message stringlengths 2 16.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 575 109k | cluster float64 16 16 | __index_level_0__ int64 1.15k 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a calculator. It has a display and two buttons.
Initially, the display shows an integer x. Snuke wants to change this value into another integer y, by pressing the following two buttons some number of times in arbitrary order:
* Button A: When pressed, the value on the display is incremented by 1.
* Button B: When pressed, the sign of the value on the display is reversed.
Find the minimum number of times Snuke needs to press the buttons to achieve his objective. It can be shown that the objective is always achievable regardless of the values of the integers x and y.
Constraints
* x and y are integers.
* |x|, |y| ≤ 10^9
* x and y are different.
Input
The input is given from Standard Input in the following format:
x y
Output
Print the minimum number of times Snuke needs to press the buttons to achieve his objective.
Examples
Input
10 20
Output
10
Input
10 -10
Output
1
Input
-10 -20
Output
12
Submitted Solution:
```
x,y=map(int,input().split())
x_=-x
y_=-y
ans=10**9+6
if x<=y:
ans=min(ans,y-x)
if x_<=y:
ans=min(ans,y-x_+1)
if x_<=y_:
ans=min(ans,y_-x_+2)
if x<=y_:
ans=min(ans,y_-x+1)
print(ans)
``` | instruction | 0 | 43,689 | 16 | 87,378 |
Yes | output | 1 | 43,689 | 16 | 87,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a calculator. It has a display and two buttons.
Initially, the display shows an integer x. Snuke wants to change this value into another integer y, by pressing the following two buttons some number of times in arbitrary order:
* Button A: When pressed, the value on the display is incremented by 1.
* Button B: When pressed, the sign of the value on the display is reversed.
Find the minimum number of times Snuke needs to press the buttons to achieve his objective. It can be shown that the objective is always achievable regardless of the values of the integers x and y.
Constraints
* x and y are integers.
* |x|, |y| ≤ 10^9
* x and y are different.
Input
The input is given from Standard Input in the following format:
x y
Output
Print the minimum number of times Snuke needs to press the buttons to achieve his objective.
Examples
Input
10 20
Output
10
Input
10 -10
Output
1
Input
-10 -20
Output
12
Submitted Solution:
```
x, y = map(int,input().split())
if x >= 0 and y >= 0:
if x <= y:
ans = abs(x - y)
else:
ans = abs(x - y) + 2
elif x >= 0 and y < 0 or x < 0 and y >= 0:
ans = abs(abs(x) - abs(y)) + 1
else:
if x <= y:
ans = abs(x - y)
else:
ans = abs(x - y) + 2
print(ans)
``` | instruction | 0 | 43,690 | 16 | 87,380 |
No | output | 1 | 43,690 | 16 | 87,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a calculator. It has a display and two buttons.
Initially, the display shows an integer x. Snuke wants to change this value into another integer y, by pressing the following two buttons some number of times in arbitrary order:
* Button A: When pressed, the value on the display is incremented by 1.
* Button B: When pressed, the sign of the value on the display is reversed.
Find the minimum number of times Snuke needs to press the buttons to achieve his objective. It can be shown that the objective is always achievable regardless of the values of the integers x and y.
Constraints
* x and y are integers.
* |x|, |y| ≤ 10^9
* x and y are different.
Input
The input is given from Standard Input in the following format:
x y
Output
Print the minimum number of times Snuke needs to press the buttons to achieve his objective.
Examples
Input
10 20
Output
10
Input
10 -10
Output
1
Input
-10 -20
Output
12
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**6)
if sys.platform in (['ios','darwin','win32']):
sys.stdin=open('Untitled.txt')
input = sys.stdin.readline
def INT(): return int(input())
def MAP(): return [int(s) for s in input().split()]
def main():
x, y = MAP()
if y>x:
if y>=0 and x>=0:
print(y-x)
elif y>=0 and x<0 and abs(x)<=abs(y):
print(1+abs(y)-abs(x))
elif y>=0 and x<0 and abs(x)>abs(y):
print(1+abs(x)-abs(y))
elif y<0:
print(y-x)
elif x>y:
if y>=0 and x>=0:
print(2+(x-y))
elif x>=0 and y<0 and abs(x)<=abs(y):
print(1+abs(y)-abs(x))
elif x>=0 and y<0 and abs(x)>abs(y):
print(1+abs(x)-abs(y))
elif x<0:
print(2+abs(y-x))
if __name__ == '__main__':
main()
``` | instruction | 0 | 43,691 | 16 | 87,382 |
No | output | 1 | 43,691 | 16 | 87,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a calculator. It has a display and two buttons.
Initially, the display shows an integer x. Snuke wants to change this value into another integer y, by pressing the following two buttons some number of times in arbitrary order:
* Button A: When pressed, the value on the display is incremented by 1.
* Button B: When pressed, the sign of the value on the display is reversed.
Find the minimum number of times Snuke needs to press the buttons to achieve his objective. It can be shown that the objective is always achievable regardless of the values of the integers x and y.
Constraints
* x and y are integers.
* |x|, |y| ≤ 10^9
* x and y are different.
Input
The input is given from Standard Input in the following format:
x y
Output
Print the minimum number of times Snuke needs to press the buttons to achieve his objective.
Examples
Input
10 20
Output
10
Input
10 -10
Output
1
Input
-10 -20
Output
12
Submitted Solution:
```
x,y=map(int,input().split())
cnt=abs(abs(y)-abs(x))
if x>=0 and y>=0:print(cnt)
else:
print(cnt+1 if (x>=0 and y<0) or (y>=0 and x<0) else cnt+2)
``` | instruction | 0 | 43,692 | 16 | 87,384 |
No | output | 1 | 43,692 | 16 | 87,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a calculator. It has a display and two buttons.
Initially, the display shows an integer x. Snuke wants to change this value into another integer y, by pressing the following two buttons some number of times in arbitrary order:
* Button A: When pressed, the value on the display is incremented by 1.
* Button B: When pressed, the sign of the value on the display is reversed.
Find the minimum number of times Snuke needs to press the buttons to achieve his objective. It can be shown that the objective is always achievable regardless of the values of the integers x and y.
Constraints
* x and y are integers.
* |x|, |y| ≤ 10^9
* x and y are different.
Input
The input is given from Standard Input in the following format:
x y
Output
Print the minimum number of times Snuke needs to press the buttons to achieve his objective.
Examples
Input
10 20
Output
10
Input
10 -10
Output
1
Input
-10 -20
Output
12
Submitted Solution:
```
x,y = map(int,input().split())
if x <= y:
if x <= -y:
print(abs(x)-abs(y)+1)
else:
print(y-x)
else:
if x *y > 0:
print(abs(x-y)+2)
else:
print(abs(x)-abs(y)+1)
``` | instruction | 0 | 43,693 | 16 | 87,386 |
No | output | 1 | 43,693 | 16 | 87,387 |
Provide a correct Python 3 solution for this coding contest problem.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15 | instruction | 0 | 43,694 | 16 | 87,388 |
"Correct Solution:
```
A = sorted(list(map(int, input().split())))
print(0 if any([a % 2 == 0 for a in A]) else A[0] * A[1])
``` | output | 1 | 43,694 | 16 | 87,389 |
Provide a correct Python 3 solution for this coding contest problem.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15 | instruction | 0 | 43,695 | 16 | 87,390 |
"Correct Solution:
```
a,b,c=map(int,input().split())
if(a%2==0 or b%2==0 or c%2==0):
print(0)
else:
print(min(a*b,b*c,c*a))
``` | output | 1 | 43,695 | 16 | 87,391 |
Provide a correct Python 3 solution for this coding contest problem.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15 | instruction | 0 | 43,696 | 16 | 87,392 |
"Correct Solution:
```
a,s,d=map(int,input().split())
if a*s*d%2:
print(a*s*d//max(a,s,d))
else:print(0)
``` | output | 1 | 43,696 | 16 | 87,393 |
Provide a correct Python 3 solution for this coding contest problem.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15 | instruction | 0 | 43,697 | 16 | 87,394 |
"Correct Solution:
```
*N,=map(int,input().split())
N.sort()
print(((0--N[2]//2)-N[2]//2)*N[0]*N[1])
``` | output | 1 | 43,697 | 16 | 87,395 |
Provide a correct Python 3 solution for this coding contest problem.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15 | instruction | 0 | 43,698 | 16 | 87,396 |
"Correct Solution:
```
a,b,c=map(int,input().split())
if not (a%2==b%2==c%2==1):
print(0)
else:
print(a*b*c//max(a,b,c))
``` | output | 1 | 43,698 | 16 | 87,397 |
Provide a correct Python 3 solution for this coding contest problem.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15 | instruction | 0 | 43,699 | 16 | 87,398 |
"Correct Solution:
```
A,B,C = map(int,input().split())
if (A%2)*(B%2)*(C%2)==0:
print(0)
else:
print(min(A*B,B*C,C*A))
``` | output | 1 | 43,699 | 16 | 87,399 |
Provide a correct Python 3 solution for this coding contest problem.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15 | instruction | 0 | 43,700 | 16 | 87,400 |
"Correct Solution:
```
A = sorted(map(int,input().split()))
a = A[-1]//2
if A[-1] % 2 == 0:
print(0)
else:
print(1*A[0]*A[1])
``` | output | 1 | 43,700 | 16 | 87,401 |
Provide a correct Python 3 solution for this coding contest problem.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15 | instruction | 0 | 43,701 | 16 | 87,402 |
"Correct Solution:
```
A = list(map(int, input().split()))
A.sort()
if A[2] % 2 == 1:
print(A[0] * A[1])
else:
print('0')
``` | output | 1 | 43,701 | 16 | 87,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15
Submitted Solution:
```
num = list(map(int,input().split()))
num = sorted(num)
if num[2]%2 == 0:
print('0')
else:
print(num[0]*num[1])
``` | instruction | 0 | 43,702 | 16 | 87,404 |
Yes | output | 1 | 43,702 | 16 | 87,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15
Submitted Solution:
```
l = list(map(int, input().split()))
l.sort()
c = l[2]
diff = c - (c >> 1) * 2
print(l[0] * l[1] * diff)
``` | instruction | 0 | 43,703 | 16 | 87,406 |
Yes | output | 1 | 43,703 | 16 | 87,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15
Submitted Solution:
```
A, B, C = map(int, input().split())
print(min((A%2)*B*C, A*(B%2)*C, A*B*(C%2)))
``` | instruction | 0 | 43,704 | 16 | 87,408 |
Yes | output | 1 | 43,704 | 16 | 87,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15
Submitted Solution:
```
a=list(map(int,input().split()))
if a[0]*a[1]*a[2]%2==0:
print(0)
else:
a.sort()
print(a[0]*a[1])
``` | instruction | 0 | 43,705 | 16 | 87,410 |
Yes | output | 1 | 43,705 | 16 | 87,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15
Submitted Solution:
```
str = input()
num = list(map(int,str.split(' ')))
if (num[0]%2)*(num[1]%2)*(num[2])==1:
num.sort()
print(num[0]*num[1])
else:
print(0)
``` | instruction | 0 | 43,706 | 16 | 87,412 |
No | output | 1 | 43,706 | 16 | 87,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15
Submitted Solution:
```
from math import floor,ceil,sqrt,factorial,log
from collections import Counter, deque
from functools import reduce
import numpy as np
import itertools
def S(): return input()
def I(): return int(input())
def MS(): return map(str,input().split())
def MI(): return map(int,input().split())
def FLI(): return [int(i) for i in input().split()]
def LS(): return list(MS())
def LI(): return list(MI())
def LLS(): return [list(map(str, l.split() )) for l in input()]
def LLI(): return [list(map(int, l.split() )) for l in input()]
def LLSN(n: int): return [LS() for _ in range(n)]
def LLIN(n: int): return [LI() for _ in range(n)]
A,B,C = MI()
max = A*B*C
min = 10**4
for i in range(1,A+1):
v = i*B*C
if (max - v) < v:
continue
n = abs((max - v) - v)
if min > n:
min = n
print(min)
``` | instruction | 0 | 43,707 | 16 | 87,414 |
No | output | 1 | 43,707 | 16 | 87,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15
Submitted Solution:
```
a, b, c = [int(i) for i in input().split()]
if a*b*c %2==0:
print(0)
print(min(a*b, a*c, b*c))
``` | instruction | 0 | 43,708 | 16 | 87,416 |
No | output | 1 | 43,708 | 16 | 87,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:
* There is at least one red block and at least one blue block.
* The union of all red blocks forms a rectangular parallelepiped.
* The union of all blue blocks forms a rectangular parallelepiped.
Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.
Constraints
* 2≤A,B,C≤10^9
Input
The input is given from Standard Input in the following format:
A B C
Output
Print the minimum possible difference between the number of red blocks and the number of blue blocks.
Examples
Input
3 3 3
Output
9
Input
2 2 4
Output
0
Input
5 3 5
Output
15
Submitted Solution:
```
def resolve():
A, B, C = list(map(int, input().split()))
ans = 10 * 10
for h in range(1, B):
ans = min(ans, abs(2 * h * A * C - A * B*C))
for w in range(1, A):
ans = min(ans, abs(2 * h * A * B - A * B*C))
print(ans)
return
resolve()
``` | instruction | 0 | 43,709 | 16 | 87,418 |
No | output | 1 | 43,709 | 16 | 87,419 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output | instruction | 0 | 44,437 | 16 | 88,874 |
"Correct Solution:
```
from collections import defaultdict
N = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
A = []
B = []
ans = []
for i in range(N):
A.append(a[i]^a[(i+1)%N])
B.append(b[i]^b[(i+1)%N])
if A == B == [A[0]]*N:
for i in range(N):
ans.append((i,a[i]^b[0]))
else:
A += A
d = defaultdict(list)
for i in range(N):
d[B[i]].append(i)
i = 0
ans = []
while i < N:
ok = True
for j in range(N):
if B[N-j-1] != A[N-j-1+i]:
if d[A[N-j-1]] == []:
i += N
ok = False
break
else:
i += N - d[A[N-j-1+i]][-1] - 1
ok = False
break
if ok:
ans.append((i,a[i]^b[0]))
if len(d[A[N-1+i]]) == 1:
i += N
else:
i += N - d[A[N-1+i]][-2] - 1
for x in ans:
print(*x)
``` | output | 1 | 44,437 | 16 | 88,875 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output | instruction | 0 | 44,438 | 16 | 88,876 |
"Correct Solution:
```
import sys
from itertools import combinations, permutations, product, combinations_with_replacement, accumulate
from heapq import heapify, heappop, heappush, heappushpop
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from math import sqrt, log, floor, ceil, factorial, cos, sin, pi#, gcd
from fractions import gcd
from operator import mul
from functools import reduce
sys.setrecursionlimit(10**8)
input = sys.stdin.readline
INF = float('inf')
LINF = 2**63-1
NIL = -LINF
MOD = 10**9+7
MGN = 4
def AST(exp: bool, msg: str = ""): assert exp, msg
def TAST(exp: bool, msg = ""):
if exp is False: print("TAssertionError:", msg)
while exp is False:
pass
def II(): return int(input())
def IF(): return float(input())
def IS(): return input().replace('\n', '')
def ILCI(n: int): return [II() for _ in range(n)]
def ILCF(n: int): return [IF() for _ in range(n)]
def ILI(): return list(map(int, input().split()))
def ILLI(n: int): return [[int(j) for j in input().split()] for i in range(n)]
def ILF(): return list(map(float, input().split()))
def ILLF(n: int): return [[float(j) for j in input().split()] for i in range(n)]
def LTOS(lst: list, sep: str = ' '): return sep.join(map(str, lst))
def DEC(lst: list): return list(map(lambda x: x-1, lst))
def INC(lst: list): return list(map(lambda x: x+1, lst))
def z_algo(S: str) -> list:
N = len(S)
if N == 0:
return []
ln = [0] * N
ln[0] = N
i = 1
w = 0
while i < N:
while i + w < N and S[w] == S[i + w]:
w += 1
ln[i] = w
if w == 0:
i += 1
else:
j = 1
while i + j < N and j + ln[j] < w:
ln[i+j] = ln[j]
j += 1
i += j
w -= j
return ln
def z_algo_search(S: str, W: str) -> list:
lens = z_algo(W+S)
if len(lens) == 0:
return []
SL = len(S)
WL = len(W)
slens = lens[WL:]
res = []
for i in range(SL):
if slens[i] >= WL:
res.append(i)
return res
def xor_next(A: list) -> list:
N = len(A)
res = []
for i in range(N-1):
res.append(A[i]^A[i+1])
res.append(A[-1]^A[0])
return res
def main():
N = II()
A = ILI()
B = ILI()
C = xor_next(A)
D = xor_next(B)
# C の中から D を周期的にを探す
pos = z_algo_search(C*2, D)
pos = list(filter(lambda x: x < N, pos))
ans = []
for i in pos:
ans.append((i, A[i] ^ B[0]))
for e in ans:
print(e[0], e[1])
if __name__ == '__main__':
main()
``` | output | 1 | 44,438 | 16 | 88,877 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output | instruction | 0 | 44,439 | 16 | 88,878 |
"Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
f = [0]*n
g = [0]*n
for i in range(n):
f[i] = a[i] ^ a[(i+1)%n]
g[i] = b[i] ^ b[(i+1)%n]
def KMP(S, W):
ls, lw = len(S), len(W)
m, i = 0, 0
T = _KMP_table(W)
res = list()
while m + i < ls:
if W[i] == S[m + i]:
i += 1
if i == lw:
res.append(m)
m += i - T[i]
if i > 0:
i = T[i]
else:
m += i - T[i]
if i > 0:
i = T[i]
return res
def _KMP_table(W):
W += ['$']
lw = len(W)
T = [0] * lw
T[0] = -1
i, j = 2, 0
while i < lw:
if W[i - 1] == W[j]:
T[i] = j + 1
i += 1
j += 1
elif j > 0:
j = T[j]
else:
T[i] = 0
i += 1
return T
res = KMP(f*2, g)
for x in res:
if x < n:
print(x, a[x]^b[0])
``` | output | 1 | 44,439 | 16 | 88,879 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output | instruction | 0 | 44,440 | 16 | 88,880 |
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from itertools import permutations
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
res = list(sys.stdin.readline())
if res[-1] == "\n":
return res[:-1]
return res
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 1000000007
def solve():
def Z_algorithm(s):
n = len(s)
z = [0]*n
z[0] = n
c = 1
for i in range(1,n):
if i+z[i-c] < c+z[c]:
z[i] = z[i-c]
else:
j = max(0,c+z[c]-i)
while i+j < n and s[j] == s[i+j]:
j += 1
z[i] = j
c = i
return z
n = I()
a = LI()
b = LI()
c = [a[i]^a[(i+1)%n] for i in range(n)]
d = [b[i]^b[(i+1)%n] for i in range(n)]
z = Z_algorithm(d+c*2)[n:2*n]
for k in range(n):
if z[k] >= n:
print(k,a[k]^b[0])
return
#Solve
if __name__ == "__main__":
solve()
``` | output | 1 | 44,440 | 16 | 88,881 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output | instruction | 0 | 44,441 | 16 | 88,882 |
"Correct Solution:
```
def KMP_table(WW):
TT = [-1] * (len(WW)+2)
ii = 2
jj = 0
TT[0] = -1
TT[1] = 0
while ii < len(WW)+1:
if WW[ii - 1] == WW[jj]:
TT[ii] = jj + 1
ii += 1
jj += 1
elif jj > 0:
jj = TT[jj]
else:
TT[ii] = 0
ii += 1
return TT
def KMP(T):
ret = -1
m = 0
i = 0
RE = []
while m + i < len(S):
if W[i] == S[m + i]:
i += 1
if i == len(W):
RE.append(m)
m = m + i - T[i]
if i > 0:
i = T[i]
else:
m = m + i - T[i]
if i > 0:
i = T[i]
return RE
N = int(input())
A = [int(a) for a in input().split()]
B = [int(a) for a in input().split()]
S = ([A[i] ^ A[i-1] for i in range(N)] * 2)[:-1]
W = [B[i] ^ B[i-1] for i in range(N)]
T = KMP_table(W)
for i in KMP(T):
print(i, A[i] ^ B[0])
``` | output | 1 | 44,441 | 16 | 88,883 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output | instruction | 0 | 44,442 | 16 | 88,884 |
"Correct Solution:
```
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
X = []
for a,b in zip(A,A[1:]):
X.append(a ^ b)
X.append(A[-1] ^ A[0])
Y = []
for a,b in zip(B,B[1:]):
Y.append(a ^ b)
Y.append(B[-1] ^ B[0])
def z_algorithm(s):
N = len(s)
ret = [0]*N
ret[0] = N
i,j = 1,0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
ret[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + ret[k]<j:
ret[i+k] = ret[k]
k += 1
i += k
j -= k
return ret
za = z_algorithm(X + [-1] + Y + Y)
for i in range(N):
if za[-N-i] == N:
print(i, A[i]^B[0])
``` | output | 1 | 44,442 | 16 | 88,885 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output | instruction | 0 | 44,443 | 16 | 88,886 |
"Correct Solution:
```
num = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = []
d = []
for i in range(num):
c.append(a[i] ^ a[(i+1)%num])
d.append(b[i] ^ b[(i+1)%num])
c += c
# print(c)
# print(d)
# テーブル作成
table = [0]
k = 0
for i in range(1, num+1):
table.append(k)
if i == num:
continue
if d[i] == d[k]:
k += 1
else:
k = 0
# print(table)
i = 0 # 探索される文字列の位置
j = 0 # 探索する文字列の位置
k = 0 # kの初期値
while k < num:
while True:
# print(i, j, k)
if c[i] == d[j]:
if j == num - 1:
print(k, a[k] ^ b[0])
i += 1
j = table[num]
k = i - j
break
i += 1
j += 1
elif j != 0:
j -= 1
else:
i += 1
j = table[j]
break
k = i - j
``` | output | 1 | 44,443 | 16 | 88,887 |
Provide a correct Python 3 solution for this coding contest problem.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output | instruction | 0 | 44,444 | 16 | 88,888 |
"Correct Solution:
```
#https://tjkendev.github.io/procon-library/python/string/rolling_hash.html
class RollingHash():
def __init__(self, s, base, mod):
self.mod = mod
self.pw = pw = [1]*(len(s)+1)
l = len(s)
self.h = h = [0]*(l+1)
v = 0
for i in range(l):
h[i+1] = v = (v * base + s[i]) % mod
v = 1
for i in range(l):
pw[i+1] = v = v * base % mod
def get(self, l, r):
return (self.h[r] - self.h[l] * self.pw[r-l]) % self.mod
N=int(input())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
c=[a[i%N]^a[(i+1)%N] for i in range(2*N)]
d=[b[i]^b[(i+1)%N] for i in range(N)]
P=RollingHash(c,97,10**9+7)
Q=RollingHash(d,97,10**9+7)
for k in range(N):
if P.get(k,k+N)==Q.get(0,N):
x=b[0]^a[k]
print(k,x)
``` | output | 1 | 44,444 | 16 | 88,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output
Submitted Solution:
```
import sys
BASE, MOD1, MOD2 = (1<<30), (1<<61)-1, (1<<31)-1
class RollingHash():
def __init__(self, s, base, mod):
self.mod = mod
self.pw = pw = [1]*(len(s)+1)
l = len(s)
self.h = h = [0]*(l+1)
v = 0
for i in range(l):
h[i+1] = v = (v * base + s[i]) % mod
v = 1
for i in range(l):
pw[i+1] = v = v * base % mod
# [l. r)
def get(self, l, r):
return (self.h[r] - self.h[l] * self.pw[r-l]) % self.mod
def concatenate(self, l1, r1, l2, r2):
return (self.get(l1, r1) * self.pw[r2-l2] + self.get(l2, r2)) % self.mod
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if n == 1:
print(0, a[0]^b[0])
sys.exit()
x = [a[n-1]^a[0]]
y = [b[n-1]^b[0]]
for i in range(n-1):
x.append(a[i]^a[i+1])
y.append(b[i]^b[i+1])
rh1_x, rh1_y = RollingHash(x, BASE, MOD1), RollingHash(y, BASE, MOD1)
rh2_x, rh2_y = RollingHash(x, BASE, MOD2), RollingHash(y, BASE, MOD2)
for k in range(n):
if rh1_y.concatenate(n-k, n, 0, n-k) == rh1_x.get(0, n) and rh2_y.concatenate(n-k, n, 0, n-k) == rh2_x.get(0, n):
print(k, a[0]^b[(n-k)%n])
``` | instruction | 0 | 44,445 | 16 | 88,890 |
Yes | output | 1 | 44,445 | 16 | 88,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output
Submitted Solution:
```
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from itertools import permutations
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
res = list(sys.stdin.readline())
if res[-1] == "\n":
return res[:-1]
return res
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 1000000007
def solve():
def Z_algorithm(s):
n = len(s)
z = [0]*n
z[0] = n
c = 1
for i in range(1,n):
if i+z[i-c] < c+z[c]:
z[i] = z[i-c]
else:
j = max(0,c+z[c]-i)
while i+j < n and s[j] == s[i+j]:
j += 1
z[i] = j
c = i
return z
n = I()
a = LI()
b = LI()
a_xor = [a[i%n]^a[(i+1)%n] for i in range(2*n)]
b_xor = [b[i%n]^b[(i+1)%n] for i in range(n)]
b_xor += a_xor
z = Z_algorithm(b_xor)
for k in range(n):
i = n+k
x = b[0]^a[k]
if z[i] >= n:
print(k,x)
return
#Solve
if __name__ == "__main__":
solve()
``` | instruction | 0 | 44,446 | 16 | 88,892 |
Yes | output | 1 | 44,446 | 16 | 88,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output
Submitted Solution:
```
class KMP:
def __init__(self, P):
self.P = P
self.N = len(P)
self.T = [0] * (self.N + 1)
self._compile()
def _compile(self):
j = 0
self.T[0] = -1
for i in range(1, self.N):
self.T[i] = j
j += 1 if self.P[i] == self.P[j] else -j
self.T[self.N] = j
def search(self, S):
NS = len(S)
i = m = 0
A = []
while m + i < NS:
if self.P[i] == S[m + i]:
i += 1
if i != self.N:
continue
A.append(m)
m += i - self.T[i]
i = max(0, self.T[i])
return A
def resolve():
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
dA = [A[i] ^ A[(i + 1) % N] for i in range(N)]
dB = [B[i] ^ B[(i + 1) % N] for i in range(N)]
dA = dA + dA[:-1]
# C string D pattern
K = KMP(dB).search(dA)
for k in K:
print(k, A[k] ^ B[0])
if __name__ == "__main__":
resolve()
``` | instruction | 0 | 44,447 | 16 | 88,894 |
Yes | output | 1 | 44,447 | 16 | 88,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output
Submitted Solution:
```
import sys,queue,math,copy,itertools,bisect,collections,heapq
def main():
MOD = 2**89-1
LI = lambda : [int(x) for x in sys.stdin.readline().split()]
NI = lambda : int(sys.stdin.readline())
N = NI()
a = LI()
b = LI()
ax = []
bx = []
for i in range(N):
ax.append(a[i-1] ^ a[i])
bx.append(b[i-1] ^ b[i])
az = 0
bz = 0
m = 2**30
for i in range(N):
az = (az * m + ax[i]) % MOD
bz = (bz * m + bx[i]) % MOD
for i in range(N):
if az == bz:
print(i,a[i] ^ b[0])
az = ((az - ax[i] * pow(m,N-1,MOD)) * m + ax[i]) % MOD
if __name__ == '__main__':
main()
``` | instruction | 0 | 44,448 | 16 | 88,896 |
Yes | output | 1 | 44,448 | 16 | 88,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output
Submitted Solution:
```
n = int(input())
*a, = map(int, input().split())
*b, = map(int, input().split())
a_bits = [0] * (30 + 1)
for _a in a:
for j in range(30 + 1):
if _a & (1 << j) > 0:
a_bits[j] += 1
b_bits = [0] * (30 + 1)
for _b in b:
for j in range(30 + 1):
if _b & (1 << j) > 0:
b_bits[j] += 1
counter = [0] * (2 * 10 ** 5 + 1)
for _bb in b_bits:
counter[_bb] += 1
for k in range(n):
x = b[0] ^ a[k]
c = [0] * (2 * 10 ** 5 + 1)
flg = False
for j in range(30 + 1):
if x & (1 << j) > 0:
t = n - a_bits[j]
else:
t = a_bits[j]
if c[t] < counter[t]:
c[t] += 1
else:
flg = True
break
if flg:
continue
if any(_a ^ x != b[i] for i, _a in enumerate(a[k + 1:n], 1)):
continue
if any(_a ^ x != b[i] for i, _a in enumerate(a[:k], n - k)):
continue
print(k, x)
# aのk個手前の要素 ^ x = b
``` | instruction | 0 | 44,449 | 16 | 88,898 |
No | output | 1 | 44,449 | 16 | 88,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output
Submitted Solution:
```
N = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
A = []
B = []
ans = []
for i in range(N):
A.append(a[i]^a[(i+1)%N])
B.append(b[i]^b[(i+1)%N])
for k in range(N):
C = A[k:]+A[:k]
if B == C:
ans.append((k,a[k]^b[0]))
for x in ans:
print(*x)
``` | instruction | 0 | 44,450 | 16 | 88,900 |
No | output | 1 | 44,450 | 16 | 88,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output
Submitted Solution:
```
def main():
mod = 2**61-1
pow3 = [1]*200001
p = 1
i3 = pow(3, mod-2, mod)
for i in range(1, 200001):
p = p*3 % mod
pow3[i] = p
def rolling_hash(seq):
h = rolling_hash2(seq)
seq_size = len(seq)
H0s = [0]*seq_size
for i, j in enumerate(seq):
H0s[i] = h
h = (h-j-1+(j+1)*pow3[n])*i3 % mod
return H0s
def rolling_hash2(seq):
p = 0
for i, j in enumerate(seq):
p += (j+1)*pow3[i] % mod
return p % mod
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m = max(a+b)
a1 = [0]*n
a2 = [0]*n
b1 = [0]*n
memo = [0]*n
cnt = n
for k in range(len(bin(m))-2):
k2 = 2**k
for i, j in enumerate(a):
p = j & k2
a1[i] = p
a2[i] = p ^ k2
for i, j in enumerate(b):
b1[i] = j & k2
b_hash = rolling_hash2(b1)
A1 = rolling_hash(a1)
A2 = rolling_hash(a2)
for i, j in enumerate(memo):
if j is None:
continue
elif A1[i] == b_hash:
pass
elif A2[i] == b_hash:
memo[i] += k2
else:
memo[i] = None
cnt -= 1
if cnt == 0:
return
for i, j in enumerate(memo):
if j is not None:
print(i, j)
main()
``` | instruction | 0 | 44,451 | 16 | 88,902 |
No | output | 1 | 44,451 | 16 | 88,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are two sequences a=\\{a_0,\ldots,a_{N-1}\\} and b=\\{b_0,\ldots,b_{N-1}\\} of N non-negative integers each.
Snuke will choose an integer k such that 0 \leq k < N and an integer x not less than 0, to make a new sequence of length N, a'=\\{a_0',\ldots,a_{N-1}'\\}, as follows:
* a_i'= a_{i+k \mod N}\ XOR \ x
Find all pairs (k,x) such that a' will be equal to b.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq N \leq 2 \times 10^5
* 0 \leq a_i,b_i < 2^{30}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{N-1}
Output
Print all pairs (k, x) such that a' and b will be equal, using one line for each pair, in ascending order of k (ascending order of x for pairs with the same k).
If there are no such pairs, the output should be empty.
Examples
Input
3
0 2 1
1 2 3
Output
1 3
Input
5
0 0 0 0 0
2 2 2 2 2
Output
0 2
1 2
2 2
3 2
4 2
Input
6
0 1 3 7 6 4
1 5 4 6 2 3
Output
2 2
5 5
Input
2
1 2
0 0
Output
Submitted Solution:
```
N=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
A=[a[i]^a[i-1] for i in range(0,N)]
B=[b[i]^b[i-1] for i in range(0,N)]
ca=0
for i in range(0,N):
ca=ca|A[i]
ca=ca<<35
ca=ca>>35
cb=0
for i in range(0,N):
cb=cb|B[i]
cb=cb<<35
cb=cb>>35
for k in range(0,N):
if ca==cb:
print(k,a[k]^b[0])
x=0
for i in range(0,35):
if cb &1==1:
x+=pow(2,i)
cb=cb>>1
x=x<<35*(N-1)
cb=cb|x
``` | instruction | 0 | 44,452 | 16 | 88,904 |
No | output | 1 | 44,452 | 16 | 88,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke got a grid from his mother, as a birthday present. The grid has H rows and W columns. Each cell is painted black or white. All black cells are 4-connected, that is, it is possible to traverse from any black cell to any other black cell by just visiting black cells, where it is only allowed to move horizontally or vertically.
The color of the cell at the i-th row and j-th column (1 ≦ i ≦ H, 1 ≦ j ≦ W) is represented by a character s_{ij}. If s_{ij} is `#`, the cell is painted black. If s_{ij} is `.`, the cell is painted white. At least one cell is painted black.
We will define fractals as follows. The fractal of level 0 is a 1 × 1 grid with a black cell. The fractal of level k+1 is obtained by arranging smaller grids in H rows and W columns, following the pattern of the Snuke's grid. At a position that corresponds to a black cell in the Snuke's grid, a copy of the fractal of level k is placed. At a position that corresponds to a white cell in the Snuke's grid, a grid whose cells are all white, with the same dimensions as the fractal of level k, is placed.
You are given the description of the Snuke's grid, and an integer K. Find the number of connected components of black cells in the fractal of level K, modulo 10^9+7.
Constraints
* 1 ≦ H,W ≦ 1000
* 0 ≦ K ≦ 10^{18}
* Each s_{ij} is either `#` or `.`.
* All black cells in the given grid are 4-connected.
* There is at least one black cell in the given grid.
Input
The input is given from Standard Input in the following format:
H W K
s_{11} .. s_{1W}
:
s_{H1} .. s_{HW}
Output
Print the number of connected components of black cells in the fractal of level K, modulo 10^9+7.
Examples
Input
3 3 3
.#.
###
#.#
Output
20
Input
3 3 3
.#.
.#
Output
20
Input
3 3 3
.#
Output
1
Input
11 15 1000000000000000000
.....#.........
....###........
....####.......
...######......
...#######.....
..##.###.##....
..##########...
.###.....####..
.####...######.
.##..##..##..#
Output
301811921
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import numpy as np
H, W, K = map(int, readline().split())
S = np.frombuffer(read(),'S1').reshape(H,-1)[:,:W] == b'#'
MOD = 10 ** 9 + 7
if K in [0,1]:
print(1)
exit()
B = S.sum()
if H == 1 or W == 1:
if B == H * W:
answer = 1
else:
answer = pow(B, K-1, MOD)
print(answer)
exit()
concat_h = S[:,0] & S[:,-1]
concat_w = S[0] & S[-1]
bl_h = np.any(concat_h)
bl_w = np.any(concat_w)
if bl_h and bl_w:
print(1)
exit()
if (not bl_h) and (not bl_w):
answer = pow(B, K-1, MOD)
print(answer)
exit()
if bl_w:
H, W = W, H
S = S.T
concat_h, concat_w = concat_w, concat_h
def power_mat(A,n,MOD):
k = A.shape[0]
if n == 0:
return np.eye(k,dtype=np.int64)
B = power_mat(A,n//2,MOD)
B = np.dot(B,B) % MOD
return np.dot(A,B) % MOD if n & 1 else B
h_edge = (S[:,:-1] & S[:,1:]).sum()
A = np.array([[B, -h_edge], [0, np.count_nonzero(concat_h)]], np.int64)
answer = power_mat(A, K-1, MOD)[0].sum() % MOD
print(answer)
``` | instruction | 0 | 44,549 | 16 | 89,098 |
No | output | 1 | 44,549 | 16 | 89,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke got a grid from his mother, as a birthday present. The grid has H rows and W columns. Each cell is painted black or white. All black cells are 4-connected, that is, it is possible to traverse from any black cell to any other black cell by just visiting black cells, where it is only allowed to move horizontally or vertically.
The color of the cell at the i-th row and j-th column (1 ≦ i ≦ H, 1 ≦ j ≦ W) is represented by a character s_{ij}. If s_{ij} is `#`, the cell is painted black. If s_{ij} is `.`, the cell is painted white. At least one cell is painted black.
We will define fractals as follows. The fractal of level 0 is a 1 × 1 grid with a black cell. The fractal of level k+1 is obtained by arranging smaller grids in H rows and W columns, following the pattern of the Snuke's grid. At a position that corresponds to a black cell in the Snuke's grid, a copy of the fractal of level k is placed. At a position that corresponds to a white cell in the Snuke's grid, a grid whose cells are all white, with the same dimensions as the fractal of level k, is placed.
You are given the description of the Snuke's grid, and an integer K. Find the number of connected components of black cells in the fractal of level K, modulo 10^9+7.
Constraints
* 1 ≦ H,W ≦ 1000
* 0 ≦ K ≦ 10^{18}
* Each s_{ij} is either `#` or `.`.
* All black cells in the given grid are 4-connected.
* There is at least one black cell in the given grid.
Input
The input is given from Standard Input in the following format:
H W K
s_{11} .. s_{1W}
:
s_{H1} .. s_{HW}
Output
Print the number of connected components of black cells in the fractal of level K, modulo 10^9+7.
Examples
Input
3 3 3
.#.
###
#.#
Output
20
Input
3 3 3
.#.
.#
Output
20
Input
3 3 3
.#
Output
1
Input
11 15 1000000000000000000
.....#.........
....###........
....####.......
...######......
...#######.....
..##.###.##....
..##########...
.###.....####..
.####...######.
.##..##..##..#
Output
301811921
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import numpy as np
H, W, K = map(int, readline().split())
S = np.frombuffer(read(),'S1').reshape(H,-1)[:,:W] == b'#'
MOD = 10 ** 9 + 7
if K in [0,1]:
print(1)
exit()
B = S.sum()
if H == 1 or W == 1:
if B == H * W:
answer = 1
else:
answer = pow(int(B), K-1, MOD)
print(answer)
exit()
concat_h = S[:,0] & S[:,-1]
concat_w = S[0] & S[-1]
bl_h = np.any(concat_h)
bl_w = np.any(concat_w)
if bl_h and bl_w:
print(1)
exit()
if (not bl_h) and (not bl_w):
answer = pow(B, K-1, MOD)
print(answer)
exit()
if bl_w:
H, W = W, H
S = S.T
concat_h, concat_w = concat_w, concat_h
def power_mat(A,n,MOD):
k = A.shape[0]
if n == 0:
return np.eye(k,dtype=np.int64)
B = power_mat(A,n//2,MOD)
B = np.dot(B,B) % MOD
return np.dot(A,B) % MOD if n & 1 else B
h_edge = (S[:,:-1] & S[:,1:]).sum()
A = np.array([[B, -h_edge], [0, np.count_nonzero(concat_h)]], np.int64)
answer = power_mat(A, K-1, MOD)[0].sum() % MOD
print(answer)
``` | instruction | 0 | 44,550 | 16 | 89,100 |
No | output | 1 | 44,550 | 16 | 89,101 |
Provide a correct Python 3 solution for this coding contest problem.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727 | instruction | 0 | 45,247 | 16 | 90,494 |
"Correct Solution:
```
import math
from functools import reduce
n = int(input())
xy = []
for i in range(n):
x, y = [int(item) for item in input().split()]
xy.append([x, y, i])
def convex_hull_graham(points):
TURN_LEFT, TURN_RIGHT, TURN_NONE = (1, -1, 0)
def cmp(a, b):
return (a > b) - (a < b)
def turn(p, q, r):
return cmp((q[0] - p[0])*(r[1] - p[1]) - (r[0] - p[0])*(q[1] - p[1]), 0)
def _keep_left(hull, r):
while len(hull) > 1 and turn(hull[-2], hull[-1], r) == TURN_RIGHT:
hull.pop()
if not len(hull) or hull[-1] != r:
hull.append(r)
return hull
points = sorted(points)
l = reduce(_keep_left, points, [])
u = reduce(_keep_left, reversed(points), [])
return l.extend(u[i] for i in range(1, len(u) - 1)) or l
ret = convex_hull_graham(xy)
ret.append(ret[0])
ret.append(ret[1])
thetas = [0.0] * n
for i in range(1, len(ret)-1):
v1 = [-ret[i][1] + ret[i-1][1], ret[i][0] - ret[i-1][0]]
v2 = [-ret[i+1][1] + ret[i][1], ret[i+1][0] - ret[i][0]]
v1_norm = math.sqrt(v1[0]**2.0 + v1[1]**2.0)
v2_norm = math.sqrt(v2[0]**2.0 + v2[1]**2.0)
thetas[ret[i][2]] = math.acos((v1[0]*v2[0] + v1[1]*v2[1]) / (v1_norm * v2_norm))
for item in thetas:
print(item / (math.pi * 2.0))
``` | output | 1 | 45,247 | 16 | 90,495 |
Provide a correct Python 3 solution for this coding contest problem.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727 | instruction | 0 | 45,248 | 16 | 90,496 |
"Correct Solution:
```
from math import atan2,pi
N=int(input())
points=[[int(i) for i in input().split()] for i in range(N)]
p=[0]*N
for i in range(N):
angle_p=[]
angle_n=[]
for j in range(N):
if i!=j:
c=atan2(points[i][1]-points[j][1],points[i][0]-points[j][0])
if c>=0:
angle_p.append(c)
else:
angle_n.append(c)
if len(angle_n)==0:
d=max(angle_p)-min(angle_p)
elif len(angle_p)==0:
d=max(angle_n)-min(angle_n)
elif max(angle_p)-min(angle_n)<pi:
d=max(angle_p)-min(angle_n)
elif pi-min(angle_p)+max(angle_n)<0:
d=2*pi-min(angle_p)+max(angle_n)
else:
d=pi
p[i]=(pi-d)/(2*pi)
for i in range(N):
print(p[i])
``` | output | 1 | 45,248 | 16 | 90,497 |
Provide a correct Python 3 solution for this coding contest problem.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727 | instruction | 0 | 45,249 | 16 | 90,498 |
"Correct Solution:
```
import math
from collections import defaultdict
def norm2(vec):
return math.sqrt(vec[0]**2 + vec[1]**2)
# any 2 points must have different position.
def ConvexHull(point_list):
pos2idx = {point_list[i]: i for i in range(len(point_list))}
y_val = defaultdict(list)
x_list = sorted(list(set([p[0] for p in point_list])))
for x, y in point_list:
y_val[x].append(y)
upper = [(x_list[0], max(y_val[x_list[0]]))]
lower = [(x_list[0], min(y_val[x_list[0]]))]
prev = float('inf')
for xi in x_list[1:]:
x0, y0 = upper[-1]
x1, y1 = xi, max(y_val[xi])
if (y1 - y0) / (x1 - x0) < prev:
upper.append((x1, y1))
prev = (y1 - y0) / (x1 - x0)
else:
while True:
x0, y0 = upper[-1]
if len(upper) == 1:
upper.append((x1, y1))
break
x00, y00 = upper[-2]
if (y1 - y0) / (x1 - x0) > (y1 - y00) / (x1 - x00):
upper.pop()
else:
prev = (y1 - y0) / (x1 - x0)
upper.append((x1, y1))
break
prev = -float('inf')
for xi in x_list[1:]:
x0, y0 = lower[-1]
x1, y1 = xi, min(y_val[xi])
if (y1 - y0) / (x1 - x0) > prev:
lower.append((x1, y1))
prev = (y1 - y0) / (x1 - x0)
else:
while True:
x0, y0 = lower[-1]
if len(lower) == 1:
lower.append((x1, y1))
break
x00, y00 = lower[-2]
if (y1 - y0) / (x1 - x0) < (y1 - y00) / (x1 - x00):
lower.pop()
else:
prev = (y1 - y0) / (x1 - x0)
lower.append((x1, y1))
break
# return upper, lower
# return [pos2idx[xy] for xy in upper], [pos2idx[xy] for xy in lower]
upper_idx, lower_idx = [pos2idx[xy] for xy in upper], [pos2idx[xy] for xy in lower]
if upper_idx[-1] == lower_idx[-1]:
upper_idx.pop()
CH_idx = upper_idx
CH_idx.extend(reversed(lower_idx))
if CH_idx[0] == CH_idx[-1] and len(CH_idx) > 1:
CH_idx.pop()
return CH_idx
N = int(input())
point_list = []
for _ in range(N):
x, y = map(int, input().split())
point_list.append((x, y))
CH = ConvexHull(point_list)
if len(CH) == 2:
ans = [0] * N
ans[CH[0]] = 0.5
ans[CH[1]] = 0.5
for i in range(N):
print(ans[i])
exit()
ans = [0] * N
for i in range(len(CH)):
s, t, u = CH[i-1], CH[i], CH[(i+1)%len(CH)]
x0, y0 = point_list[s]
x1, y1 = point_list[t]
x2, y2 = point_list[u]
vec0 = (y0-y1, x1-x0)
vec1 = (y1-y2, x2-x1)
ans[t] = math.acos((vec0[0] * vec1[0] + vec0[1] * vec1[1]) / (norm2(vec0) * norm2(vec1))) / (2 * math.pi)
for i in range(N):
print(ans[i])
``` | output | 1 | 45,249 | 16 | 90,499 |
Provide a correct Python 3 solution for this coding contest problem.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727 | instruction | 0 | 45,250 | 16 | 90,500 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def ccw(a, b, c):
ax = b[0] - a[0]
ay = b[1] - a[1]
bx = c[0] - a[0]
by = c[1] - a[1]
t = ax*by - ay*bx;
if t > 0:
return 1
if t < 0:
return -1
if ax*bx + ay*by < 0:
return 2
if ax*ax + ay*ay < bx*bx + by*by:
return -2
return 0
def convex_hull(ps):
n = len(ps)
k = 0
ps.sort()
ch = [None] * (n*2)
for i in range(n):
while k >= 2 and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
t = k + 1
for i in range(n-2,-1,-1):
while k >= t and ccw(ch[k-2], ch[k-1], ps[i]) <= 0:
k -= 1
ch[k] = ps[i]
k += 1
return ch[:k-1]
def radian(a, b):
return math.atan2(b[1]-a[1], b[0]-a[0])
def main():
n = I()
a = [LI() + [_] for _ in range(n)]
t = convex_hull(a)
l = len(t)
r = [0] * n
if l == 2:
r[t[0][2]] = 0.5
r[t[1][2]] = 0.5
else:
for i in range(l):
r1 = radian(t[i], t[i-1])
r2 = radian(t[i], t[(i+1) % l])
k = r2 - r1
while k < 0:
k += math.pi * 2
e = (k-math.pi) / math.pi / 2
r[t[i][2]] = e
return '\n'.join(map(str,r))
print(main())
``` | output | 1 | 45,250 | 16 | 90,501 |
Provide a correct Python 3 solution for this coding contest problem.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727 | instruction | 0 | 45,251 | 16 | 90,502 |
"Correct Solution:
```
import math
import fractions
#import sys
#input = sys.stdin.readline
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
# divisors.sort()
return divisors
def ValueToBits(x,digit):
res = [0 for i in range(digit)]
now = x
for i in range(digit):
res[i]=now%2
now = now >> 1
return res
def BitsToValue(arr):
n = len(arr)
ans = 0
for i in range(n):
ans+= arr[i] * 2**i
return ans
def ZipArray(a):
aa = [[a[i],i]for i in range(n)]
aa.sort(key = lambda x : x[0])
for i in range(n):
aa[i][0]=i+1
aa.sort(key = lambda x : x[1])
b=[aa[i][0] for i in range(len(a))]
return b
def ValueToArray10(x, digit):
ans = [0 for i in range(digit)]
now = x
for i in range(digit):
ans[digit-i-1] = now%10
now = now //10
return ans
def Zeros(a,b):
if(b<=-1):
return [0 for i in range(a)]
else:
return [[0 for i in range(b)] for i in range(a)]
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
'''
def cmb(n, r, p):
if (r < 0) or (n < r):
return 0
r = min(r, n - r)
return fact[n] * factinv[r] * factinv[n-r] % p
p = 10**9 + 7
N = 10 ** 6 + 2
fact = [1, 1] # fact[n] = (n! mod p)
factinv = [1, 1] # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1] # factinv 計算用
for i in range(2, N + 1):
fact.append((fact[-1] * i) % p)
inv.append((-inv[p % i] * (p // i)) % p)
factinv.append((factinv[-1] * inv[-1]) % p)
'''
#a = list(map(int, input().split()))
#################################################
#################################################
#################################################
#################################################
#11:00
n = int(input())
xy = []
for i in range(n):
xy.append(list(map(int, input().split())))
xy[-1].append(i)
origin = list(xy)
xy.sort(key = lambda x : -x[1])
xy.sort(key = lambda x : x[0])
#print(xy)
down = [0]
while(True):
now = down[-1]
nowxy = xy[down[-1]]
#print(now,nowxy)
if(now==n-1):
break
shouldBreak = 0
for j in range(now+1,n):
if(nowxy[0] == xy[j][0]):
down.append(j)
shouldBreak=1
break
if(shouldBreak):
continue
gradmin = 10**18
nxt = -1
for j in range(now+1,n):
grad = (xy[j][1]-nowxy[1])/(xy[j][0]-nowxy[0])
#print(grad)
if(gradmin > grad):
gradmin = grad
nxt = j
down.append(nxt)
#print(down)
down = [xy[i][2] for i in down]
down = down[0:len(down)-1]
#print(down)
xy.sort(key = lambda x : x[1])
xy.sort(key = lambda x : -x[0])
#print(xy)
up = [0]
while(True):
now = up[-1]
nowxy = xy[up[-1]]
#print(now,nowxy)
if(now==n-1):
break
shouldBreak = 0
for j in range(now+1,n):
if(nowxy[0] == xy[j][0]):
up.append(j)
#print(j)
shouldBreak=1
break
if(shouldBreak):
continue
gradmin = 10**18
nxt = -1
for j in range(now+1,n):
grad = (xy[j][1]-nowxy[1])/(xy[j][0]-nowxy[0])
if(gradmin > grad):
#print("g",j,grad)
gradmin = grad
nxt = j
up.append(nxt)
#print(up)
#print(up)
up = [xy[i][2] for i in up]
up = up[:len(up)-1]
#print(up)
seq = list(down)
for i in up:
seq.append(i)
#print(seq)
ans = [0.0 for i in range(n)]
def u2(a):
return math.sqrt(a[0]*a[0]+a[1]*a[1])
def dot(a,b):
return a[0]*b[0]+a[1]*b[1]
m = len(seq)
for i in range(len(seq)):
num1,num2,num3 = seq[(i-1)%m],seq[i],seq[(i+1)%m]
p1,p2,p3 = origin[num1],origin[num2], origin[num3]
left = [p1[0]-p2[0],p1[1]-p2[1]]
right = [p3[0]-p2[0],p3[1]-p2[1]]
cos = dot(left,right)/u2(left)/u2(right)
#print(cos,left,right)
deg = math.pi - math.acos(cos)
ans[num2] = deg/(2.0*math.pi)
for i in ans:
print(i)
``` | output | 1 | 45,251 | 16 | 90,503 |
Provide a correct Python 3 solution for this coding contest problem.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727 | instruction | 0 | 45,252 | 16 | 90,504 |
"Correct Solution:
```
import math
n = int(input())
ps = [tuple(map(int, input().split())) for _ in range(n)]
mp = {tuple(p): i for i, p in enumerate(ps)}
# ベクトルp0p1に対する点p2の位置（反時計回り, もしくは直線上ならTrue, 時計回りならFalse）を返す。
def ccw(p0, p1, p2):
return (p1[0] - p0[0]) * (p2[1] - p0[1]) - (p1[1] - p0[1]) * (p2[0] - p0[0]) >= 0 # (ベクトルp0p1とベクトルp0p2の外積) >= 0
# 凸包を求める。psは点(x, y)の集合。
def convex_hull(ps):
upper = []
lower = []
ps.sort()
for p in ps:
while len(lower) > 1 and ccw(lower[-2], lower[-1], p):
lower.pop()
lower.append(p)
for p in ps[::-1]:
while len(upper) > 1 and ccw(upper[-2], upper[-1], p):
upper.pop()
upper.append(p)
return lower[:-1] + upper[:-1]
ans = [0] * n
ch = convex_hull(ps)
ch += [ch[0]]
ch += [ch[1]]
for pre, now, next in zip(ch, ch[1:], ch[2:]):
x0 = now[0] - pre[0]
y0 = now[1] - pre[1]
x1 = next[0] - now[0]
y1 = next[1] - now[1]
ans[mp[tuple(now)]] = math.atan2(x1*y0 - x0*y1, x0*x1 + y0*y1) / (2 * math.pi) # atan2(外積, 内積), 外積: |u||v|sinθ, 内積: |u||v|cosθ
for a in ans:
print(a)
``` | output | 1 | 45,252 | 16 | 90,505 |
Provide a correct Python 3 solution for this coding contest problem.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727 | instruction | 0 | 45,253 | 16 | 90,506 |
"Correct Solution:
```
from math import pi, atan2
def andrange(a, b):
if a == (0, 0) or b == (0, 0): return (0, 0)
if a == (0, 1): return b
if b == (0, 1): return a
if a[0] > b[0]: a, b = b, a
if b[1] < a[1]: return b
if b[0] < a[1]: return (b[0], a[1])
if b[1] < a[0] + 1: return (0, 0)
if b[1] < a[1] + 1: return (a[0], b[1] - 1)
return a
def calcrange(P, Q):
c = atan2(P[1]-Q[1], P[0]-Q[0]) / (2*pi)
if c > 0: return (c, c+0.5)
else: return (c+1, c+1.5)
N = int(input())
X = []
for _ in range(N):
x, y = map(int, input().split())
X.append((x, y))
A = [(0, 1) for _ in range(N)]
for i in range(N):
for j in range(N):
if i != j:
A[i] = andrange(A[i], calcrange(X[i], X[j]))
for a in A:
print("{:.15f}".format(a[1] - a[0]))
``` | output | 1 | 45,253 | 16 | 90,507 |
Provide a correct Python 3 solution for this coding contest problem.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727 | instruction | 0 | 45,254 | 16 | 90,508 |
"Correct Solution:
```
import math
from collections import defaultdict
n = int(input())
ipt = [list(map(int,input().split())) for i in range(n)]
xy = sorted(ipt)
convh1 = [xy[0],xy[1]]
def slope(xy1,xy2,const = 1):
x1,y1 = xy1
x2,y2 = xy2
dx,dy = x2-x1,y2-y1
if dx == 0:
return 10**18*(y2-y1)*const//abs(y2-y1)
else:
return dy/dx
def convex(xy1,xy2,xy3,const = 1):
x1,y1 = xy1
x2,y2 = xy2
x3,y3 = xy3
if slope([x1,y1],[x2,y2],const) < slope([x2,y2],[x3,y3],const):
return True
return False
for i in range(2,n):
x,y = xy[i]
while len(convh1) >= 2 and not convex(convh1[-2],convh1[-1],[x,y]):
convh1.pop()
convh1.append(xy[i])
convh2 = [xy[-1],xy[-2]]
for i in range(n-2)[::-1]:
x,y = xy[i]
while len(convh2) >= 2 and not convex(convh2[-2],convh2[-1],[x,y],-1):
convh2.pop()
convh2.append(xy[i])
convh = convh1+convh2[1:-1]
prob = [0 for i in convh]
dc = defaultdict(int)
for i,xy in enumerate(convh):
x,y = xy
px,py = convh[i-1]
nx,ny = convh[(i+1)%len(convh)]
px -= x
py -= y
nx -= x
ny -= y
rad = math.atan2(ny,nx)-math.atan2(py,px)
dc[(x,y)] = (math.pi-min(abs(rad),2*math.pi-abs(rad)))/(2*math.pi)
for x,y in ipt:
print(dc[(x,y)])
``` | output | 1 | 45,254 | 16 | 90,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727
Submitted Solution:
```
import sys
stdin = sys.stdin
def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())
from math import acos
def ccw(p1,p2,p3):
return (p2[0]-p1[0])*(p3[1]-p1[1]) - (p2[1]-p1[1])*(p3[0]-p1[0])
def ConvexHull(points):
u = []
l = []
points.sort()
for p in points:
while len(u)>1 and ccw(u[-2], u[-1], p)>=0:
u.pop()
while len(l)>1 and ccw(l[-2], l[-1], p)<=0:
l.pop()
u.append(p)
l.append(p)
return u + l[-2:0:-1]
def angle(v1, v2):
inner = v1[0]*v2[0] + v1[1]*v2[1]
normv1 = (v1[0]**2 + v1[1]**2) ** 0.5
normv2 = (v2[0]**2 + v2[1]**2) ** 0.5
return acos(inner / (normv1*normv2))
n = ni()
if n == 1:
print(1)
elif n == 2:
print(0.5)
print(0.5)
else:
points = [list(li()) + [i] for i in range(n)]
ch = ConvexHull(points)
bsc = []
for i in range(len(ch)):
x1,y1,i1 = ch[i]
x2,y2,i2 = ch[i-1]
bsc.append((y1-y2, x2-x1))
if len(ch) == 2:
angles = [1,1]
else:
angles = [angle(bsc[i], bsc[i-1]) for i in range(len(ch))]
sm = sum(angles)
ans = [0]*n
for i in range(len(ch)):
x,y,ind = ch[i-1]
ans[ind] = angles[i] / sm
for x in ans:
print(x)
``` | instruction | 0 | 45,255 | 16 | 90,510 |
Yes | output | 1 | 45,255 | 16 | 90,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727
Submitted Solution:
```
from math import acos
pi = 3.14159265358979323846264338
N = int(input())
A = [[0, 0, 0, _] for _ in range(N)]
flg = [False for _ in range(N)]
for i in range(N):
A[i][1], A[i][2] = map(int, input().split())
A[i][0] = A[i][1] * A[i][1] + A[i][2] * A[i][2]
X = [[A[i][1], A[i][2]] for i in range(N)]
A.sort(reverse=True)
stx, sty = A[0][1], A[0][2]
px, py = stx, sty
flg[A[0][3]] = True
prev_point = A[0][3]
convex_hull = [A[0][3]]
while True:
min_prod2 = [[1, 1], [N+1, N+1], 1]
for i in range(N):
for j in range(N):
if i != prev_point and j != prev_point:
ffflg = True
dx1 = X[i][0] - px
dx2 = X[j][0] - px
dy1 = X[i][1] - py
dy2 = X[j][1] - py
inner_prod = dx1 * dx2 + dy1 * dy2
norm1 = dx1 * dx1 + dy1 * dy1
norm2 = dx2 * dx2 + dy2 * dy2
sign = (inner_prod > 0) - (inner_prod < 0)
if sign < min_prod2[2]:
min_prod2[0] = [inner_prod*inner_prod, norm1*norm2]
min_prod2[1] = [i, j]
min_prod2[2] = sign
elif sign == min_prod2[2]:
if sign == 1:
if min_prod2[0][0] * norm1 * norm2 > inner_prod * inner_prod * min_prod2[0][1]:
min_prod2[0] = [inner_prod*inner_prod, norm1*norm2]
min_prod2[1] = [i, j]
elif min_prod2[0][0] * norm1 * norm2 == inner_prod * inner_prod * min_prod2[0][1]:
if min_prod2[0][1] > norm1 * norm2 or min_prod2[1] == [N+1, N+1]:
min_prod2[0] = [inner_prod*inner_prod, norm1*norm2]
min_prod2[1] = [i, j]
elif sign == 0:
if min_prod2[0][1] > norm1 * norm2 or min_prod2[1] == [N+1, N+1]:
min_prod2[0] = [inner_prod*inner_prod, norm1*norm2]
min_prod2[1] = [i, j]
else:
if min_prod2[0][0] * norm1 * norm2 < inner_prod * inner_prod * min_prod2[0][1]:
min_prod2[0] = [inner_prod*inner_prod, norm1*norm2]
min_prod2[1] = [i, j]
elif min_prod2[0][0] * norm1 * norm2 == inner_prod * inner_prod * min_prod2[0][1]:
if min_prod2[0][1] > norm1 * norm2 or min_prod2[1] == [N+1, N+1]:
min_prod2[0] = [inner_prod*inner_prod, norm1*norm2]
min_prod2[1] = [i, j]
if flg[min_prod2[1][0]] and flg[min_prod2[1][1]]:
break
else:
if flg[min_prod2[1][0]] and not flg[min_prod2[1][1]]:
prev_point = min_prod2[1][1]
elif not flg[min_prod2[1][0]] and flg[min_prod2[1][1]]:
prev_point = min_prod2[1][0]
elif not flg[min_prod2[1][0]] and not flg[min_prod2[1][1]]:
prev_point = min_prod2[1][0]
convex_hull.append(prev_point)
flg[prev_point] = True
px, py = X[prev_point][0], X[prev_point][1]
M = len(convex_hull)
prob = [0 for _ in range(N)]
for i in range(M):
center = convex_hull[i]
ii, jj = convex_hull[(i-1)%M], convex_hull[(i+1)%M]
dx1 = X[ii][0] - X[center][0]
dx2 = X[jj][0] - X[center][0]
dy1 = X[ii][1] - X[center][1]
dy2 = X[jj][1] - X[center][1]
ip = dx1 * dx2 + dy1 * dy2
n1 = dx1 * dx1 + dy1 * dy1
n2 = dx2 * dx2 + dy2 * dy2
n1 = n1**0.5
n2 = n2**0.5
prob[center] = 0.5 - (acos((ip/n1)/n2))/(2*pi)
for i in range(N):
print(prob[i])
``` | instruction | 0 | 45,256 | 16 | 90,512 |
Yes | output | 1 | 45,256 | 16 | 90,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).
Let R=10^{10^{10^{10}}}. Ringo performs the following operation:
* Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
* Pick two real numbers x and y independently according to uniform distribution on [-R,R].
* If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
* 2 \leq N \leq 100
* |x_i|,|y_i| \leq 10^6(1\leq i\leq N)
* All given points are pairwise distinct.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.
Examples
Input
2
0 0
1 1
Output
0.5
0.5
Input
5
0 0
2 8
4 5
2 6
3 10
Output
0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727
Submitted Solution:
```
def main():
n = int(input())
ab = [list(map(int, input().split())) for _ in [0]*n]
def det(x, y, z):
a = [x[0]-y[0], x[1]-y[1]]
b = [z[0]-x[0], z[1]-x[1]]
return a[0]*b[1]-a[1]*b[0]
def convex_hull(ab):
q, k = [], 0
xy = sorted(ab) # ここをいじることで初期位置を変えられる。
for i in range(n):
while k > 1 and det(q[k-1], q[k-2], xy[i]) <= 0:
k -= 1
q.pop()
q.append(xy[i])
k += 1
t = k
for i in range(n-2, -1, -1):
while k > t and det(q[k-1], q[k-2], xy[i]) <= 0:
k -= 1
q.pop()
q.append(xy[i])
k += 1
return q[:k-1]
ans = convex_hull(ab)
l = len(ans)
d = dict()
from math import acos
from math import pi
if l == 2:
for a, b in ans:
d[(a, b)] = 0.5
else:
b = 0
ans += ans
for i in range(l):
a1 = ans[i][0]-ans[i-1][0]
b1 = ans[i][1]-ans[i-1][1]
a2 = ans[i+1][0]-ans[i][0]
b2 = ans[i+1][1]-ans[i][1]
a = acos((a1*a2+b1*b2)/((a1**2+b1**2)*(a2**2+b2**2))**0.5)
d[tuple(ans[i])] = a/2/pi
for a, b in ab:
if (a, b) in d:
print(d[(a, b)])
else:
print(0)
main()
``` | instruction | 0 | 45,257 | 16 | 90,514 |
Yes | output | 1 | 45,257 | 16 | 90,515 |
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