between those two points and plant new flowers, all of them being the same type.
You will be given Q queries and each of those queries will be of the following two types:
1. You will be given three integers L, R, X describing that Sarah has planted flowers of type X between positions L and R inclusive. Formally collection is changed such that A[i]=X for all i in range [L.. R].
2. You will be given integer K, width of the resident's view and you have to determine the beautiness value B resident has associated with the collection
For each query of second type print the result β beautiness B of the collection.
Input
First line contains two integers N and Q \;(1 β€ N, Q β€ 10^5) β number of flowers and the number of queries, respectively.
The second line contains N integers A_1, A_2, ..., A_N\;(1 β€ A_i β€ 10^9) β where A_i represents type of the i-th flower.
Each of the next Q lines describe queries and start with integer Tβ\{1, 2\}.
* If T = 1, there will be three more integers in the line L, R, X\;(1 β€ L, R β€ N;\; 1 β€ X β€ 10^9) β L and R describing boundaries and X describing the flower type
* If T = 2, there will be one more integer in the line K\;(1 β€ K β€ N) β resident's width of view
Output
For each query of the second type print the beautiness B of the collection.
Example
Input
5 5
1 2 3 4 5
2 3
1 1 2 5
2 4
1 2 4 5
2 2
Output
9
6
4
Note
Let's look at the example.
Initially the collection is [1, 2, 3, 4, 5]. In the first query K = 3, we consider sections of three flowers with the first being [1, 2, 3]. Since beautiness of the section is the number of distinct flower types in that section, B_1 = 3. Second section is [2, 3, 4] and B_2 = 3. Third section is [3, 4, 5] and B_3 = 3, since the flower types are all distinct. The beautiness value resident has associated with the collection is B = B_1 + B_2 + B_3 = 3 + 3 + 3 = 9.
After the second query, the collection becomes [5, 5, 3, 4, 5].
For the third query K = 4, so we consider sections of four flowers with the first being [5, 5, 3, 4]. There are three distinct flower types [5, 3, 4] in this section, so B_1 = 3. Second section [5, 3, 4, 5] also has 3 distinct flower types, so B_2 = 3. The beautiness value resident has associated with the collection is B = B_1 + B_2 = 3 + 3 = 6
After the fourth query, the collection becomes [5, 5, 5, 5, 5].
For the fifth query K = 2 and in this case all the four sections are same with each of them being [5, 5]. Beautiness of [5, 5] is 1 since there is only one distinct element in this section [5]. Beautiness of the whole collection is B = B_1 + B_2 + B_3 + B_4 = 1 + 1 + 1 + 1 = 4
Submitted Solution:
```
p = input()
q = input()
arr = []
arr1 = []
for i in range(len(p)):
for j in range(i + 1, len(p) + 1):
arr.append(p[i:j])
arr = list(set(arr))
for i in range(len(q)):
for j in range(i + 1, len(q) + 1):
arr1.append(q[i:j])
arr1 = list(set(arr1))
arr3 = []
# arr4=[]
for i in range(len(arr)):
n1 = int(len(arr[i]) / 2)
count = 0
if len(arr[i]) % 2 == 0:
for j in range(n1):
if arr[i][n1 - 1 - j] == arr[i][n1 + j]:
count += 1
else:
for j in range(n1):
if arr[i][n1 - 1 - j] == arr[i][n1 + 1 + j]:
count += 1
if count == n1:
arr3.append(arr[i])
arr5 = []
for i in range(len(arr1)):
n1 = int(len(arr1[i]) / 2)
count = 0
if len(arr1[i]) % 2 == 0:
for j in range(n1):
if arr1[i][n1 - 1 - j] == arr1[i][n1 + j]:
count += 1
else:
for j in range(n1):
if arr1[i][n1 - 1 - j] == arr1[i][n1 + 1 + j]:
count += 1
if count == n1:
# arr4.append(arr1[i])
for k in range(len(arr3)):
arr5.append(arr3[k] + arr1[i])
"""
for i in range(len(arr3)):
for j in range(len(arr4)):
arr5.append(arr3[i]+arr4[j])
"""
print(len(list(set(arr5))))
```
|
instruction
| 0
| 5,235
| 3
| 10,470
|
No
|
output
| 1
| 5,235
| 3
| 10,471
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
|
instruction
| 0
| 5,323
| 3
| 10,646
|
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
input=sys.stdin.readline
from collections import defaultdict as dc
from collections import Counter
from bisect import bisect_right, bisect_left
import math
from operator import itemgetter
from heapq import heapify, heappop, heappush
from queue import PriorityQueue as pq
n,k=map(int,input().split())
if k>=n*(n-1)//2:
print("no solution")
else:
for i in range(n):
print(0,i)
```
|
output
| 1
| 5,323
| 3
| 10,647
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
|
instruction
| 0
| 5,324
| 3
| 10,648
|
Tags: constructive algorithms, implementation
Correct Solution:
```
n, k = map(int, input().split())
up = (int)(n * (n - 1) / 2);
if k >= up:
print ("no solution")
else:
for i in range (0, n):
print (0, i)
```
|
output
| 1
| 5,324
| 3
| 10,649
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
|
instruction
| 0
| 5,325
| 3
| 10,650
|
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k = map(int ,input().split())
if((n&1)==0) :
time = (n//2)*(n-1)
else :
time = ((n-1)//2)*n
if(k>=time):
print("no solution")
else :
for i in range(n):
print("0 "+str(i))
```
|
output
| 1
| 5,325
| 3
| 10,651
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
|
instruction
| 0
| 5,326
| 3
| 10,652
|
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k = map(int,input().split())
if n*(n-1) <= k*2:
print('no solution')
else:
x = 11
for i in range(n-1):
print(1,x)
x+=3
print(1,x-5)
```
|
output
| 1
| 5,326
| 3
| 10,653
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
|
instruction
| 0
| 5,327
| 3
| 10,654
|
Tags: constructive algorithms, implementation
Correct Solution:
```
n, k = map(int, input().split())
if (k >= n * (n - 1) // 2):
print("no solution")
else:
for i in range(n):
print(0, i)
```
|
output
| 1
| 5,327
| 3
| 10,655
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
|
instruction
| 0
| 5,328
| 3
| 10,656
|
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k=map(int, input().split())
if(n*(n-1)//2 <=k):
print('no solution')
else:
for i in range(n):
print(0,i)
```
|
output
| 1
| 5,328
| 3
| 10,657
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
|
instruction
| 0
| 5,329
| 3
| 10,658
|
Tags: constructive algorithms, implementation
Correct Solution:
```
n, k = map(int, input().split())
tot = 0
for i in range(n):
for j in range(i + 1, n):
tot += 1
if tot <= k:
print('no solution')
else:
for i in range(n):
print(1, i)
```
|
output
| 1
| 5,329
| 3
| 10,659
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
|
instruction
| 0
| 5,330
| 3
| 10,660
|
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k = map(int,input().split())
if n*(n-1) <= k*2:
print('no solution')
else:
for i in range(n):
print(0,i)
```
|
output
| 1
| 5,330
| 3
| 10,661
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
n,k=map(int,input().split())
if n*(n-1)/2<=k:
print("no solution")
else:
for i in range(0,n):print(0,i)
# Made By Mostafa_Khaled
```
|
instruction
| 0
| 5,331
| 3
| 10,662
|
Yes
|
output
| 1
| 5,331
| 3
| 10,663
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
n,k=list(map(int,input().split()))
if (n*n-n)//2<=k:
print("no solution")
else:
x=0
y=0
store=[]
count=0
store.append(str(x)+' '+str(y))
while len(store)<n:
y+=1
store.append(str(x)+' '+str(y))
for j in store:
print(j)
```
|
instruction
| 0
| 5,332
| 3
| 10,664
|
Yes
|
output
| 1
| 5,332
| 3
| 10,665
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
s=input().split()
n=int(s[0])
k=int(s[1])
ans=[]
a=0
for i in range(n):
for j in range(i+1,n):
a+=1
if(a<=k):
print("no solution")
else:
for i in range(-10**9,-10**9+n):
print("0 "+str(i))
```
|
instruction
| 0
| 5,333
| 3
| 10,666
|
Yes
|
output
| 1
| 5,333
| 3
| 10,667
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
n, k = map(int, input().split())
if k >= n * (n - 1) // 2:
print("no solution")
else:
for i in range(n):
print(i, i * (n + 1))
```
|
instruction
| 0
| 5,334
| 3
| 10,668
|
Yes
|
output
| 1
| 5,334
| 3
| 10,669
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
s=input().split()
n=int(s[0])
k=int(s[0])
ans=[]
a=0
for i in range(1,n+1):
for j in range(i+1,n+1):
a+=1
if(a<=k):
print("no solution")
else:
for i in range(2):
for j in range(n//2):
ans.append(str(i)+" "+str(j))
if(len(ans)<n):
ans.append("2 0")
for i in range(n):
print(ans[i])
```
|
instruction
| 0
| 5,335
| 3
| 10,670
|
No
|
output
| 1
| 5,335
| 3
| 10,671
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
s=input().split()
n=int(s[0])
k=int(s[0])
ans=[]
a=0
for i in range(n):
for j in range(i+1,n):
a+=1
if(a<k):
print("no solution")
else:
for i in range(2):
for j in range(n//2):
ans.append(str(i)+" "+str(j))
if(len(ans)<n):
ans.append("2 0")
for i in range(n):
print(ans[i])
```
|
instruction
| 0
| 5,336
| 3
| 10,672
|
No
|
output
| 1
| 5,336
| 3
| 10,673
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
s=input().split()
n=int(s[0])
k=int(s[0])
ans=[]
a=0
for i in range(1,n+1):
for j in range(i+1,n+1):
a+=1
if(a<k):
print("no solution")
else:
for i in range(2):
for j in range(n//2):
ans.append(str(i)+" "+str(j))
if(len(ans)<n):
ans.append("2 0")
for i in range(n):
print(ans[i])
```
|
instruction
| 0
| 5,337
| 3
| 10,674
|
No
|
output
| 1
| 5,337
| 3
| 10,675
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 β€ n β€ 2000, 1 β€ k β€ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| β€ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| β€ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
n,k = map(int,input().split())
if n*(n-1) < k*2:
print('no solution')
else:
for i in range(1,n+1):
print(1,i)
```
|
instruction
| 0
| 5,338
| 3
| 10,676
|
No
|
output
| 1
| 5,338
| 3
| 10,677
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wilbur the pig really wants to be a beaver, so he decided today to pretend he is a beaver and bite at trees to cut them down.
There are n trees located at various positions on a line. Tree i is located at position xi. All the given positions of the trees are distinct.
The trees are equal, i.e. each tree has height h. Due to the wind, when a tree is cut down, it either falls left with probability p, or falls right with probability 1 - p. If a tree hits another tree while falling, that tree will fall in the same direction as the tree that hit it. A tree can hit another tree only if the distance between them is strictly less than h.
For example, imagine there are 4 trees located at positions 1, 3, 5 and 8, while h = 3 and the tree at position 1 falls right. It hits the tree at position 3 and it starts to fall too. In it's turn it hits the tree at position 5 and it also starts to fall. The distance between 8 and 5 is exactly 3, so the tree at position 8 will not fall.
As long as there are still trees standing, Wilbur will select either the leftmost standing tree with probability 0.5 or the rightmost standing tree with probability 0.5. Selected tree is then cut down. If there is only one tree remaining, Wilbur always selects it. As the ground is covered with grass, Wilbur wants to know the expected total length of the ground covered with fallen trees after he cuts them all down because he is concerned about his grass-eating cow friends. Please help Wilbur.
Input
The first line of the input contains two integers, n (1 β€ n β€ 2000) and h (1 β€ h β€ 108) and a real number p (0 β€ p β€ 1), given with no more than six decimal places.
The second line of the input contains n integers, x1, x2, ..., xn ( - 108 β€ xi β€ 108) in no particular order.
Output
Print a single real number β the expected total length of the ground covered by trees when they have all fallen down. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
2 2 0.500000
1 2
Output
3.250000000
Input
4 3 0.4
4 3 1 2
Output
6.631200000
Note
Consider the first example, we have 2 trees with height 2.
<image> There are 3 scenarios:
1. Both trees falls left. This can either happen with the right tree falling left first, which has <image> probability (also knocking down the left tree), or the left tree can fall left and then the right tree can fall left, which has <image> probability. Total probability is <image>.
2. Both trees fall right. This is analogous to (1), so the probability of this happening is <image>.
3. The left tree fall left and the right tree falls right. This is the only remaining scenario so it must have <image> probability.
Cases 1 and 2 lead to a total of 3 units of ground covered, while case 3 leads to a total of 4 units of ground covered. Thus, the expected value is <image>.
Submitted Solution:
```
def expected(h, p, x):
n = len(x)
q = 1 - p
lh = [min(h, (x[i]-x[i-1] if i > 0 else h)) for i in range(n)]
rh = [min(h, (x[i+1]-x[i] if i+1 < n else h)) for i in range(n)]
A = [[None for _ in range(n)] for _ in range(n)]
for i in range(n):
A[i][i] = lh[i], rh[i], p*lh[i] + q*rh[i]
# print("A[{}][{}] = {}".format(i, i, A[i][i]))
for l in range(1, n):
for i in range(n-l):
j = i+l
eil = lh[i] + A[i+1][j][2]
eir = rh[i] + (A[i+1][j][2] if rh[i] == h else A[i+1][j][1])
ejl = lh[j] + (A[i][j-1][2] if lh[j] == h else A[i][j-1][0])
ejr = rh[j] + A[i][j-1][2]
A[i][j] = ejl, eir, 0.5*(p*eil + q*eir) + 0.5*(p*ejl + q*ejr)
# print("A[{}][{}] = {}".format(i, j, A[i][j]))
return A[0][n-1][2]
if __name__ == '__main__':
sn, sh, sp = input().split()
n = int(sn)
h = int(sh)
p = float(sp)
xs = sorted(map(int, input().split()))
print('{:0.7f}'.format(expected(h, p, xs)))
```
|
instruction
| 0
| 5,438
| 3
| 10,876
|
No
|
output
| 1
| 5,438
| 3
| 10,877
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wilbur the pig really wants to be a beaver, so he decided today to pretend he is a beaver and bite at trees to cut them down.
There are n trees located at various positions on a line. Tree i is located at position xi. All the given positions of the trees are distinct.
The trees are equal, i.e. each tree has height h. Due to the wind, when a tree is cut down, it either falls left with probability p, or falls right with probability 1 - p. If a tree hits another tree while falling, that tree will fall in the same direction as the tree that hit it. A tree can hit another tree only if the distance between them is strictly less than h.
For example, imagine there are 4 trees located at positions 1, 3, 5 and 8, while h = 3 and the tree at position 1 falls right. It hits the tree at position 3 and it starts to fall too. In it's turn it hits the tree at position 5 and it also starts to fall. The distance between 8 and 5 is exactly 3, so the tree at position 8 will not fall.
As long as there are still trees standing, Wilbur will select either the leftmost standing tree with probability 0.5 or the rightmost standing tree with probability 0.5. Selected tree is then cut down. If there is only one tree remaining, Wilbur always selects it. As the ground is covered with grass, Wilbur wants to know the expected total length of the ground covered with fallen trees after he cuts them all down because he is concerned about his grass-eating cow friends. Please help Wilbur.
Input
The first line of the input contains two integers, n (1 β€ n β€ 2000) and h (1 β€ h β€ 108) and a real number p (0 β€ p β€ 1), given with no more than six decimal places.
The second line of the input contains n integers, x1, x2, ..., xn ( - 108 β€ xi β€ 108) in no particular order.
Output
Print a single real number β the expected total length of the ground covered by trees when they have all fallen down. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
2 2 0.500000
1 2
Output
3.250000000
Input
4 3 0.4
4 3 1 2
Output
6.631200000
Note
Consider the first example, we have 2 trees with height 2.
<image> There are 3 scenarios:
1. Both trees falls left. This can either happen with the right tree falling left first, which has <image> probability (also knocking down the left tree), or the left tree can fall left and then the right tree can fall left, which has <image> probability. Total probability is <image>.
2. Both trees fall right. This is analogous to (1), so the probability of this happening is <image>.
3. The left tree fall left and the right tree falls right. This is the only remaining scenario so it must have <image> probability.
Cases 1 and 2 lead to a total of 3 units of ground covered, while case 3 leads to a total of 4 units of ground covered. Thus, the expected value is <image>.
Submitted Solution:
```
def expected(h, p, x):
n = len(x)
q = 1 - p
lh = [min(h, (x[i]-x[i-1] if i > 0 else h)) for i in range(n)]
rh = [min(h, (x[i+1]-x[i] if i+1 < n else h)) for i in range(n)]
A = [[None for _ in range(n)] for _ in range(n)]
for i in range(n):
A[i][i] = lh[i], rh[i], p*rh[i] + q*lh[i]
for l in range(1, n):
for i in range(n-l):
j = i+l
eil = lh[i] + A[i+1][j][2]
eir = rh[i] + (A[i+1][j][2] if rh[i] == h else A[i+1][j][1])
ejl = lh[j] + (A[i][j-1][2] if lh[j] == h else A[i][j-1][0])
ejr = rh[j] + A[i][j-1][2]
A[i][j] = ejl, eir, 0.5*(p*eir + q*eil) + 0.5*(p*ejr + q*ejl)
return A[0][n-1][2]
if __name__ == '__main__':
sn, sh, sp = input().split()
n = int(sn)
h = int(sh)
p = float(sp)
xs = sorted(map(int, input().split()))
print('{:0.7f}'.format(expected(h, p, xs)))
```
|
instruction
| 0
| 5,439
| 3
| 10,878
|
No
|
output
| 1
| 5,439
| 3
| 10,879
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wilbur the pig really wants to be a beaver, so he decided today to pretend he is a beaver and bite at trees to cut them down.
There are n trees located at various positions on a line. Tree i is located at position xi. All the given positions of the trees are distinct.
The trees are equal, i.e. each tree has height h. Due to the wind, when a tree is cut down, it either falls left with probability p, or falls right with probability 1 - p. If a tree hits another tree while falling, that tree will fall in the same direction as the tree that hit it. A tree can hit another tree only if the distance between them is strictly less than h.
For example, imagine there are 4 trees located at positions 1, 3, 5 and 8, while h = 3 and the tree at position 1 falls right. It hits the tree at position 3 and it starts to fall too. In it's turn it hits the tree at position 5 and it also starts to fall. The distance between 8 and 5 is exactly 3, so the tree at position 8 will not fall.
As long as there are still trees standing, Wilbur will select either the leftmost standing tree with probability 0.5 or the rightmost standing tree with probability 0.5. Selected tree is then cut down. If there is only one tree remaining, Wilbur always selects it. As the ground is covered with grass, Wilbur wants to know the expected total length of the ground covered with fallen trees after he cuts them all down because he is concerned about his grass-eating cow friends. Please help Wilbur.
Input
The first line of the input contains two integers, n (1 β€ n β€ 2000) and h (1 β€ h β€ 108) and a real number p (0 β€ p β€ 1), given with no more than six decimal places.
The second line of the input contains n integers, x1, x2, ..., xn ( - 108 β€ xi β€ 108) in no particular order.
Output
Print a single real number β the expected total length of the ground covered by trees when they have all fallen down. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
2 2 0.500000
1 2
Output
3.250000000
Input
4 3 0.4
4 3 1 2
Output
6.631200000
Note
Consider the first example, we have 2 trees with height 2.
<image> There are 3 scenarios:
1. Both trees falls left. This can either happen with the right tree falling left first, which has <image> probability (also knocking down the left tree), or the left tree can fall left and then the right tree can fall left, which has <image> probability. Total probability is <image>.
2. Both trees fall right. This is analogous to (1), so the probability of this happening is <image>.
3. The left tree fall left and the right tree falls right. This is the only remaining scenario so it must have <image> probability.
Cases 1 and 2 lead to a total of 3 units of ground covered, while case 3 leads to a total of 4 units of ground covered. Thus, the expected value is <image>.
Submitted Solution:
```
"""
Codeforces Round #331 (Div. 2)
Problem 596 D. Wilbur and Trees
@author yamaton
@date 2015-11-15
"""
import itertools as it
import functools
import operator
import collections
import math
import sys
def is_overlapping(s1, s2):
return not (max(s1) < min(s2) or max(s2) < min(s1))
def merge_intervals(segs):
segs.sort()
stack = []
stack = [segs[0]]
for s in segs[1:]:
r = stack[-1]
if is_overlapping(s, r):
r = stack.pop()
interval = (min(s[0], r[0]), max(s[1], r[1]))
stack.append(interval)
else:
stack.append(s)
return stack
def interval_sum(segs):
return sum(b - a for (a, b) in merge_intervals(segs))
def add(segs, s):
return segs + [s]
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = it.tee(iterable)
next(b, None)
return zip(a, b)
def cascade_right(trees, h):
return sum(1 for _ in it.takewhile(lambda tup: tup[1] - tup[0] < h, pairwise(trees)))
def cascade_left(trees, h):
return cascade_right(reversed(trees), h)
def solve(xs, n, h, p):
def covered(segments, trees):
if not trees:
return interval_sum(segments)
else :
nn = len(trees)
llseg = add(segments, (trees[0] - h, trees[0]))
ll = covered(llseg, trees[1:])
rrseg = add(segments, (trees[-1], trees[-1] + h))
rr = covered(rrseg, trees[:-1])
if nn == 1:
return p * ll + (1-p) * rr
idx = cascade_right(trees, h)
lrseg = add(segments, (trees[0], trees[idx] + h))
lr = covered(lrseg, trees[idx+1:])
idx = cascade_left(trees, h)
rlseg = add(segments, (trees[nn-idx-1] - h, trees[-1]))
rl = covered(rlseg, trees[:(nn-idx-1)])
return 0.5 * (p * (ll + rl) + (1 - p) * (lr + rr))
xs.sort()
return covered([], xs)
def pp(*args, **kwargs):
return print(*args, file=sys.stderr, **kwargs)
def main():
[n, h, p] = input().strip().split()
[n, h] = map(int, [n, h])
p = float(p)
xs = [int(_c) for _c in input().strip().split()]
result = solve(xs, n, h, p)
print("%.9f" % result)
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 5,440
| 3
| 10,880
|
No
|
output
| 1
| 5,440
| 3
| 10,881
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
|
instruction
| 0
| 5,555
| 3
| 11,110
|
Tags: constructive algorithms, data structures, geometry, implementation, sortings, two pointers
Correct Solution:
```
# http://codeforces.com/problemset/problem/848/B
from collections import defaultdict
def get_dest(start, w, h):
if start[0] == 1:
return (str(start[1]), str(h))
else:
return (str(w), str(start[1]))
n, w, h = [int(x) for x in input().split()]
dancers = []
groups = defaultdict(list)
destinations = [None for x in range(n)]
for ii in range(n):
g, p, t = [int(x) for x in input().split()]
dancers.append((g, p, t))
groups[p-t].append(ii)
for gg in groups.values():
V, H = [], []
for ii in gg:
dancer = dancers[ii]
if dancer[0] == 1:
V.append(dancer)
else:
H.append(dancer)
V.sort(key=lambda x: -x[1])
H.sort(key=lambda x: x[1])
table = {orig: get_dest(new, w, h) for orig, new in zip(V+H, H+V)}
for ii in gg:
destinations[ii] = table[dancers[ii]]
# print(destinations)
for dd in destinations:
print(" ".join(dd))
```
|
output
| 1
| 5,555
| 3
| 11,111
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
|
instruction
| 0
| 5,556
| 3
| 11,112
|
Tags: constructive algorithms, data structures, geometry, implementation, sortings, two pointers
Correct Solution:
```
# https://codeforces.com/problemset/problem/848/B
def push(d, val, type_, arr):
# pos index
type_ %= 2
if val not in d:
d[val] = [[],[]]
d[val][type_].append(arr)
d = {}
n, w, h = map(int, input().split())
for index in range(n):
g, p, t = map(int, input().split())
push(d, p-t, g, [p ,index])
for k, v in d.items():
v[0]=sorted(v[0], key = lambda x: x[0], reverse=True)
v[1]=sorted(v[1], key = lambda x: x[0], reverse=False)
ans = [0] * n
for v in d.values():
cur=0
bound = len(v[1])
step = len(v[0])
merge = v[0]+v[1]
n_ = len(merge)
for pos, index in merge:
if cur<bound:
ans[index]=str(merge[(step+cur)%n_][0])+' '+str(h)
else:
ans[index]=str(w)+' '+str(merge[(step+cur)%n_][0])
cur+=1
print('\n'.join(ans))
```
|
output
| 1
| 5,556
| 3
| 11,113
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
|
instruction
| 0
| 5,557
| 3
| 11,114
|
Tags: constructive algorithms, data structures, geometry, implementation, sortings, two pointers
Correct Solution:
```
n,w,h = map(int,input().split())
D = []
original = []
for i in range(n):
g,p,t = map(int,input().split())
a = p-t
p = p if g == 1 else -p
original.append(())
D.append((a,p,i))
D.sort()
from bisect import bisect
res = [None]*n
i = 0
while i < len(D):
a = D[i][0]
j = bisect(D, (a+1,-n,0), lo=i)
m = bisect(D, (a,0,0), lo=i,hi=j)
L = D[i:j]
R = D[m:j]+D[i:m]
for t in range(len(L)):
_,_,d = L[t]
_,p,_ = R[t]
if p > 0:
res[d] = (p,h)
else:
res[d] = (w,-p)
i = j
print('\n'.join(str(x)+' '+str(y) for x,y in res))
```
|
output
| 1
| 5,557
| 3
| 11,115
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
|
instruction
| 0
| 5,558
| 3
| 11,116
|
Tags: constructive algorithms, data structures, geometry, implementation, sortings, two pointers
Correct Solution:
```
import sys
def main():
n, w, h = map(int, sys.stdin.readline().split())
gv = {}
gh = {}
for i in range(n):
g, p, t = map(int, sys.stdin.readline().split())
c = p - t
t = (g, p, -t, i)
if g == 1:
if c in gv:
gv[c].append(t)
else:
gv[c] = [t]
else:
if c in gh:
gh[c].append(t)
else:
gh[c] = [t]
a = [0] * n
u = {}
for k in gv:
g = gv[k]
u[k] = True
if k in gh:
g = sorted(g, key=lambda x: x[1])
g2 = sorted(gh[k], key=lambda x: x[1], reverse=True)
l1 = len(g)
l2 = len(g2)
q = [(j[1], h) for j in g] + [(w, j[1]) for j in g2]
for i in range(l2):
a[g2[i][3]] = q[i]
for i in range(l2, l2 + l1):
a[g[i - l2][3]] = q[i]
else:
for i in g:
a[i[3]] = (i[1], h)
for k in gh:
if k not in u:
for i in gh[k]:
a[i[3]] = (w, i[1])
for i in range(n):
print(a[i][0], a[i][1])
main()
```
|
output
| 1
| 5,558
| 3
| 11,117
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
|
instruction
| 0
| 5,559
| 3
| 11,118
|
Tags: constructive algorithms, data structures, geometry, implementation, sortings, two pointers
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n,w,h=map(int,input().split())
dic={}
ansx=[0]*n
ansy=[0]*n
for i in range(n):
g,x,t=map(int,input().split())
if x-t in dic:
dic[x-t][g-1].append([x,i])
else:
dic[x-t]=[]
dic[x-t].append([])
dic[x-t].append([])
dic[x-t][g-1].append([x,i])
for i in dic:
dic[i][0].sort()
dic[i][1].sort()
dic[i][1].reverse()
people=[]
for j in dic[i][1]:
people.append(j[1])
for j in dic[i][0]:
people.append(j[1])
pointer=0
for j in dic[i][0]:
person=people[pointer]
ansx[person]=j[0]
ansy[person]=h
pointer+=1
for j in dic[i][1]:
person=people[pointer]
ansx[person]=w
ansy[person]=j[0]
pointer+=1
for i in range(n):
print(ansx[i],ansy[i])
```
|
output
| 1
| 5,559
| 3
| 11,119
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
|
instruction
| 0
| 5,560
| 3
| 11,120
|
Tags: constructive algorithms, data structures, geometry, implementation, sortings, two pointers
Correct Solution:
```
from collections import namedtuple
Dancer = namedtuple('Dancer', ['category', 'x', 'y', 'idx', 'group'])
def read_dancer(idx):
group, pos, time = [int(x) for x in input().split(' ')]
x, y = None, None
if group == 1:
x, y = pos, 0
else:
x, y = 0, pos
return Dancer(time-pos, x, y, idx, group)
n, w, h = [int(x) for x in input().split(' ')]
dancers = [read_dancer(idx) for idx in range(n)]
dancers_in = sorted(dancers, key = lambda d: (d.category, -d.group, d.x, -d.y))
dancers_out = sorted(dancers, key = lambda d: (d.category, d.group, d.x, -d.y))
end_pos = [None for _ in range(n)]
def get_end_pos(dancer):
x, y = None, None
if dancer.x == 0:
x, y = w, dancer.y
else:
x, y = dancer.x, h
return (x, y)
for i in range(n):
end_pos[dancers_in[i].idx] = get_end_pos(dancers_out[i])
for i in range(n):
print(*end_pos[i])
```
|
output
| 1
| 5,560
| 3
| 11,121
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
|
instruction
| 0
| 5,561
| 3
| 11,122
|
Tags: constructive algorithms, data structures, geometry, implementation, sortings, two pointers
Correct Solution:
```
def read_ints():
return [int(i) for i in input().split()]
n, w, h = read_ints()
tanc = [read_ints() for i in range(n)]
vert = [(k[0], k[1], k[2], i) for i, k in enumerate(tanc) if k[0] == 1]
hor = [(k[0], k[1], k[2], i) for i, k in enumerate(tanc) if k[0] == 2]
vert_st = dict()
for v in vert:
st = v[1] - v[2]
if st not in vert_st:
vert_st[st] = []
vert_st[st].append(v)
hor_st = dict()
for v in hor:
st = v[1] - v[2]
if st not in hor_st:
hor_st[st] = []
hor_st[st].append(v)
result = [None] * n
def process_group(hor, vert):
global result, w, h
hor.sort(key=lambda x: x[1])
vert.sort(key = lambda x: x[1], reverse=True)
indices = [i[3] for i in vert] + [i[3] for i in hor]
coords = [(w, i[1]) for i in hor] + [(i[1], h) for i in vert]
for i, c in zip(indices, coords):
result[i] = c
for st in set(list(vert_st.keys()) + list(hor_st.keys())):
process_group(hor_st.get(st, []), vert_st.get(st, []))
for c in result:
print(*c)
```
|
output
| 1
| 5,561
| 3
| 11,123
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
|
instruction
| 0
| 5,562
| 3
| 11,124
|
Tags: constructive algorithms, data structures, geometry, implementation, sortings, two pointers
Correct Solution:
```
n,w,h = map(int,input().split())
def it(n):
for i in range(n):
g,p,t = map(int,input().split())
yield (p-t, p if g==1 else -p, i)
D = sorted(it(n))
from bisect import bisect
from itertools import chain
res = [None]*n
i = 0
while i < len(D):
k = D[i][0]
j = bisect(D, (k+1,-n,0), lo=i)
m = bisect(D, (k,0,0), lo=i,hi=j)
for s,t in zip(range(i,j),chain(range(m,j),range(i,j))):
p,d = D[t][1],D[s][2]
res[d] = (p,h) if p > 0 else (w,-p)
i = j
print('\n'.join(' '.join(map(str,r)) for r in res))
```
|
output
| 1
| 5,562
| 3
| 11,125
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
Submitted Solution:
```
n,w,h = map(int,input().split())
D = []
original = []
for i in range(n):
g,p,t = map(int,input().split())
a = p-t
p = p if g == 1 else -p
original.append(())
D.append((a,p,i))
D.sort()
from bisect import bisect
res = [None]*n
i = 0
while i < len(D):
a = D[i][0]
j = bisect(D, (a+1,-n,0), lo=i)
m = bisect(D, (a,0,0), lo=i,hi=j)
L = D[i:j]
R = D[m:j]+D[i:m]
for t in range(len(L)):
_,_,d = L[t]
_,p,_ = R[t]
if p > 0:
res[d] = (p,h)
else:
res[d] = (w,-p)
i = j
print('\n'.join(str(x)+' '+str(y) for x,y in res))
# Made By Mostafa_Khaled
```
|
instruction
| 0
| 5,563
| 3
| 11,126
|
Yes
|
output
| 1
| 5,563
| 3
| 11,127
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
Submitted Solution:
```
import sys
def main():
n, w, h = map(int, sys.stdin.readline().split())
x = [0] * n
gv = {}
gh = {}
for i in range(n):
g, p, t = map(int, sys.stdin.readline().split())
c = p - t
if g == 1:
x[i] = (g, p, -t)
if c in gv:
gv[c].append(i)
else:
gv[c] = [i]
else:
x[i] = (g, -t, p)
if c in gh:
gh[c].append(i)
else:
gh[c] = [i]
a = [0] * n
u = {}
for k in gv:
g = gv[k]
u[k] = True
if k in gh:
g2 = gh[k]
g = sorted(g)
g2 = sorted(g2)
l1 = len(g)
l2 = len(g2)
for i in range(min(l1, l2)):
a[g2[i]] = (x[g[i]][1], h)
if l2 > l1:
for i in range(l1, l2):
a[g2[i]] = (w, x[g2[i - l1]][2])
for i in range(l1):
a[g[i]] = (w, x[g2[i + l2 - l1]][2])
elif l1 > l2:
for i in range(l2, l1):
a[g[i - l2]] = (x[g[l2]][1], h)
for i in range(l2):
a[g[i + l1 - l2]] = (w, x[g2[i]][2])
else:
for i in g:
a[i] = (x[i][1], h)
for k in gh:
if k not in u:
for i in gh[k]:
a[i] = (w, x[i][2])
for i in range(n):
print(a[i][0], a[i][1])
main()
```
|
instruction
| 0
| 5,564
| 3
| 11,128
|
No
|
output
| 1
| 5,564
| 3
| 11,129
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
Submitted Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n,w,h=map(int,input().split())
dic={}
ansx=[0]*n
ansy=[0]*n
for i in range(n):
g,x,t=map(int,input().split())
if x-t in dic:
dic[x-t][g-1].append([x,i])
else:
dic[x-t]=[]
dic[x-t].append([])
dic[x-t].append([])
dic[x-t][g-1].append([x,i])
for i in dic:
dic[i][0].sort()
dic[i][1].sort()
dic[i][1].reverse()
people=[]
for j in dic[i][1]:
people.append(j[1])
for j in dic[i][0]:
people.append(j[1])
pointer=0
for j in dic[i][0]:
person=people[pointer]
ansx[person]=j[0]
ansy[person]=h
pointer+=1
for j in dic[i][1]:
person=people[pointer]
ansx[person]=w
ansy[person]=j[0]
for i in range(n):
print(ansx[i],ansy[i])
```
|
instruction
| 0
| 5,565
| 3
| 11,130
|
No
|
output
| 1
| 5,565
| 3
| 11,131
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
Submitted Solution:
```
import sys
def main():
n, w, h = map(int, sys.stdin.readline().split())
x = [0] * n
gv = {}
gh = {}
for i in range(n):
g, p, t = map(int, sys.stdin.readline().split())
c = p - t
if g == 1:
x[i] = (g, p, -t)
if c in gv:
gv[c].append(i)
else:
gv[c] = [i]
else:
x[i] = (g, -t, p)
if c in gh:
gh[c].append(i)
else:
gh[c] = [i]
a = [0] * n
u = {}
for k in gv:
g = gv[k]
u[k] = True
if k in gh:
g2 = gh[k]
g = sorted(g)
g2 = sorted(g2)
l1 = len(g)
l2 = len(g2)
q = [(x[j][1], h) for j in g] + [(w, x[j][2]) for j in g2]
for i in range(l2):
a[g2[i]] = q[i]
for i in range(l2, l2 + l1):
a[g[i - l2]] = q[i]
else:
for i in g:
a[i] = (x[i][1], h)
for k in gh:
if k not in u:
for i in gh[k]:
a[i] = (w, x[i][2])
for i in range(n):
print(a[i][0], a[i][1])
main()
```
|
instruction
| 0
| 5,566
| 3
| 11,132
|
No
|
output
| 1
| 5,566
| 3
| 11,133
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w Γ h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of the stage stand n dancers. The i-th of them falls into one of the following groups:
* Vertical: stands at (xi, 0), moves in positive y direction (upwards);
* Horizontal: stands at (0, yi), moves in positive x direction (rightwards).
<image>
According to choreography, the i-th dancer should stand still for the first ti milliseconds, and then start moving in the specified direction at 1 unit per millisecond, until another border is reached. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
When two dancers collide (i.e. are on the same point at some time when both of them are moving), they immediately exchange their moving directions and go on.
<image>
Dancers stop when a border of the stage is reached. Find out every dancer's stopping position.
Input
The first line of input contains three space-separated positive integers n, w and h (1 β€ n β€ 100 000, 2 β€ w, h β€ 100 000) β the number of dancers and the width and height of the stage, respectively.
The following n lines each describes a dancer: the i-th among them contains three space-separated integers gi, pi, and ti (1 β€ gi β€ 2, 1 β€ pi β€ 99 999, 0 β€ ti β€ 100 000), describing a dancer's group gi (gi = 1 β vertical, gi = 2 β horizontal), position, and waiting time. If gi = 1 then pi = xi; otherwise pi = yi. It's guaranteed that 1 β€ xi β€ w - 1 and 1 β€ yi β€ h - 1. It is guaranteed that no two dancers have the same group, position and waiting time at the same time.
Output
Output n lines, the i-th of which contains two space-separated integers (xi, yi) β the stopping position of the i-th dancer in the input.
Examples
Input
8 10 8
1 1 10
1 4 13
1 7 1
1 8 2
2 2 0
2 5 14
2 6 0
2 6 1
Output
4 8
10 5
8 8
10 6
10 2
1 8
7 8
10 6
Input
3 2 3
1 1 2
2 1 1
1 1 5
Output
1 3
2 1
1 3
Note
The first example corresponds to the initial setup in the legend, and the tracks of dancers are marked with different colours in the following figure.
<image>
In the second example, no dancers collide.
Submitted Solution:
```
n, w, h = map(int, input().split(" "))
x = []
y = []
for _ in range(n):
g, p, t = map(int, input().split(" "))
if g == 1:
x.append([p,t])
else:
y.append([p,t])
xf = []
yf = []
for i in x:
for j in y:
p1 = i[0]
t1 = i[1]
p2 = j[0]
t2 = j[1]
if (p1+t1 == p2+t2):
y.append(x.remove(i))
x.append(y.remove(j))
else:
continue
for i in x:
print(i[0], h)
for i in y:
print(w, i[0])
```
|
instruction
| 0
| 5,567
| 3
| 11,134
|
No
|
output
| 1
| 5,567
| 3
| 11,135
|
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two small spaceship, surrounded by two groups of enemy larger spaceships. The space is a two-dimensional plane, and one group of the enemy spaceships is positioned in such a way that they all have integer y-coordinates, and their x-coordinate is equal to -100, while the second group is positioned in such a way that they all have integer y-coordinates, and their x-coordinate is equal to 100.
Each spaceship in both groups will simultaneously shoot two laser shots (infinite ray that destroys any spaceship it touches), one towards each of the small spaceships, all at the same time. The small spaceships will be able to avoid all the laser shots, and now want to position themselves at some locations with x=0 (with not necessarily integer y-coordinates), such that the rays shot at them would destroy as many of the enemy spaceships as possible. Find the largest numbers of spaceships that can be destroyed this way, assuming that the enemy spaceships can't avoid laser shots.
Input
The first line contains two integers n and m (1 β€ n, m β€ 60), the number of enemy spaceships with x = -100 and the number of enemy spaceships with x = 100, respectively.
The second line contains n integers y_{1,1}, y_{1,2}, β¦, y_{1,n} (|y_{1,i}| β€ 10 000) β the y-coordinates of the spaceships in the first group.
The third line contains m integers y_{2,1}, y_{2,2}, β¦, y_{2,m} (|y_{2,i}| β€ 10 000) β the y-coordinates of the spaceships in the second group.
The y coordinates are not guaranteed to be unique, even within a group.
Output
Print a single integer β the largest number of enemy spaceships that can be destroyed.
Examples
Input
3 9
1 2 3
1 2 3 7 8 9 11 12 13
Output
9
Input
5 5
1 2 3 4 5
1 2 3 4 5
Output
10
Note
In the first example the first spaceship can be positioned at (0, 2), and the second β at (0, 7). This way all the enemy spaceships in the first group and 6 out of 9 spaceships in the second group will be destroyed.
In the second example the first spaceship can be positioned at (0, 3), and the second can be positioned anywhere, it will be sufficient to destroy all the enemy spaceships.
|
instruction
| 0
| 5,580
| 3
| 11,160
|
Tags: bitmasks, brute force, geometry
Correct Solution:
```
n, m = map(int, input().strip().split())
y1 = list(map(int, input().strip().split()))
y2 = list(map(int, input().strip().split()))
y1.sort()
y2.sort()
u1 = list()
u2 = list()
p = 0
while p < n:
q = p
while q < n and y1[q] == y1[p]:
q += 1
u1.append((y1[p], q - p))
p = q
p = 0
while p < m:
q = p
while q < m and y2[q] == y2[p]:
q += 1
u2.append((y2[p], q - p))
p = q
# print(u1)
# print(u2)
n = len(u1)
m = len(u2)
res = 0
for i in range(n):
for j in range(m):
ya = u1[i][0] + u2[j][0] # first small ship
y1_stat = [True] * n
y2_stat = [True] * m
for ii in range(n):
for jj in range(m):
if u1[ii][0] + u2[jj][0] == ya:
y1_stat[ii] = False # right large ship destroyed
y2_stat[jj] = False # left large ship destroyed
f = dict()
for ii in range(n):
for jj in range(m):
yb = u1[ii][0] + u2[jj][0] # second small ship
inc = 0
if y1_stat[ii]:
inc += u1[ii][1]
if y2_stat[jj]:
inc += u2[jj][1]
if yb in f:
f[yb] += inc
else:
f[yb] = inc
yb = -1
if f:
yb = max(f, key=f.get)
for ii in range(n):
for jj in range(m):
if u1[ii][0] + u2[jj][0] == yb:
y1_stat[ii] = False
y2_stat[jj] = False
cur = 0
cur += sum(u1[ii][1] for ii in range(n) if not y1_stat[ii])
cur += sum(u2[jj][1] for jj in range(m) if not y2_stat[jj])
res = max(res, cur)
print(res)
```
|
output
| 1
| 5,580
| 3
| 11,161
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
instruction
| 0
| 6,080
| 3
| 12,160
|
Tags: math
Correct Solution:
```
def gh(a,b):
t=1
while b>0:
if b&1:t=(t*a)%1000000007
b=b>>1
a=(a*a)%1000000007
return t
a=int(input())
if a==0:print(1);exit()
print((((gh(2,2*a-1)+gh(2,a-1))%1000000007)))
```
|
output
| 1
| 6,080
| 3
| 12,161
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
instruction
| 0
| 6,081
| 3
| 12,162
|
Tags: math
Correct Solution:
```
def modularExponentiation(x,n,M):
res=1
while n>0:
if n%2==1:
res=(res*x)%M
x=(x*x)%M
n=n//2
return res
n=int(input())
s=modularExponentiation(2,n,10**9+7)
print(s*(s+1)//2%(10**9+7))
```
|
output
| 1
| 6,081
| 3
| 12,163
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
instruction
| 0
| 6,082
| 3
| 12,164
|
Tags: math
Correct Solution:
```
MOD = 10**9 + 7
def mat_mul(a, b):
result = [[0, 0], [0, 0]]
for i in range(2):
for j in range(2):
tmp = 0
for k in range(2):
tmp = (tmp + a[i][k] * b[k][j]) % MOD
result[i][j] = tmp
return result
def mat_pow(mat, exp):
if exp == 0:
return [[1, 0], [0, 1]]
elif exp % 2 == 0:
tmp = mat_pow(mat, exp // 2)
return mat_mul(tmp, tmp)
else:
return mat_mul(mat, mat_pow(mat, exp - 1))
n = int(input())
transform = mat_pow([[3, 1], [1, 3]], n)
print(transform[0][0])
```
|
output
| 1
| 6,082
| 3
| 12,165
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
instruction
| 0
| 6,083
| 3
| 12,166
|
Tags: math
Correct Solution:
```
sa=int(input())
mod=10**9+7
if sa==0:
print(1)
else:
print((pow(2, 2*sa-1, mod)+pow(2, sa-1, mod))%mod)
```
|
output
| 1
| 6,083
| 3
| 12,167
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
instruction
| 0
| 6,084
| 3
| 12,168
|
Tags: math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
mod=int(1e9)+7
def lcm(a, b):
return (a * b) / gcd(a, b)
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def main():
# for n in range(1, 385599125):
# mod=int(1e9)+7
# #n=int(input())
# dp=[[0,0] for i in range(n+1)]
# dp[0][0]=1 # up facing
# dp[0][1]=0 # down facing
# for i in range(1, n+1):
# dp[i][0]=(3*dp[i-1][0]+dp[i-1][1])%mod
# dp[i][1]=(dp[i-1][0]+3*dp[i-1][1])%mod
# print((dp[n][0]+dp[n][1]),(dp[n][0])%mod)
#
n = int(input())
if n==0:
print(1)
return
m = power(4, n-1)
#print(m / 2)
ans=(2*m + power(2, n - 1)) % mod
print(int(ans))
return
#
# # ans= (4^n)/2+(4^n-1)
if __name__ == "__main__":
main()
```
|
output
| 1
| 6,084
| 3
| 12,169
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
instruction
| 0
| 6,085
| 3
| 12,170
|
Tags: math
Correct Solution:
```
n= int(input())
m= 10**9 + 7
if n==0:
print(1)
else:
print((pow(2,2*n-1,m)+pow(2,n-1,m))%m)
```
|
output
| 1
| 6,085
| 3
| 12,171
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
instruction
| 0
| 6,086
| 3
| 12,172
|
Tags: math
Correct Solution:
```
N = int(input())
result = (pow(2, N, int(1e9 + 7)) * (pow(2, N, int(1e9 + 7)) + 1)) % int(1e9+7)
result = (result * pow(2, int(1e9+5), int(1e9+7))) % int(1e9+7)
print(result)
```
|
output
| 1
| 6,086
| 3
| 12,173
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
|
instruction
| 0
| 6,087
| 3
| 12,174
|
Tags: math
Correct Solution:
```
n=int(input())
e=10**9+7
print(1if n==0else(pow(2,n-1,e)+pow(2,n*2-1,e))%e)
# Made By Mostafa_Khaled
```
|
output
| 1
| 6,087
| 3
| 12,175
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
Submitted Solution:
```
n=int(input())
m=pow(10,9)+7
if n==0:
print(1)
else:
print((pow(2,2*n-1,m)+pow(2,n-1,m))%m)
```
|
instruction
| 0
| 6,088
| 3
| 12,176
|
Yes
|
output
| 1
| 6,088
| 3
| 12,177
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
Submitted Solution:
```
def power(a, b, mod, c):
if b == 0:
return c
if b % 2 == 1:
return power(a, b - 1, mod, c) * a % mod
else:
return power(a, b // 2, mod, c) ** 2 % mod
return c
n = int(input())
n = power(2, n, 1000000007, 1)
print((n * (n + 1)) // 2 % 1000000007)
```
|
instruction
| 0
| 6,089
| 3
| 12,178
|
Yes
|
output
| 1
| 6,089
| 3
| 12,179
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
mod=int(1e9)+7
def lcm(a, b):
return (a * b) / gcd(a, b)
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def main():
# for n in range(1, 10):
# mod=int(1e9)+7
# #n=int(input())
# dp=[[0,0] for i in range(n+1)]
# dp[0][0]=1 # up facing
# dp[0][1]=0 # down facing
# for i in range(1, n+1):
# dp[i][0]=(3*dp[i-1][0]+dp[i-1][1])%mod
# dp[i][1]=(dp[i-1][0]+3*dp[i-1][1])%mod
# print((dp[n][0]+dp[n][1]),(dp[n][0])%mod)
#
n = int(input())
m = power(4, n)
#print(m / 2)
ans=(m / 2 + power(2, n - 1)) % mod
print(int(ans))
return
# ans= (4^n)/2+(4^n-1)
return
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 6,090
| 3
| 12,180
|
No
|
output
| 1
| 6,090
| 3
| 12,181
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
mod=int(1e9)+7
def lcm(a, b):
return (a * b) / gcd(a, b)
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def main():
# for n in range(1, 10):
# mod=int(1e9)+7
# #n=int(input())
# dp=[[0,0] for i in range(n+1)]
# dp[0][0]=1 # up facing
# dp[0][1]=0 # down facing
# for i in range(1, n+1):
# dp[i][0]=(3*dp[i-1][0]+dp[i-1][1])%mod
# dp[i][1]=(dp[i-1][0]+3*dp[i-1][1])%mod
# print((dp[n][0]+dp[n][1]),(dp[n][0])%mod)
#
n = int(input())
m = power(4, n)
#print(m / 2)
print((m / 2 + power(2, n - 1)) % mod)
return
# ans= (4^n)/2+(4^n-1)
return
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 6,091
| 3
| 12,182
|
No
|
output
| 1
| 6,091
| 3
| 12,183
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
Submitted Solution:
```
N = int(input())
print(int(pow(2, N, 1000000007) * (pow(2, N, 1000000007) + 1) / 2 % 1000000007))
```
|
instruction
| 0
| 6,092
| 3
| 12,184
|
No
|
output
| 1
| 6,092
| 3
| 12,185
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
<image>
Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.
Input
The first line contains a single integer n (0 β€ n β€ 1018) β the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Output
Print a single integer β the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
Examples
Input
1
Output
3
Input
2
Output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
Submitted Solution:
```
sa=int(input())
mod=10**9+7
print(pow(2, 2*sa-1, mod)+pow(2, sa-1, mod))
```
|
instruction
| 0
| 6,093
| 3
| 12,186
|
No
|
output
| 1
| 6,093
| 3
| 12,187
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A coder cannot sit and code all day. Sometimes it is a good idea to rise from the desk, have a rest, have small talk with colleagues and even play. The coders of the F company have their favorite ball game.
Let's imagine the game on the plane with a cartesian coordinate system. The point (0, 0) contains the player who chooses an arbitrary direction and throws a ball in that direction. The ball hits the plane at distance d from the player's original position and continues flying in the same direction. After the ball hits the plane for the first time, it flies on and hits the plane again at distance 2Β·d from the player's original position and so on (it continue flying in the chosen direction and hitting the plane after each d units). All coders in the F company are strong, so the ball flies infinitely far away.
The plane has n circles painted on it. If a ball hits the plane and hits a circle that is painted on the plane (including its border), then the player gets one point. The ball can hit multiple circles at once and get one point for each of them (if the ball hits some circle x times during the move, the player also gets x points). Count the maximum number of points a player can get if he throws a ball in the arbitrary direction. Note that the direction may have real cooridinates.
Input
The first line contains two space-separated integers β n ΠΈ d (1 β€ n β€ 2Β·104; 5 β€ d β€ 10). Next n lines contain the circles' description. The i-th line contains three space-separated integers xi, yi, ri ( - 10000 β€ xi, yi β€ 10000; 1 β€ r β€ 50), where (xi, yi, ri) are the coordinates of the center and the radius of the circle, correspondingly. The point (0, 0) is not inside or on the border of some circle.
Output
Print a single integer β the maximum number of points you can get.
Examples
Input
2 5
1 1 1
5 0 1
Output
1
Input
2 5
4 0 3
5 3 1
Output
2
Input
1 10
20 0 10
Output
3
Submitted Solution:
```
import math
class Point:
def __init__(self, x1, y1):
self.x = x1
self.y = y1
class Circle:
def __init__(self, x1, y1, r1):
self.A = Point(0, 0)
self.B = Point(0, 0)
self.x = float(x1)
self.y = float(y1)
self.r = float(r1)
self.A.x = self.x - self.r #A is bottom left vertex
self.A.y = self.y - self.r
self.B.x = self.x + self.r #B is top right vertex
self.B.y = self.y + self.r
#special class that has 5 lists to store circles in
class SpecialCircleList:
def append(self, c):
if(c.x >= 0):
if(c.x - c.r <= 0):
self.crosses_axis.append(c)
return
if(c.x <= 0):
if(c.x + c.r >= 0):
self.crosses_axis.append(c)
return
if(c.y >= 0):
if(c.y - c.r <= 0):
self.crosses_axis.append(c)
return
if(c.y <= 0):
if(c.y + c.r >= 0):
self.crosses_axis.append(c)
return
if(c.x > 0 and c.y > 0):
self.first_quadrant.append(c)
elif(c.x < 0 and c.y > 0):
self.second_quadrant.append(c)
elif(c.x < 0 and c.y < 0):
self.thrid_quadrant.append(c)
else:
self.fourth_quadrant.append(c)
#pass in a bounce point or unit vector, and i'll return a list of circles to check
def getRelevantList(self, p):
retList = list()
retList.extend(self.crosses_axis)
if(p.x >= 0 and p.y >= 0):
retList.extend(self.first_quadrant)
elif(p.x < 0 and p.y >= 0):
retList.extend(self.second_quadrant)
elif(p.x < 0 and p.y < 0):
retList.extend(self.thrid_quadrant)
else:
retList.extend(self.fourth_quadrant)
return(retList)
def __init__(self):
self.crosses_axis = list() #for any circle that even remotely crosses and axis
self.first_quadrant = list() #for any circle that exclusively resides in quad
self.second_quadrant = list()
self.thrid_quadrant = list()
self.fourth_quadrant = list()
circle_list = SpecialCircleList()
d = 0
bound_up = 0
bound_down = 0
bound_left = 0
bound_right = 0
max_score = 0
def main():
global circle_list
global max_score
global d
global bound_up
global bound_down
global bound_left
global bound_right
line_in = input("")
(n, d) = line_in.split()
n = int(n)
d = int(d)
for i in range(n):
line_in = input("")
(x, y, r) = line_in.split()
x = int(x)
y = int(y)
r = int(r)
newCircle = Circle(x, y, r)
circle_list.append(newCircle)
#determine checking bounds
if(y+r > bound_up):
bound_up = y + r
if(y-r < bound_down):
bound_down = y - r
if(x + r > bound_right):
bound_right = x + r
if(x - r < bound_left):
bound_left = x - r
#brute force for angles 0->360
for angle in range(0, 360, 5):
#generate unit vector with direction
unit_vector = Point(math.cos(angle*math.pi/180),math.sin(angle*math.pi/180))
#get a list of points where the ball will bounce for a given direction/line
bounce_list = getBouncePoints(unit_vector, bound_up, bound_right,\
bound_down, bound_left)
if(len(bounce_list) <= max_score):
continue #no point doing this one if i've already got a higher score
score_for_line = checkIfInAnyCircles(bounce_list, unit_vector)
if(score_for_line > max_score):
max_score = score_for_line
print(max_score)
#returns the score of a line/trajectory
def checkIfInAnyCircles(bounce_list, unit_vector):
global circle_list
list_of_relavent_circles = circle_list.getRelevantList(unit_vector)
score = 0
if(len(list_of_relavent_circles) == 0 or len(bounce_list) == 0):
return(0)
for b in bounce_list:
for c in list_of_relavent_circles:
if(b.x >= c.A.x and b.y >= c.A.y and b.x <= c.B.x and b.y <= c.B.y):
if(thoroughCheck(b, c)):
score += 1
return(score)
#after passing a basic bounded box collision check, do a more thorough check with
#floating point arithmatic to see if a bounce (b) lands within a given circle, (c)
def thoroughCheck(b, c):
distanceFromCircleCentre = (b.x - c.x)**2 + (b.y - c.y)**2
distanceFromCircleCentre = math.sqrt(distanceFromCircleCentre)
if(distanceFromCircleCentre <= c.r):
return(True)
else:
return(False)
#returns a list of points which a ball will bounce at given a direction/trajectory
#takes into consideration bounding box, so doens't calculate more than necessary
def getBouncePoints(unit_vector, bound_up, bound_right, bound_down, bound_left ):
global d
bounce_list = list()
bounce_vector = Point(unit_vector.x * d, unit_vector.y * d)
newBounce = Point(0, 0)
if(unit_vector.x >= 0 and unit_vector.y >= 0):
#first quadrant
while(newBounce.x <= bound_right and newBounce.y <= bound_up):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
elif(unit_vector.x < 0 and unit_vector.y >= 0):
#second quadrant
while(newBounce.x >= bound_left and newBounce.y <= bound_up):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
elif(unit_vector.x < 0 and unit_vector.y < 0):
#thrid quadrant
while(newBounce.x >= bound_left and newBounce.y >= bound_down):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
else:
#fourth quadrant
while(newBounce.x <= bound_left and newBounce.y >= bound_down):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
return(bounce_list)
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 6,168
| 3
| 12,336
|
No
|
output
| 1
| 6,168
| 3
| 12,337
|
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