TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
N, M = list(map(int, input().split()))

A = [input() for _ in range(N)]
B = [input() for _ in range(M)]

result = True
for i in range(N - M + 1):
    for j in range(N - M + 1):
        for x in range(M):
            for y in range(M):
                s1 = A[i+x][j+y]
                s2 = B[x][y]
                if s1 != s2:
                    result = False
                    break
        if result is False:
            break

if result is True:
    print("Yes")
else:
    print("No")


N, M = list(map(int, input().split()))

A = [input() for _ in range(N)]
B = [input() for _ in range(M)]

result = True
for i in range(N - M + 1):
    for j in range(N - M + 1):
        for x in range(M):
            for y in range(M):
                s1 = A[i+x][j+y]
                s2 = B[x][y]
                if s1 != s2:
                    result = False
                    break
            if result is False:
                break

if result is True:
    print("Yes")
else:
    print("No")


N, M = list(map(int, input().split()))

A = [input() for _ in range(N)]
B = [input() for _ in range(M)]



for i in range(N - M + 1):
    for j in range(N - M + 1):
        for k in range(M):
            s1 = A[i + k][j : j + M]
            s2 = B[k]
            if s1 != s2:
                break
        else:
            print("Yes")

print("No")


N, M = list(map(int, input().split()))

A = [input() for _ in range(N)]
B = [input() for _ in range(M)]

for i in range(N - M + 1):
    for j in range(N - M + 1):
        result = True
        for x in range(M):
            for y in range(M):
                sA = A[i + x][j + y]
                sB = B[x][y]
                if sA != sB:
                    result = False
                    break
            if result is False:
                break

        if result is True:
            print("Yes")
            exit()

print("No")


Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
n, m = map(int, input().split())

a = [input() for x in range(n)]
b = [input() for x in range(m)]

ball = ''.join(b)

def check():
    for i in range(n):
        if i + m > n:
            continue

        for j in range(n):
            if j + m > n:
                continue

            aall = ""
            for k in range(m):
                aall += a[i+k][j:j+m]

            if aall == ball:
                return True

    return False

ret = "YES" if check() else "NO"
print(ret)


n, m = map(int, input().split())

a = [input() for x in range(n)]
b = [input() for x in range(m)]

ball = ''.join(b)

def check():
    for i in range(n):
        if i + m > n:
            continue

        for j in range(n):
            if j + m > n:
                continue

            aall = ""
            for k in range(m):
                aall += a[i+k][j:j+m]

            if aall == ball:
                return True

    return False

ret = "Yes" if check() else "No"
print(ret)


Verdicts: Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
import functools 

def find_indices(a, b, L):
    
    return [i for i in range(L) if b in a[i:]]

def solve():
    N,M = map(int, input().split())
    A = [input() for _ in range(N)]
    B = [input() for _ in range(M)]
    idx = 0
    start_pos = []
    for S_a in A:
        indices = find_indices(S_a, B[idx], N)
        
        if indices:
            start_pos.append(set(indices))
            idx += 1
        else:
            start_pos = []
            idx = 0
            
        if idx == M:
            break
    
    # print(start_pos)
    print('Yes' if indices and functools.reduce(lambda a,b:a&b, start_pos) else 'No')

def solve():
    N,M = map(int, input().split())
    A = [input() for _ in range(N)]
    B = [input() for _ in range(M)]
    idx = 0
    start_pos = {}
    for S_a in A:
        indices = set(find_indices(S_a, B[idx], N))
        if not start_pos:
            start_pos = indices
        else:
            start_pos &= indices
        
        if start_pos:
            idx += 1
        else:
            start_pos = {}
            idx = 0
            
        if idx == M:
            break
    
    print('Yes' if start_pos else 'No')
    
solve()

def find_indices(a, b, L):
    
    return [i for i in range(L) if a[i:].startswith(b)]

def solve():
    N,M = map(int, input().split())
    A = [input() for _ in range(N)]
    B = [input() for _ in range(M)]
    idx = 0
    start_pos = {}
    for S_a in A:
        indices = set(find_indices(S_a, B[idx], N))
        if not start_pos:
            start_pos = indices
        else:
            start_pos = start_pos&indices
        
        if start_pos:
            idx += 1
        else:
            start_pos = {}
            idx = 0
            
        if idx == M:
            break
    
    print('Yes' if start_pos else 'No')
    
solve()

Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
N, M = map(int, input().split())
A = [input() for _ in range(N)]
B = [input() for _ in range(N)]

for i in range(N-M+1):
  for j in range(N-M+1):
    
    for k in range(M):
      if A[i+k][j:j+M] != B[k]:
        break
    else:
      print('Yes')
      exit()
      
    break

print('No')

N, M = map(int, input().split())
A = [input() for _ in range(N)]
B = [input() for _ in range(M)]

for i in range(N-M+1):
  for j in range(N-M+1):
    
    for k in range(M):
      if A[i+k][j:j+M] != B[k]:
        break
    else:
      print('Yes')
      exit()

print('No')


Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
import sys
n, m = map(int, input().split())
n_list = [list(input()) for i in range(n)]
m_list = [list(input()) for i in range(m)]

def bunkatu(x, y):
    global n_list
    lis = []
    for i in range(y, y + m):
        lis.append(n_list[i][x: x + m])
    return lis

for j in range(n - m + 1):
    for i in range(n - m + 1):
        tmp = bunkatu(i, j)
        if m_list == tmp:
            print('YES')
            sys.exit(0)
print('NO')

import sys
n, m = map(int, input().split())
n_list = [list(input()) for i in range(n)]
m_list = [list(input()) for i in range(m)]

def bunkatu(x, y):
    global n_list
    lis = []
    for i in range(y, y + m):
        lis.append(n_list[i][x: x + m])
    return lis

for j in range(n - m + 1):
    for i in range(n - m + 1):
        tmp = bunkatu(i, j)
        if m_list == tmp:
            print('Yes')
            sys.exit(0)
print('No')

Verdicts: Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
# -*- coding: utf-8 -*-

def solve(N, M, A, B):
    
    include = False
    for h in range(N-M+1):
        for w in range(N-M+1):
            match = True
            for i in range(M):
                for j in range(M):
                    if(A[h+i][w+j] != B[i][j]):
                        match = False
            if(match):
                include = True
    
    print("Yes" if include else "No")
    
    
if __name__ == '__main__':
    N, M = map(int, input().split())
    A, B = [], []
    for i in range(N):
        A.append(input())
    for j in range(M):
        B.append(input())

    solve(N, M, A, B

# -*- coding: utf-8 -*-

def solve(N, M, A, B):
    include = False
    for h in range(N-M+1):
        for w in range(N-M+1):
            match = True
            for i in range(M):
                for j in range(M):
                    if(A[h+i][w+j] != B[i][j]):
                        match = False
            if(match):
                include = True

    print("Yes" if include else "No")

if __name__ == '__main__':
    N, M = map(int, input().split())
    A, B = [], []
    for i in range(N):
        A.append(input())
    for j in range(M):
        B.append(input())

    solve(N, M, A, B)

Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
n,  m = map(int, input().split())
np = [input() for _ in range(n)]
mp = [input() for _ in range(m)]
for i in range(n-m-1):
  for  j in range(n-m-1):
    f = True
    for k in range(m):
      if np[i+k][j:j+m] != mp[i+k]:
        f = False
    if f:
      print("Yes")
      exit(0)
else:
  print("No")

n,  m = map(int, input().split())
np = [input() for _ in range(n)]
mp = [input() for _ in range(m)]
for i in range(n-m+1):
  for j in range(n-m+1):
    f = True
    for k in range(m):
      if np[i+k][j:j+m] != mp[k]:
        f = False
    if f:
      print("Yes")
      exit(0)
else:
  print("No")


Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
n,m=[int(i) for i in input().split()]
a=[input() for i in range(n)]
b=[input() for i in range(m)]
for i in range(m):
    for j in range(len(b[i])):
        if b[i][j]!=a[i][j]:
            result=True
            break
print("No" if result else "Yes")

n,m=[int(i) for i in input().split()]
a=[input() for i in range(n)]
b=[input() for i in range(m)]
result=True
for i in range(m):
    if (b[i] in a[i])==False:
        result=False
        break
print("Yes" if result else "No")

Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
# -*- coding: utf-8 -*-
import sys
import math
import numpy as np


def main():
    
    # M = int(input())
    N, M = map(int, input().split())
    sA = ''.join([input() for i in range(N)])
    sB = ''.join([input() for i in range(M)])
    A = np.array([s for s in sA]).reshape(N,N)
    B = np.array([s for s in sB]).reshape(M,M)

    
    result = np.zeros((N-M,N-M), dtype=bool)
    for i in range(N-M+1):
        for j in range(N-M+1):
            result[i,j] = (A[i:i+M, j:j+M] == B).all()

    
    if result.any():
        print('Yes')
    else:
        print('No')


if __name__ == '__main__':
    main()


# -*- coding: utf-8 -*-
import sys
import math
import numpy as np


def main():
    
    # M = int(input())
    N, M = map(int, input().split())
    sA = ''.join([input() for i in range(N)])
    sB = ''.join([input() for i in range(M)])
    A = np.array([s for s in sA]).reshape(N,N)
    B = np.array([s for s in sB]).reshape(M,M)

    
    result = np.zeros((N-M+1,N-M+1), dtype=bool)
    for i in range(N-M+1):
        for j in range(N-M+1):
            result[i,j] = (A[i:i+M, j:j+M] == B).all()

    
    if result.any():
        print('Yes')
    else:
        print('No')


if __name__ == '__main__':
    main()


Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
#n = int(input())
#n,k = map(int,input().split())
#x = list(map(int,input().split()))


N,M = map(int,input().split())
A = [list(input()) for _ in range(N)]
B = [list(input()) for _ in range(M)]

flg1 = 1
flg2 = 0
for i in range(N-M):
    if flg2:
        break
    for j in range(N-M):
        if flg2:
            break
        flg1 = 1
        
        for k in range(M):
            if not(flg1):
                break
            for l in range(M):
                if A[i+k][j+l]==B[k][l]:
                    continue
                else:
                    flg1 = 0
        else:
            if flg1:
                print("Yes")
                flg2 = 1
else:
    print("No")

#n = int(input())
#n,k = map(int,input().split())
#x = list(map(int,input().split()))


N,M = map(int,input().split())
A = [list(input()) for _ in range(N)]
B = [list(input()) for _ in range(M)]

flg1 = 1
flg2 = 0
for i in range(N-M+1):
    if flg2:
        break
    for j in range(N-M+1):
        if flg2:
            break
        flg1 = 1
        
        for k in range(M):
            if not(flg1):
                break
            for l in range(M):
                if A[i+k][j+l]==B[k][l]:
                    continue
                else:
                    flg1 = 0
        else:
            if flg1:
                print("Yes")
                flg2 = 1
else:
    if not(flg2):
        print("No")

Verdicts: Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time

sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7

def LI(): return list(map(int, input().split()))
def II(): return int(input())
def LS(): return input().split()
def S(): return input()


def main():
    n,m = LI()
    a = []
    for i in range(n):
        t = 0
        for c in S():
            t <<= 1
            if c == '#':
                t += 1
        a.append(t)

    b = []
    for i in range(m):
        t = 0
        for c in S():
            t <<= 1
            if c == '#':
                t += 1
        b.append(t)

    mask = 2**m-1
    for i in range(n-m+1):
        for j in range(n-m+1):
            f = True
            for k in range(m):
                ak = (a[k+j] >> i) & mask
                if ak != b[k]:
                    f = False
                    break
            if f:
                return 'YES'
    return 'NO'


print(main())


import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time

sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7

def LI(): return list(map(int, input().split()))
def II(): return int(input())
def LS(): return input().split()
def S(): return input()


def main():
    n,m = LI()
    a = []
    for i in range(n):
        t = 0
        for c in S():
            t <<= 1
            if c == '#':
                t += 1
        a.append(t)

    b = []
    for i in range(m):
        t = 0
        for c in S():
            t <<= 1
            if c == '#':
                t += 1
        b.append(t)

    mask = 2**m-1
    for i in range(n-m+1):
        for j in range(n-m+1):
            f = True
            for k in range(m):
                ak = (a[k+j] >> i) & mask
                if ak != b[k]:
                    f = False
                    break
            if f:
                return 'Yes'
    return 'No'


print(main())


Verdicts: Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
N, M = map(int, input().split())
A = [input() for i in range(N)]
B = [input() for i in range(M)]
i = 0
while(i <= N - M):
    x = A[i].find(B[0])
    while(N - M >= x >= 0):
        for j in range(1,M):
            if A[i+j][x-1:x-1+M] != B[j]:
                x = A[i].find(B[0], x+1)
                break
        else:
            print("Yes")
            exit()
    i += 1
print("No")


N, M = map(int, input().split())
A = [input() for i in range(N)]
B = [input() for i in range(M)]
i = 0
while(i <= N - M):
    x = A[i].find(B[0])
    while(N - M >= x and x >= 0):
        for j in range(1,M):
            if A[i+j][x:x+M] != B[j]:
                x = A[i].find(B[0], x+1)
                break
        else:
            print("Yes")
            exit()
    i += 1
print("No")


Verdicts: Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
def solve(n, m, NUM):
    for i in range(len(m)):
        if n[NUM + i] is not m[i]:
            return -1
    return 1


N, M = [int(i) for i in input().split()]
n_pic = [input() for i in range(N)]
m_pic = [input() for i in range(M)]
num = 0
if M is 1:
    for n in n_pic:
        if -1 is not n.find(m_pic[0]):
            print('Yes')
            exit()
    print('No')
    exit()

for n in range(len(n_pic)):
    for n_len in range(N - M + 1):
        while -1 is not solve(n_pic[n+num], m_pic[num], n_len):
            num += 1
            print(num)
            if num is M:
                print('Yes')
                exit()
        num = 0
print('No')


def solve(n, m, NUM):
    for i in range(len(m)):
        if n[NUM + i] is not m[i]:
            return -1
    return 1


N, M = [int(i) for i in input().split()]
n_pic = [input() for i in range(N)]
m_pic = [input() for i in range(M)]
num = 0
if M is 1:
    for n in n_pic:
        if -1 is not n.find(m_pic[0]):
            print('Yes')
            exit()
    print('No')
    exit()

for n in range(N - M + 1):
    for n_len in range(N - M + 1):
        while -1 is not solve(n_pic[n + num], m_pic[num], n_len):
            num += 1
            if num is M:
                print('Yes')
                exit()
        num = 0
print('No')


Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
import numpy as np
import sys
N = int(input())
M = int(input())
A = []
B = []
def change(n):
  if(n=='#'):
    return True;
  return False;

for i in range(N):
  temp = input()
  A.append(list(map(change, temp)))
for i in range(M):
  temp = input()
  B.append(list(map(change, temp)))
A = np.array(A)
B = np.array(B)

for i in range(N-M+1):
  for j in range(N-M+1):
    if (B== A[i:M+j, i:M+j]).all():
      print("Yes")
      sys.exit()

print("No")


import numpy as np
import sys
temp = input().split()
N = int(temp[0])
M = int(temp[1])
A = []
B = []
check = False
def change(n):
  if(n=='#'):
    return True;
  return False;

for i in range(N):
  temp = input()
  A.append(list(map(change, temp)))
for i in range(M):
  temp = input()
  B.append(list(map(change, temp)))
A = np.array(A)
B = np.array(B)

for i in range(N-M+1):
  for j in range(N-M+1):
    if (B== A[i:M+i, j:M+j]).all():
      check = True
      break
  if check:
    break

if check:
  print("Yes")
else:
  print("No")


Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
N,M=map(int,input().split())
A=[input() for _ in range(N)]
B=[input() for _ in range(M)]
ans="No"
for i in range(N-M+1):
  for j in range(len(A[0])-len(B[0])+1):
    t = 0
    for k in range(len(B[0])):
      for l in range(M):
        if A[k+q][l+p] == B[q][p]:
          t += 1
    if t==len(B[0])*M:
      ans="Yes"
      exit()
print(ans)

N,M=map(int,input().split())
A=[input() for _ in range(N)]
B=[input() for _ in range(M)]
ans="No"
for i in range(N-M+1):
  for j in range(len(A[0])-len(B[0])+1):
    t = 0
    for k in range(len(B[0])):
      for l in range(M):
        if A[k+q][l+p] == B[q][p]:
          t += 1
    if t==len(B[0])*M:
      ans="Yes"
      break
print(ans)

N,M=map(int,input().split())
A=[input() for _ in range(N)]
B=[input() for _ in range(M)]
ans="No"
for i in range(N-M+1):
  for j in range(len(A[0])-len(B[0])+1):
    t = 0
    for k in range(len(B[0])):
      for l in range(M):
        if A[i+l][j+k] == B[l][k]:
          t += 1
    if t==len(B[0])*M:
      ans="Yes"
      exit()
print(ans)


N,M=map(int,input().split())
A=[input() for _ in range(N)]
B=[input() for _ in range(M)]
ans="No"
for i in range(N-M+1):
  for j in range(N-M+1):
    if [a[j:j+M] for a in A[i:i+M]]==B:    
      ans="Yes"
print(ans)

Verdicts: Runtime Error

Runtime Error

Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
t,*a=open(0);n,m=map(int,t.split());r=range(n-m+1)
print('YNeos'[all(a[n:]!=[t[j:j+m]for t in a[i:i+m]]for i in r for j in r)::2])

x,*c=open(0);n,m=map(int,x.split());o=n-m+1;print("YNeos"[all(any(c[p//o+q//m][p%o+q%m]!=c[n+q//m][q%m]for q in range(m*m))for p in range(o*o))::2])

Verdicts: Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
n,m = map(int,input().split())
A = [list(input()) for _ in range(n)]
B = [list(input()) for _ in range(m)]
ans = False
for i in range(n-m+1):
    for j in range(n-m+1):
        flg = True
        for e, k in enumerate(range(m)):
            if A[i+e][j:j+m+1] != B[k]:
                flg = False
        if flg:
            ans = True
            break
print('YES' if ans else 'NO')

n,m = map(int,input().split())
A = [list(input()) for _ in range(n)]
B = [list(input()) for _ in range(m)]
ans = False
for i in range(n-m+1):
    for j in range(n-m+1):
        flg = True
        for e, k in enumerate(range(m)):
            if A[i+e][j:j+m] != B[k]:
                flg = False
        if flg:
            ans = True
            break
print('Yes' if ans else 'No')


Verdicts: Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
N, M = map(int, input().split())

A = [ input() for _ in range(N) ]
B = [ input() for _ in range(M) ]

for i in range(N-M+1):
  for _ in range(A[i].count(B[0])):
    l = A[i].index(B[0], l)
    result = True
    for k in range(1, M):
      if B[k] != A[i+k][l:l+len(B[k])]:
        result = False
        break
    if result:
      print("Yes")
      exit()
    l += 1
      
print("No")

N, M = map(int, input().split())

A = [ input() for _ in range(N) ]
B = [ input() for _ in range(M) ]

for i in range(N-M):
  for j in range(A[i].count(B[0])-1):
    l = A[i].index(B[0], j)
    result = True
    for k in range(1, M):
      if B[k] != A[i+k][l:l+len(B[k])]:
        result = False
        break
    if result:
      print("Yes")
      exit()
      
print("No")

N, M = map(int, input().split())

A = [ input() for _ in range(N) ]
B = [ input() for _ in range(M) ]

for i in range(N-M+1):
  l = 0
  for _ in range(A[i].count(B[0])):
    l = A[i].index(B[0], l)
    result = True
    for k in range(1, M):
      if B[k] != A[i+k][l:l+len(B[k])]:
        result = False
        break
    if result:
      print("Yes")
      exit()
    l += 1
      
print("No")

Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
import sys

N, M = map(int, input().split())
A = [input() for _ in range(N)]
B = [input() for _ in range(M)]

for i in range(N - M + 1):
    if i >= M:
        break

    matched = True
    for j in range(N - M + 1):
        if A[i][j:j+M] == B[i][j:j+M]:
            continue
        else:
            matched = False
            break
    if matched:
        print('Yes')
        sys.exit()

print('No')

import sys

N, M = map(int, input().split())
A = [input() for _ in range(N)]
B = [input() for _ in range(M)]

for i in range(N - M + 1):
    for j in range(N - M + 1):
        matched = True

        for k in range(M):
            if A[i+k][j:j+M] == B[k]:
                continue
            else:
                matched = False
                break
    
        if matched:
            print('Yes')
            sys.exit()

print('No')

Verdicts: Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
N, M = map(int, input().split())
a = []
b = []
for i in range(N):
    a.append(input())
c = 0
for i in range(M):
    b.append(input())
for i in range(N-M+1):
    for j in range(N-M+1):
        for k in range(M):
            s1 = a[i][j:j+M] 
            s2 = b[k]
            if s1 != s2:
                break
        else:
            print("Yes")
print("No")

N, M = map(int, input().split())
list = []
for i in range(H):
    list.append(input())
count = 0
for i in range(M):
    x = input()
    for j in list:
    if x in j:
        count+=1
print("No" if count=0 else "Yes")

N, M = map(int, input().split())
a = []
b = []
for i in range(N):
    a.append(input())
c = 0
for i in range(M):
    b.append(input())
for i in range(N-M+1):
    for j in range(N-M+1):
        for k in range(M):
            if a[i+k][j:j+M] == b[k]:
                c += 1
        if c == M:
            print("Yes")
            break
print("No")

N, M = map(int, input().split())
a = []
b = []
for i in range(N):
    a.append(input())
c = 0
ans = "No"
for i in range(M):
    b.append(input())
for i in range(N-M+1):
    for j in range(N-M+1):
        for k in range(M):
            if a[i+k][j:j+M] == b[k]:
                c += 1
        if c == M:
            ans = "Yes"
            break
        c = 0
print(ans)

Verdicts: Wrong Answer

Runtime Error

Wrong Answer

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
import numpy as np

N,M = map(int, input().split())

A=[]
for i in range(N):
    A.append(list(input()))

B=[]
for j in range(M):
    B.append(list(input()))

A=np.asarray(A)
B=np.asarray(B)
flg=False
if (A == B).all():
    flg=True
  
for x in range(N-M):
    for y in range(N-M):
        if (A[x:x+M,y:y+M] == B).all():
            flg=True
if flg:
    print("Yes")
else:
    print("No")

import numpy as np

N,M = map(int, input().split())

A=[]
for i in range(N):
    A.append(list(input()))

B=[]
for j in range(M):
    B.append(list(input()))

A=np.asarray(A)
B=np.asarray(B)
flg=False

  
for x in range(N-M):
    for y in range(N-M):
        if (A[x:x+M,y:y+M] == B).all():
            flg=True
if flg:
    print("Yes")
else:
    try:
        if (A == B).all():
            print("Yes")
        else:
            print("No")
    except:
        print("No")

Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
def f():
  n,m=map(int,input().split())
  o=[input() for _ in range(n)]
  p=[input() for _ in range(m)]
  for i in range(n-m+1):
    for j in range(n-m+1):
      for k in range(i,n-m+1):
        for l in range(j,n-m+1):
          if p[k-i][l-j]!=o[k][l]:
            return False
  return True

if f():
  print("Yes")
else:
  print("No")

def func():
  n,m=map(int,input().split())
  o=[input() for _ in range(n)]
  p=[input() for _ in range(m)]
  q=[]
  for i in range(n-m+1):
    for j in range(n-m+1):
      fl=True
      for k in range(m):
        for l in range(m):
          if p[k][l]!=o[i+k][j+l]:
            fl=False
      q.extend([fl])
  return q

a=func()
if sum(a):
  print("Yes")
else:
  print("No")

Verdicts: Runtime Error

Accepted
Problem: You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.   A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.   Also, the given images are binary images, and the color of each pixel is either white or black.  In the input, every pixel is represented by a character: `.` corresponds to a white pixel, and `#` corresponds to a black pixel.   The image A is given as N strings A_1,...,A_N.   The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).   Similarly, the template image B is given as M strings B_1,...,B_M.   The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).  Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Submission attempts (in order):
n = int(input())

a = list(map(int, input().split()))
check = False
for i in range(n):
    if a[i] % 2 != 0:
        check = True
        break
count = 0
while check == False:
    for i in range(n):
        a[i] /= 2
        if a[i] % 2 != 0:
            check = True
            break
    count += 1
print(count)


n, m = list(map(int, input().split()))
#a = list(map(int, input().split()))
a = []
b = []
for i in range(n):
    a.append(input())
for i in range(m):
    b.append(input())

for i in range(n-m+1):
    for j in range(len(a[0])-len(b[0])+1):
        count = 0
        if b[0] == a[i][j:j+len(b[0])]:
            # print("aaaa")
            for k in range(m):
                if b[k] == a[i+k][j:j+len(b[0])]:
                    count += 1
            if count == m:
                print("Yes")
                exit()
# print(count)
print("No")


Verdicts: Runtime Error

Accepted