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Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
input = sys.stdin.readline
I = lambda: list(map(int, input().split()))
n, m = I()
a = [I() for i in range(n)]
b = [I() for i in range(n)]
x = [[] for i in range(n + m - 1)]
y = [[] for i in range(n + m - 1)]
for i in range(n):
for j in range(m):
x[i + j].append(a[i][j])
y[i + j].append(b[i][j])
for i in range(n + m - 1):
if sorted(x[i]) != sorted(y[i]):
print("NO")
sys.exit()
print("YES")
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
ma = [[] for i in range(n + m)]
mb = [[] for i in range(n + m)]
w = []
for i in range(n):
w = list(map(int, input().split()))
for j in range(m):
ma[i + j].append(w[j])
w = []
for i in range(n):
w = list(map(int, input().split()))
for j in range(m):
mb[i + j].append(w[j])
for i in range(n + m):
if sorted(ma[i]) != sorted(mb[i]):
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
mat1 = []
mat2 = []
for i in range(n):
mat1.append(list(map(int, input().split())))
for i in range(n):
mat2.append(list(map(int, input().split())))
i = 0
temp1 = []
temp2 = []
while i < m:
k = i
j = 0
while k >= 0 and j < n:
temp1.append(mat1[j][k])
temp2.append(mat2[j][k])
k -= 1
j += 1
temp1.sort()
temp2.sort()
if temp1 == temp2:
temp1.clear()
temp2.clear()
else:
print("NO")
exit()
i += 1
i = 1
while i < n:
k = m - 1
j = i
while j < n and k >= 0:
temp1.append(mat1[j][k])
temp2.append(mat2[j][k])
k -= 1
j += 1
temp1.sort()
temp2.sort()
if temp1 == temp2:
temp1.clear()
temp2.clear()
else:
print("NO")
exit()
i += 1
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
def check(a, b, x, y):
i = 0
j = 0
while x + i < n and y - j >= 0:
if a[x + i][y - j] == b[x][y]:
a[x][y], a[x + i][y - j] = a[x + i][y - j], a[x][y]
return True
i += 1
j += 1
return False
a = []
for i in range(n):
a.append(list(map(int, input().split())))
b = []
for i in range(n):
b.append(list(map(int, input().split())))
for i in range(n):
for j in range(m):
if check(a, b, i, j) == False:
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER RETURN NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
num = [int(x) for x in input().split(" ")]
n, m = num[0], num[1]
table1 = list()
table2 = list()
for i in range(n):
num = [int(x) for x in input().split(" ")]
table1.append(num)
test = table1
max_col = len(test)
max_row = len(test[0])
cols = [[] for i in range(max_col)]
rows = [[] for i in range(max_row)]
fdiag = [[] for i in range(max_col + max_row - 1)]
for y in range(max_col):
for x in range(max_row):
cols[y].append(test[y][x])
rows[x].append(test[y][x])
fdiag[x + y].append(test[y][x])
dig1 = fdiag
for i in range(n):
num = [int(x) for x in input().split(" ")]
table2.append(num)
test = table2
max_col = len(test)
max_row = len(test[0])
cols = [[] for i in range(max_col)]
rows = [[] for i in range(max_row)]
fdiag = [[] for i in range(max_col + max_row - 1)]
for y in range(max_col):
for x in range(max_row):
cols[y].append(test[y][x])
rows[x].append(test[y][x])
fdiag[x + y].append(test[y][x])
dig2 = fdiag
f = 0
for i in range(m + n - 1):
if sorted(dig1[i]) != sorted(dig2[i]):
f = 1
break
if f == 1:
print("NO")
else:
print("YES")
|
ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
N, M = map(int, input().split())
A = [list(map(int, input().split())) for i in range(N)]
B = [list(map(int, input().split())) for i in range(N)]
ans = "YES"
for i in range(N):
t1 = sorted([A[i - j][j] for j in range(min(N, M)) if i - j >= 0])
t2 = sorted([B[i - j][j] for j in range(min(N, M)) if i - j >= 0])
if t1 != t2:
ans = "NO"
break
for j in range(M):
t1 = sorted([A[N - i - 1][j + i] for i in range(min(N, M)) if j + i < M])
t2 = sorted([B[N - i - 1][j + i] for i in range(min(N, M)) if j + i < M])
if t1 != t2:
ans = "NO"
break
print(ans)
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
r, c = list(map(int, input().split()))
f = [None] * (r + c - 1)
l = [None] * (r + c - 1)
for i in range(0, r + c - 1):
f[i] = []
l[i] = []
for i in range(0, r):
ip = list(map(int, input().split()))
for j in range(0, c):
f[i + j].append(ip[j])
for i in range(0, r):
ip = list(map(int, input().split()))
for j in range(0, c):
l[i + j].append(ip[j])
for i in range(0, r + c - 1):
f[i].sort()
l[i].sort()
if f[i] == l[i]:
continue
else:
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NONE BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR LIST ASSIGN VAR VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = list(map(int, input().split()))
A = []
B = []
for i in range(n):
A.append(list(map(int, input().split())))
for i in range(n):
B.append(list(map(int, input().split())))
dA = []
dB = []
for d in range(m):
for i in range(d, -1, -1):
if d - i >= n:
break
dA.append(A[d - i][i])
dB.append(B[d - i][i])
if sorted(dA) != sorted(dB):
print("NO")
exit()
dA, dB = [], []
for d in range(1, n):
for i in range(d, n):
if i - d >= m:
break
dA.append(A[i][m - (i - d) - 1])
dB.append(B[i][m - (i - d) - 1])
if sorted(dA) != sorted(dB):
print("NO")
exit()
dA, dB = [], []
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR LIST LIST EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
m1 = []
m2 = []
for i in range(n):
m1.append([int(j) for j in input().split()])
for i in range(n):
m2.append([int(j) for j in input().split()])
d1 = [[] for i in range(n + m - 1)]
d2 = [[] for i in range(n + m - 1)]
for i in range(n):
for j in range(m):
d1[i + j].append(m1[i][j])
d2[i + j].append(m2[i][j])
flag = True
for i in range(n + m - 1):
d1[i].sort()
d2[i].sort()
if d1[i][:] != d2[i][:]:
flag = False
if flag:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
ar = [0] * 2
diag = [0] * 2
for i in range(2):
ar[i] = [[int(t) for t in input().split()] for j in range(n)]
diag[i] = [{} for j in range(n + m - 1)]
for num in range(2):
for i in range(n):
for j in range(m):
cur = ar[num][i][j]
it = diag[num][j + i]
if it.get(cur) == None:
it[cur] = 0
it[cur] += 1
for i in range(n + m - 1):
if diag[0][i] != diag[1][i]:
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR DICT VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR NONE ASSIGN VAR VAR NUMBER VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = list(map(int, input().split()))
a = []
for i in range(n):
a.append(list(map(int, input().split())))
b = []
for i in range(n):
b.append(list(map(int, input().split())))
ok = "YES"
for i in range(m - 1 + n - 1 + 1):
d = dict()
for x in range(min(i + 1, m)):
if i - x < n:
if a[i - x][x] in d:
d[a[i - x][x]] += 1
else:
d[a[i - x][x]] = 1
for x in range(min(i + 1, m)):
if i - x < n:
if b[i - x][x] in d:
d[b[i - x][x]] -= 1
if d[b[i - x][x]] == 0:
d.__delitem__(b[i - x][x])
else:
print("NO")
return
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR VAR VAR IF VAR BIN_OP VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR VAR VAR IF VAR BIN_OP VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(n)]
B = [list(map(int, input().split())) for _ in range(n)]
ans = True
for dig in range(n):
i = dig
j = 0
d1 = dict()
d2 = dict()
while i >= 0 and j < m:
d1[A[i][j]] = d1.get(A[i][j], 0) + 1
d2[B[i][j]] = d2.get(B[i][j], 0) + 1
i -= 1
j += 1
if d1 != d2:
ans = False
break
for dig in range(1, m):
i = n - 1
j = dig
d1 = dict()
d2 = dict()
while i >= 0 and j < m:
d1[A[i][j]] = d1.get(A[i][j], 0) + 1
d2[B[i][j]] = d2.get(B[i][j], 0) + 1
i -= 1
j += 1
if d1 != d2:
ans = False
break
print("YES" if ans else "NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR WHILE VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR WHILE VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR STRING STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = list(map(int, input().split()))
a, b = [], []
for _ in range(n):
aux = [int(x) for x in input().split()]
a.append(aux)
for _ in range(n):
aux = [int(x) for x in input().split()]
b.append(aux)
def diagonalOrder(a, b):
for line in range(1, n + m):
start_col = max(0, line - n)
count = min(line, m - start_col, n)
aux_a = []
aux_b = []
for j in range(count):
aux_a.append(a[min(n, line) - j - 1][start_col + j])
aux_b.append(b[min(n, line) - j - 1][start_col + j])
if sorted(aux_a) != sorted(aux_b):
return "NO"
return "YES"
print(diagonalOrder(a, b))
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FUNC_DEF FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN STRING RETURN STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
data_1 = []
data_2 = []
fl = True
diagonal_1 = [[] for i in range(n + m - 1)]
diagonal_2 = [[] for i in range(n + m - 1)]
for i in range(n):
data_3 = list(map(int, input().split()))
data_1.append(data_3)
for i in range(n):
data_3 = list(map(int, input().split()))
data_2.append(data_3)
if data_1[0][0] != data_2[0][0]:
print("No")
elif data_1[0][0] != data_2[0][0]:
print("NO")
else:
for i in range(len(data_1)):
for j in range(len(data_1[i])):
diagonal_1[i + j - 1].append(data_1[i][j])
for i in range(len(data_1)):
for j in range(len(data_1[i])):
diagonal_2[i + j - 1].append(data_2[i][j])
for i in range(len(diagonal_1)):
diagonal_1[i].sort()
for i in range(len(diagonal_1)):
diagonal_2[i].sort()
if diagonal_1 == diagonal_2:
print("Yes")
else:
print("No")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
from sys import stdin, stdout
n, m = map(int, stdin.readline().split())
mata = [None] * n
matb = [None] * n
for i in range(n):
mata[i] = list(map(int, stdin.readline().split()))
for i in range(n):
matb[i] = list(map(int, stdin.readline().split()))
la = [None] * (n + m - 1)
lb = [None] * (n + m - 1)
for i in range(n + m - 1):
la[i] = []
lb[i] = []
for i in range(n):
for j in range(m):
la[i + j].append(mata[i][j])
lb[i + j].append(matb[i][j])
flag = True
for i in range(n + m - 1):
la[i].sort()
lb[i].sort()
if la[i] != lb[i]:
flag = False
if flag:
stdout.write("YES")
else:
stdout.write("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE VAR ASSIGN VAR BIN_OP LIST NONE VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NONE BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NONE BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR LIST ASSIGN VAR VAR LIST FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = [[int(x) for x in input().split()] for _ in range(n)]
b = [[int(x) for x in input().split()] for _ in range(n)]
transa = [[] for _ in range(n + m)]
transb = [[] for _ in range(n + m)]
for i in range(n):
for j in range(m):
transa[i + j].append(a[i][j])
transb[i + j].append(b[i][j])
for i in range(n + m):
if sorted(transa[i]) != sorted(transb[i]):
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
sizeI, sizeJ = map(int, input().split())
have = [list(map(int, input().split())) for _ in range(sizeI)]
need = [list(map(int, input().split())) for _ in range(sizeI)]
def solve():
for k in range(sizeI + sizeJ - 1):
haveDict, needDict = {}, {}
for j in range(k + 1):
i = k - j
if i < sizeI and j < sizeJ:
haveVal, needVal = have[i][j], need[i][j]
haveDict.setdefault(haveVal, 0)
haveDict[haveVal] += 1
needDict.setdefault(needVal, 0)
needDict[needVal] += 1
if haveDict != needDict:
return "NO"
return "YES"
print(solve())
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR DICT DICT FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER IF VAR VAR RETURN STRING RETURN STRING EXPR FUNC_CALL VAR FUNC_CALL VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = []
b = []
for i in range(n):
c = list(map(int, input().split()))
a.append(c)
for i in range(n):
c = list(map(int, input().split()))
b.append(c)
for i in range(n):
for j in range(m):
if a[i][j] == b[i][j]:
continue
else:
k = i + 1
l = j - 1
flag = 0
while k < n and 0 <= l:
if a[k][l] == b[i][j]:
flag = 1
a[i][j], a[k][l] = a[k][l], a[i][j]
break
l -= 1
k += 1
if flag == 0:
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR NUMBER VAR IF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = [[] for _ in range(n + m)]
b = [[] for _ in range(n + m)]
for i in range(n):
for j, x in enumerate(input().split()):
a[i + j].append(x)
for i in range(n):
for j, x in enumerate(input().split()):
b[i + j].append(x)
for i in range(n + m):
a[i].sort()
b[i].sort()
if a[i] != b[i]:
print("NO")
quit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
def main():
n, m = (int(x) for x in input().split(" "))
a = []
b = []
for _ in range(n):
a.append([int(x) for x in input().split(" ")])
for _ in range(n):
b.append([int(x) for x in input().split(" ")])
for s in range(n + m - 1):
ma = {}
for j in range(max(s - n + 1, 0), min(m - 1, s) + 1):
if a[s - j][j] in ma:
ma[a[s - j][j]] += 1
else:
ma[a[s - j][j]] = 1
mb = {}
for j in range(max(s - n + 1, 0), min(m - 1, s) + 1):
if b[s - j][j] in mb:
mb[b[s - j][j]] += 1
else:
mb[b[s - j][j]] = 1
if len(ma) != len(mb):
print("NO")
return
for k, v in ma.items():
if k not in mb or mb[k] != v:
print("NO")
return
print("YES")
main()
|
FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING RETURN FOR VAR VAR FUNC_CALL VAR IF VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
MOD = 10**9 + 7
I = lambda: list(map(int, input().split()))
n, m = I()
a = [[] for i in range(n + m - 1)]
b = [[] for i in range(n + m - 1)]
for i in range(n):
l = I()
for j in range(m):
a[i + j].append(l[j])
for i in range(n):
l = I()
for j in range(m):
b[i + j].append(l[j])
ans = "YES"
for i in range(n + m - 1):
if sorted(a[i]) != sorted(b[i]):
ans = "NO"
break
print(ans)
|
ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = 0, 0
a, b = [], []
def get_diags(a):
diags = []
for c in range(len(a[0])):
x, y = 0, c
diags.append([])
while x < n and y >= 0:
diags[-1].append(a[x][y])
x += 1
y -= 1
for r in range(len(a)):
x, y = r, m - 1
diags.append([])
while x < n and y >= 0:
diags[-1].append(a[x][y])
x += 1
y -= 1
return diags
def solve(a, b):
if n == 1 or m == 1:
return a == b
a_diags = get_diags(a)
b_diags = get_diags(b)
a_diags = [sorted(r) for r in a_diags]
b_diags = [sorted(r) for r in b_diags]
for i in range(len(a_diags)):
if a_diags[i] != b_diags[i]:
return False
return True
def _input():
global n, m
n, m = map(int, input().split())
for _ in range(n):
a.append(list(map(int, input().split())))
for _ in range(n):
b.append(list(map(int, input().split())))
_input()
print("YES" if solve(a, b) else "NO")
|
ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR LIST LIST FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR EXPR FUNC_CALL VAR LIST WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR LIST WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF IF VAR NUMBER VAR NUMBER RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR STRING STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = []
b = []
lol = 0
la = [[] for i in range(n * m)]
lb = [[] for i in range(n * m)]
for i in range(n):
a.append(list(map(int, input().split())))
for i in range(n):
b.append(list(map(int, input().split())))
for i in range(n):
for j in range(m):
la[i + j].append(a[i][j])
lb[i + j].append(b[i][j])
for i in range(n * m):
la[i].sort()
lb[i].sort()
if la[i] != lb[i]:
lol = 1
break
if lol == 0:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().strip().split())
matrix1 = []
matrix2 = []
ans = "YES"
for _ in range(n):
matrix1.append(list(map(int, input().strip().split())))
for _ in range(n):
matrix2.append(list(map(int, input().strip().split())))
for idx_sum in range(n + m - 1):
line1 = []
line2 = []
for raw_id in range(idx_sum + 1):
if raw_id < n and idx_sum - raw_id < m:
line1.append(matrix1[raw_id][idx_sum - raw_id])
line2.append(matrix2[raw_id][idx_sum - raw_id])
line1 = sorted(line1)
line2 = sorted(line2)
if line1 != line2:
ans = "NO"
print(ans)
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n_m = input().split(" ")
n = int(n_m[0])
m = int(n_m[1])
A = []
for i in range(n):
t = input().split(" ")
A.append(t)
B = []
for i in range(n):
t = input().split(" ")
B.append(t)
for i in range(m + n - 1):
text_1 = []
text_2 = []
for j in range(i + 1):
if i - j < n and j < m:
text_1.append(A[i - j][j])
text_2.append(B[i - j][j])
if i - j >= n and j < m:
continue
if i - j < n and j >= m:
continue
if i - j >= n and j >= m:
continue
text_1.sort()
text_2.sort()
f = True
if text_1 == text_2:
f == True
else:
f = False
break
if f:
print("YES")
else:
print("NO")
|
ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR VAR EXPR VAR NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
grid1 = [list(input().split()) for _ in range(n)]
grid2 = [list(input().split()) for _ in range(n)]
for i in range(n + m - 1):
s1, s2 = [], []
for j in range(n):
x, y = j, i - j
if 0 <= x < n and 0 <= y < m:
s1.append(grid1[x][y])
s2.append(grid2[x][y])
if sorted(s1) != sorted(s2):
print("NO")
sys.exit()
print("YES")
|
IMPORT ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR IF NUMBER VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().strip().split())
def inrange(x, y):
return x >= 0 and y >= 0 and x < n and y < m
a = []
b = []
for _ in range(n):
a.append(list(map(int, input().split())))
for _ in range(n):
b.append(list(map(int, input().split())))
s = True
for i in range(n + m - 1):
ii = i
j = 0
aa = {}
bb = {}
while ii >= 0 and j < m:
if inrange(ii, j):
if str(a[ii][j]) in aa.keys():
aa[str(a[ii][j])] += 1
else:
aa[str(a[ii][j])] = 0
if str(b[ii][j]) in bb.keys():
bb[str(b[ii][j])] += 1
else:
bb[str(b[ii][j])] = 0
ii -= 1
j += 1
if not aa == bb:
s = False
break
if s:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN VAR NUMBER VAR NUMBER VAR VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR DICT ASSIGN VAR DICT WHILE VAR NUMBER VAR VAR IF FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
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Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
mat1 = [0] * n
mat2 = [0] * n
for i in range(n):
mat1[i] = input().split()
for i in range(n):
mat2[i] = input().split()
ans = "YES"
for i in range(n + m - 1):
if i < min(n - 1, m - 1):
lst1 = [0] * (i + 1)
lst2 = [0] * (i + 1)
for j, k in enumerate(range(i + 1)):
lst1[j] = mat1[min(i, n - 1) - k][i - min(i, n - 1) + k]
lst2[j] = mat2[min(i, n - 1) - k][i - min(i, n - 1) + k]
elif min(n - 1, m - 1) <= i <= max(n - 1, m - 1):
lst1 = [0] * min(n, m)
lst2 = [0] * min(n, m)
for j, k in enumerate(range(min(n, m))):
lst1[j] = mat1[min(i, n - 1) - k][i - min(i, n - 1) + k]
lst2[j] = mat2[min(i, n - 1) - k][i - min(i, n - 1) + k]
elif max(n - 1, m - 1) < i:
lst1 = [0] * (min(n, m) - (i - max(n, m) + 1))
lst2 = [0] * (min(n, m) - (i - max(n, m) + 1))
for j, k in enumerate(range(min(n, m) - (i - max(n, m) + 1))):
lst1[j] = mat1[min(i, n - 1) - k][i - min(i, n - 1) + k]
lst2[j] = mat2[min(i, n - 1) - k][i - min(i, n - 1) + k]
lst1.sort()
lst2.sort()
if lst1 != lst2:
ans = "NO"
break
print(ans)
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR IF FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR IF FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
m1, m2 = [list(map(int, input().split())) for _ in range(n)], [
list(map(int, input().split())) for _ in range(n)
]
for i in range(n):
x, y = 0, i
d = {}
while y >= 0 and x < m:
d[m1[y][x]] = d[m1[y][x]] + 1 if m1[y][x] in d else 1
d[m2[y][x]] = d[m2[y][x]] - 1 if m2[y][x] in d else -1
x += 1
y -= 1
for k in d:
if d[k] != 0:
print("NO")
raise SystemExit()
for i in range(1, m):
x, y = i, n - 1
d = {}
while x < m and y >= 0:
d[m1[y][x]] = d[m1[y][x]] + 1 if m1[y][x] in d else 1
d[m2[y][x]] = d[m2[y][x]] - 1 if m2[y][x] in d else -1
x += 1
y -= 1
for k in d:
if d[k] != 0:
print("NO")
raise SystemExit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR ASSIGN VAR DICT WHILE VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR DICT WHILE VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = list(map(int, input().split()))
a = []
b = []
for i in range(n):
a.append(list(map(int, input().split())))
for i in range(n):
b.append(list(map(int, input().split())))
a1 = [[] for i in range(n + m - 1)]
for i in range(n):
j = 0
k = i
while k >= 0 and j < m:
a1[i].append(a[k][j])
k -= 1
j += 1
for i in range(1, m):
j = n - 1
k = i
while k < m and j >= 0:
a1[n + i - 1].append(a[j][k])
k += 1
j -= 1
b1 = [[] for i in range(n + m - 1)]
for i in range(n):
j = 0
k = i
while k >= 0 and j < m:
b1[i].append(b[k][j])
k -= 1
j += 1
for i in range(1, m):
j = n - 1
k = i
while k < m and j >= 0:
b1[n + i - 1].append(b[j][k])
k += 1
j -= 1
for i in range(n + m - 1):
q1 = a1[i]
q2 = b1[i]
q1.sort()
q2.sort()
if q1 != q2:
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
def diagonal(a):
global n, m
max_col = n
max_row = m
fdiag = [[] for i in range(max_col + max_row - 1)]
for y in range(max_col):
for x in range(max_row):
fdiag[x + y].append(a[y][x])
return fdiag
n, m = map(int, input().split())
a = []
b = []
for i in range(n):
temp = list(map(int, input().split()))
a.append(temp)
for i in range(n):
temp = list(map(int, input().split()))
b.append(temp)
ans_a = diagonal(a)
ans_b = diagonal(b)
for i in range(n + m - 1):
if sorted(ans_a[i]) != sorted(ans_b[i]):
print("NO")
exit()
print("YES")
|
FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR RETURN VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
gi = lambda: list(map(int, input().strip().split()))
n, m = gi()
A = [gi() for e in range(n)]
B = [gi() for e in range(n)]
for k in range(m):
a = sorted([A[j][k - j] for j in range(min(n, k + 1))])
b = sorted([B[j][k - j] for j in range(min(n, k + 1))])
if a != b:
print("NO")
exit()
for k in range(m):
a = sorted([A[-j - 1][j - 1 - k] for j in range(min(n, k + 1))])
b = sorted([B[-j - 1][-k + j - 1] for j in range(min(n, k + 1))])
if a != b:
print("NO")
exit()
if n <= m:
print("YES")
exit()
for k in range(n):
a = sorted([A[k - j][j] for j in range(min(m, k + 1))])
b = sorted([B[k - j][j] for j in range(min(m, k + 1))])
if a != b:
print("NO")
exit()
for k in range(n):
a = sorted([A[-k - 1 + j][-j - 1] for j in range(min(m, k + 1))])
b = sorted([B[-k - 1 + j][-j - 1] for j in range(min(m, k + 1))])
if a != b:
print("NO")
exit()
print("YES")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = []
for i in range(n):
a.append(list(map(int, input().split())))
b = []
for i in range(n):
b.append(list(map(int, input().split())))
diasA = [[] for i in range(m + n)]
diasB = [[] for i in range(m + n)]
for i in range(n):
for j in range(m):
diasA[i + j].append(a[i][j])
diasB[i + j].append(b[i][j])
result = all(sorted(da) == sorted(db) for da, db in zip(diasA, diasB))
if result:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = list(map(int, input().split()))
a = [0] * n
b = [0] * n
ok = 1
for i in range(n):
a[i] = list(map(int, input().split()))
for i in range(n):
b[i] = list(map(int, input().split()))
for i in range(m):
tmp1 = []
tmp2 = []
deph = 0
it = i
while deph < n and it >= 0:
tmp1.append(a[deph][it])
tmp2.append(b[deph][it])
deph += 1
it -= 1
tmp1.sort()
tmp2.sort()
if tmp1 != tmp2:
ok = 0
for i in range(1, n):
tmp1 = []
tmp2 = []
deph = i
it = m - 1
while deph < n and it >= 0:
tmp1.append(a[deph][it])
tmp2.append(b[deph][it])
deph += 1
it -= 1
tmp1.sort()
tmp2.sort()
if tmp1 != tmp2:
ok = 0
if ok:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
n, m = map(int, input().split())
A = []
for i in range(0, n):
A.append(input().split())
B = []
for i in range(0, n):
B.append(input().split())
for i in range(1, n + m):
x, y = min(i - 1, n - 1), max(i - n, 0)
a, b = [], []
while x >= 0 and y < m:
a.append(A[x][y])
b.append(B[x][y])
x -= 1
y += 1
a.sort()
b.sort()
for k in range(len(a)):
if a[k] != b[k]:
print("NO")
sys.exit()
print("YES")
|
IMPORT ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR LIST LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = [int(item) for item in input().split()]
a = [[] for _ in range(n)]
b = [[] for _ in range(n)]
for i in range(n):
a[i] = [int(item) for item in input().split()]
for i in range(n):
b[i] = [int(item) for item in input().split()]
ok = True
for i in range(n):
for l in range(m):
if a[i][l] == b[i][l]:
continue
flag = False
for col in range(max(0, l - n + i + 1), min(l, m)):
if b[i][l] == a[i + l - col][col]:
flag = True
a[i][l], a[i + l - col][col] = a[i + l - col][col], a[i][l]
if not flag:
ok = False
break
if ok == False:
break
print("YES" if ok else "NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR IF VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR STRING STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
from sys import *
n, m = map(int, stdin.readline().split())
a = []
for i in range(n):
x = list(map(int, stdin.readline().split()))
a.append(x)
b = []
for i in range(n):
x = list(map(int, stdin.readline().split()))
b.append(x)
c1 = []
for i in range(n + m):
c1.append([])
c2 = []
for i in range(n + m):
c2.append([])
for i in range(n):
for j in range(m):
c1[i + j].append(a[i][j])
c2[i + j].append(b[i][j])
f = 0
for i in range(n + m):
c1[i].sort()
c2[i].sort()
if c1[i] != c2[i]:
f = 1
break
if f == 1:
print("NO")
else:
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR LIST FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = list(map(int, input().split()))
a = []
for i in range(n):
a.append(list(map(int, input().split())))
b = []
flag = True
for i in range(n):
b.append(list(map(int, input().split())))
for i in range(m + n - 1):
cur = []
curb = []
for l in range(i + 1):
try:
cur.append(a[i - l][l])
curb.append(b[i - l][l])
except:
pass
cur.sort()
curb.sort()
if cur != curb:
flag = False
if flag:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
def rint():
return map(int, sys.stdin.readline().split())
n, m = rint()
a = [[] for j in range(n)]
b = [[] for j in range(n)]
for i in range(n):
a[i] = list(rint())
for i in range(n):
b[i] = list(rint())
for s in range(n + m - 1):
adict = dict()
bdict = dict()
for r in range(s + 1):
if r >= n:
break
c = s - r
if c < 0:
break
if c >= m:
continue
if a[r][c] in adict:
adict[a[r][c]] += 1
else:
adict[a[r][c]] = 1
if b[r][c] in bdict:
bdict[b[r][c]] += 1
else:
bdict[b[r][c]] = 1
if adict != bdict:
print("NO")
exit()
print("YES")
|
IMPORT FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().strip().split())
arr = []
for i in range(n):
a = list(map(int, input().strip().split()))
arr.append(a)
brr = []
for i in range(n):
a = list(map(int, input().strip().split()))
brr.append(a)
flag = 0
j = 0
for i in range(n):
a = []
b = []
p = i
q = 0
while p >= 0 and q < m:
a.append(arr[p][q])
b.append(brr[p][q])
p -= 1
q += 1
a = sorted(a)
b = sorted(b)
if a != b:
flag = 1
for j in range(m):
a = []
b = []
p = n - 1
q = j
while p >= 0 and q < m:
a.append(arr[p][q])
b.append(brr[p][q])
p -= 1
q += 1
a = sorted(a)
b = sorted(b)
if a != b:
flag = 1
if flag == 0:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a_sets = {}
b_sets = {}
for i in range(n):
ll = map(int, input().split())
for j, x in enumerate(ll):
a_sets.setdefault(i + j, []).append(x)
for i in range(n):
ll = map(int, input().split())
for j, x in enumerate(ll):
b_sets.setdefault(i + j, []).append(x)
for k in a_sets:
a_k_line = sorted(a_sets[k])
b_k_line = sorted(b_sets[k])
if a_k_line != b_k_line:
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL FUNC_CALL VAR BIN_OP VAR VAR LIST VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL FUNC_CALL VAR BIN_OP VAR VAR LIST VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
from sys import stdin, stdout
def inp():
return int(stdin.readline())
def minp():
return map(int, stdin.readline().rstrip().split())
def linp():
return list(minp())
def main():
n, m = minp()
a = []
b = []
for i in range(n):
l = linp()
a.append(l)
for i in range(n):
l = linp()
b.append(l)
p = [[] for i in range(n + m)]
q = [[] for j in range(n + m)]
f = 1
for i in range(n):
for j in range(m):
p[i + j].append(a[i][j])
q[i + j].append(b[i][j])
for i in range(n + m):
if sorted(p[i]) == sorted(q[i]):
pass
else:
print("NO")
f = 0
break
if f == 1:
print("YES")
main()
|
FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = []
b = []
d = dict()
td = dict()
for i in range(n):
a.append(list(map(int, input().split())))
for i in range(n):
b.append(list(map(int, input().split())))
for i in range(n):
for j in range(m):
idx = i + j
try:
d[idx].append(a[i][j])
except KeyError:
d[idx] = [a[i][j]]
try:
td[idx].append(b[i][j])
except KeyError:
td[idx] = [b[i][j]]
for i in d:
d[i].sort()
td[i].sort()
if d[i] != td[i]:
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR LIST VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR LIST VAR VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = [[i for i in map(int, input().split())] for i in range(n)]
b = [[i for i in map(int, input().split())] for i in range(n)]
ans = 1
op = 1
for j in range(m):
da = {}
db = {}
i = 0
while j >= 0 and i < n:
da[a[i][j]] = da.get(a[i][j], 0) + 1
db[b[i][j]] = db.get(b[i][j], 0) + 1
j -= 1
i += 1
if da != db:
ans = 0
if not ans:
break
for i in range(1, n):
da = {}
db = {}
j = m - 1
while j >= 0 and i < n:
da[a[i][j]] = da.get(a[i][j], 0) + 1
db[b[i][j]] = db.get(b[i][j], 0) + 1
j -= 1
i += 1
if da != db:
ans = 0
if not ans:
break
if ans:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR DICT ASSIGN VAR DICT ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR DICT ASSIGN VAR DICT ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
A = []
B = []
for a in range(n):
A.append(list(map(int, input().split())))
for _ in range(n):
B.append(list(map(int, input().split())))
for c in range(n + m):
rinda = 0 if c < m else c - m
kolona = c if c < m else m - 1
Ad = []
Bd = []
while kolona >= 0 and rinda < n:
Ad.append(A[rinda][kolona])
Bd.append(B[rinda][kolona])
rinda += 1
kolona -= 1
Ad.sort()
Bd.sort()
if Ad != Bd:
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
sys.setrecursionlimit(10**5 + 1)
inf = int(10**20)
max_val = inf
min_val = -inf
RW = lambda: sys.stdin.readline().strip()
RI = lambda: int(RW())
RMI = lambda: [int(x) for x in sys.stdin.readline().strip().split()]
RWI = lambda: [x for x in sys.stdin.readline().strip().split()]
nb_rows, nb_cols = RMI()
mat_A = [RMI() for _ in range(nb_rows)]
mat_B = [RMI() for _ in range(nb_rows)]
mat_AB = [[] for _ in range(nb_rows + nb_cols)]
mat_BA = [[] for _ in range(nb_rows + nb_cols)]
for i in range(nb_rows):
for j in range(nb_cols):
mat_AB[i + j].append(mat_A[i][j])
mat_BA[i + j].append(mat_B[i][j])
mat_ABS = [sorted(x) for x in mat_AB]
mat_BAS = [sorted(x) for x in mat_BA]
print(["NO", "YES"][mat_ABS == mat_BAS])
|
IMPORT EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR LIST STRING STRING VAR VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().strip().split())
A = []
B = []
for _ in range(n):
a = list(map(int, input().strip().split()))
A.append(a)
for _ in range(n):
b = list(map(int, input().strip().split()))
B.append(b)
flag = True
for i in range(n - 1):
if i == 0:
if A[0][0] != B[0][0]:
flag = False
else:
a = i
b = 0
temp = {}
temp2 = {}
while a >= 0 and b < m:
try:
temp[B[a][b]] += 1
except:
temp[B[a][b]] = 1
try:
temp2[A[a][b]] += 1
except:
temp2[A[a][b]] = 1
a -= 1
b += 1
for k in temp2.keys():
try:
if temp2[k] == temp[k]:
nothing = 1
else:
flag = False
break
except:
flag = False
break
for j in range(m):
a = n - 1
b = j
temp = {}
temp2 = {}
while a >= 0 and b < m:
try:
temp[B[a][b]] += 1
except:
temp[B[a][b]] = 1
try:
temp2[A[a][b]] += 1
except:
temp2[A[a][b]] = 1
a -= 1
b += 1
for k in temp2.keys():
try:
if temp2[k] == temp[k]:
nothing = 1
else:
flag = False
break
except:
flag = False
break
if flag:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR DICT ASSIGN VAR DICT WHILE VAR NUMBER VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR DICT ASSIGN VAR DICT WHILE VAR NUMBER VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
cnt = lambda s, x: s.count(x)
ii = lambda: int(input())
si = lambda: input()
f = lambda: map(int, input().split())
dgl = lambda: list(map(int, input()))
il = lambda: list(map(int, input().split()))
n, m = f()
m1, m2 = [], []
lm1, lm2 = [[] for _ in range(n + m)], [[] for _ in range(n + m)]
for _ in range(n):
m1.append(il())
for _ in range(n):
m2.append(il())
for i in range(n):
for j in range(m):
lm1[i + j].append(m1[i][j])
lm2[i + j].append(m2[i][j])
for i in range(n + m):
if sorted(lm1[i]) != sorted(lm2[i]):
exit(print("NO"))
print("YES")
|
ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR VAR LIST LIST ASSIGN VAR VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
pa = [{} for _ in range(n + m - 1)]
pb = [{} for _ in range(n + m - 1)]
for i in range(n):
tmp = list(map(int, input().split()))
for j in range(m):
if tmp[j] in pa[i + j]:
pa[i + j][tmp[j]] += 1
else:
pa[i + j][tmp[j]] = 1
for i in range(n):
tmp = list(map(int, input().split()))
for j in range(m):
if tmp[j] in pb[i + j]:
pb[i + j][tmp[j]] += 1
else:
pb[i + j][tmp[j]] = 1
if pa == pb:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR DICT VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = [int(s) for s in input().split()]
a = []
b = []
c = []
ans = 0
c = []
d = []
for i in range(n):
a.append([int(s) for s in input().split()])
for i in range(n):
b.append([int(s) for s in input().split()])
for i in range(0, n + m - 1):
c.append([])
d.append([])
for i in range(n):
for j in range(m):
c[i + j].append(a[i][j])
d[i + j].append(b[i][j])
for i in range(0, n + m - 1):
e = sorted(c[i])
f = sorted(d[i])
if e == f:
ans += 1
if ans == n + m - 1:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR LIST EXPR FUNC_CALL VAR LIST FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = []
b = []
for i in range(n):
a1 = [int(x) for x in input().split()]
a.append(a1)
for i in range(n):
a1 = [int(x) for x in input().split()]
b.append(a1)
ad = [[] for x in range(n * m)]
bd = [[] for x in range(n * m)]
for i in range(n):
for j in range(m):
ad[i + j].append(a[i][j])
for i in range(n):
for j in range(m):
bd[i + j].append(b[i][j])
for i in range(len(ad)):
x = sorted(ad[i])
y = sorted(bd[i])
if x != y:
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = [int(i) for i in input().split()]
a = [[] for i in range(n + m + 1)]
b = [[] for i in range(n + m + 1)]
s = []
st = []
for i in range(n):
p = [int(i) for i in input().split()]
s.append(p)
for i in range(n):
p = [int(i) for i in input().split()]
st.append(p)
for i in range(n):
for j in range(m):
a[i + j].append(s[i][j])
b[i + j].append(st[i][j])
l = 1
for i in range(len(a)):
a[i].sort()
b[i].sort()
if a[i] != b[i]:
l = 0
print("No")
break
if l == 1:
print("Yes")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = [int(x) for x in input().split(" ")]
ROW = n
COL = m
data = []
def diagonalOrder(matrix):
d = []
for line in range(1, ROW + COL):
start_col = max(0, line - ROW)
count = min(line, COL - start_col, ROW)
row = []
for j in range(0, count):
row.append(matrix[min(ROW, line) - j - 1][start_col + j])
row.sort()
d.append(row)
return d
for _ in range(n):
data.append([int(x) for x in input().split(" ")])
d1 = diagonalOrder(data)
data = []
for _ in range(n):
data.append([int(x) for x in input().split(" ")])
d2 = diagonalOrder(data)
if d1 == d2:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = list(map(int, input().split()))
A = [list(map(int, input().split())) for i in range(n)]
B = [list(map(int, input().split())) for i in range(n)]
C = [
sorted([A[j][i - j] for j in range(i + 1) if 0 <= j < n and i - j < m])
for i in range(n + m - 1)
]
D = [
sorted([B[j][i - j] for j in range(i + 1) if 0 <= j < n and i - j < m])
for i in range(n + m - 1)
]
if C == D:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
def ria():
return [int(i) for i in input().split()]
h, w = ria()
mt1 = []
mt2 = []
for y in range(h):
mt1.append(ria())
for y in range(h):
mt2.append(ria())
for x in range(w):
st1 = []
st2 = []
y = 0
while w > x >= 0 and h > y >= 0:
st1.append(mt1[y][x])
st2.append(mt2[y][x])
x -= 1
y += 1
st1.sort()
st2.sort()
if st1 != st2:
print("NO")
exit(0)
for y in range(h):
st1 = []
st2 = []
x = w - 1
while w > x >= 0 and h > y >= 0:
st1.append(mt1[y][x])
st2.append(mt2[y][x])
x -= 1
y += 1
st1.sort()
st2.sort()
if st1 != st2:
print("NO")
exit(0)
print("YES")
|
FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
def e(x, y, d):
while y >= 0 and x < m:
d[p[y][x]] = d[p[y][x]] + 1 if p[y][x] in d else 1
d[q[y][x]], x, y = d[q[y][x]] - 1 if q[y][x] in d else -1, x + 1, y - 1
for k in d:
if d[k] != 0:
print("NO")
raise SystemExit()
n, m = map(int, input().split())
p, q = [list(map(int, input().split())) for _ in range(n)], [
list(map(int, input().split())) for _ in range(n)
]
for i in range(n):
e(0, i, {})
for i in range(m):
a, b = (e(i, n - 1, {}), print("YES")) if i == m - 1 else (e(i, n - 1, {}), None)
|
FUNC_DEF WHILE VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER VAR DICT FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER DICT FUNC_CALL VAR STRING FUNC_CALL VAR VAR BIN_OP VAR NUMBER DICT NONE
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
l1 = []
l2 = []
for i in range(n):
ka = list(map(int, input().split()))
l1.append(ka)
for i in range(n):
ka = list(map(int, input().split()))
l2.append(ka)
ba = []
ha = []
for i in range(n):
for j in range(m):
ha.append(l1[i][j])
ba.append(l2[i][j])
if sorted(ha) != sorted(ba):
print("NO")
exit()
for i in range(n + m - 1):
ha = []
ba = []
x = i
y = 0
while x >= 0 and y < m:
try:
ha.append(l1[x][y])
ba.append(l2[x][y])
except:
pass
x -= 1
y += 1
if sorted(ha) != sorted(ba):
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
A = []
B = []
[N, M] = [int(x) for x in input().split()]
for i in range(N):
A.append([int(x) for x in input().split()])
for i in range(N):
B.append([int(x) for x in input().split()])
a_t = [[] for x in range(1010)]
b_t = [[] for x in range(1010)]
for n in range(N):
for m in range(M):
a_t[n + m].append(A[n][m])
b_t[n + m].append(B[n][m])
for i in range(0, 1000):
a_t[i] = sorted(a_t[i])
b_t[i] = sorted(b_t[i])
if a_t[i] != b_t[i]:
print
print("NO")
exit()
print("YES")
|
ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN LIST VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR EXPR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
[n, m] = input().split(" ")
n = int(n)
m = int(m)
matrixA = []
matrixB = []
for i in range(n):
matrixA.append(list(map(int, input().split(" "))))
for i in range(n):
matrixB.append(list(map(int, input().split(" "))))
def buildDiagonals(matrix):
r = [None] * (n + m - 1)
for i in range(m):
for j in range(n):
if not r[i + j]:
r[i + j] = []
r[i + j].append(matrix[j][i])
for x in range(len(r)):
r[x].sort()
return r
diagonalsA = buildDiagonals(matrixA)
diagonalsB = buildDiagonals(matrixB)
if str(diagonalsA) == str(diagonalsB):
print("YES")
else:
print("NO")
|
ASSIGN LIST VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING FUNC_DEF ASSIGN VAR BIN_OP LIST NONE BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR LIST EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
mat1 = [[(0) for i in range(m)] for j in range(n)]
mat2 = [[(0) for i in range(m)] for j in range(n)]
for i in range(n):
mat1[i] = list(map(int, input().split()))
for i in range(n):
mat2[i] = list(map(int, input().split()))
l1 = [dict() for i in range(n + m - 1)]
l2 = [dict() for i in range(n + m - 1)]
for i in range(n):
for j in range(m):
if mat1[i][j] in l1[i + j]:
l1[i + j][mat1[i][j]] += 1
else:
l1[i + j][mat1[i][j]] = 1
if mat2[i][j] in l2[i + j]:
l2[i + j][mat2[i][j]] += 1
else:
l2[i + j][mat2[i][j]] = 1
flag = 0
for i in range(n + m - 1):
if l1[i] != l2[i]:
flag = 1
print(flag * "NO" + (1 - flag) * "YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR STRING BIN_OP BIN_OP NUMBER VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
def main():
n, m = map(int, input().split())
a = [[] for i in range(n)]
b = [[] for i in range(n)]
aDicts = [dict() for i in range(n + m - 1)]
bDicts = [dict() for i in range(n + m - 1)]
for i in range(n):
a[i] = list(map(int, input().split()))
for i in range(n):
b[i] = list(map(int, input().split()))
for i in range(n):
for j in range(m):
if a[i][j] in aDicts[i + j].keys():
aDicts[i + j][a[i][j]] += 1
else:
aDicts[i + j][a[i][j]] = 1
if b[i][j] in bDicts[i + j].keys():
bDicts[i + j][b[i][j]] += 1
else:
bDicts[i + j][b[i][j]] = 1
for i in range(n + m - 1):
if aDicts[i] != bDicts[i]:
print("NO")
return 0
print("YES")
return 0
main()
|
FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING RETURN NUMBER EXPR FUNC_CALL VAR STRING RETURN NUMBER EXPR FUNC_CALL VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
lst = []
lst2 = []
for c in range(n):
lst.append(list(map(int, input().split())))
for c in range(n):
lst2.append(list(map(int, input().split())))
for c in range(n):
for r in range(len(lst[c])):
p = 0
if lst[c][r] != lst2[c][r]:
a = c
b = r
while a != n and b != -1:
if lst2[a][b] == lst[c][r]:
p = 1
break
a += 1
b -= 1
if p == 1:
if True:
lst2[a][b], lst2[c][r] = lst2[c][r], lst2[a][b]
else:
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER IF NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
n, m = tuple(map(int, input().split(" ")))
A = []
for i in range(n):
A.append(list(map(int, input().split(" "))))
B = []
for i in range(n):
B.append(list(map(int, input().split(" "))))
if A[0][0] != B[0][0] or A[-1][-1] != B[-1][-1]:
print("NO")
sys.exit(0)
for i in range(n):
for j in range(m):
if i == 0 and j == 0 or i == n - 1 and j == m - 1:
continue
sum1 = i + j
found = False
for k in range(sum1 + 1):
if k < n and sum1 - k < m and A[i][j] == B[k][sum1 - k]:
found = True
B[k][sum1 - k] = -1
break
if not found:
print("NO")
sys.exit(0)
print("YES")
|
IMPORT ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
A = dict()
B = dict()
for i in range(n):
a = list(map(int, input().split()))
for j in range(m):
if i + j in A:
if a[j] in A[i + j]:
A[i + j][a[j]] += 1
else:
A[i + j][a[j]] = 1
else:
A[i + j] = {a[j]: 1}
for i in range(n):
b = list(map(int, input().split()))
for j in range(m):
if i + j in B:
if b[j] in B[i + j]:
B[i + j][b[j]] += 1
else:
B[i + j][b[j]] = 1
else:
B[i + j] = {b[j]: 1}
yes = True
for d in A:
if A[d] != B[d]:
yes = False
break
print("YES") if yes else print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR DICT VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR DICT VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER EXPR VAR FUNC_CALL VAR STRING FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
n, m = list(map(int, input().split()))
m1 = [list(map(int, input().split())) for i in range(n)]
m2 = [list(map(int, input().split())) for i in range(n)]
for j in range(m + n - 1):
x1 = {}
x2 = {}
for i in range(j + 1):
if j - i < n and i < m:
x1[m1[j - i][i]] = 0
x2[m2[j - i][i]] = 0
for i in range(j + 1):
if j - i < n and i < m:
x1[m1[j - i][i]] += 1
x2[m2[j - i][i]] += 1
if x1 != x2:
print("NO")
return
print("YES")
|
IMPORT ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, k = map(int, input().split())
a = [[] * k] * n
for i in range(n):
a[i] = list(map(int, input().split()))
b = [[] * k] * n
for i in range(n):
b[i] = list(map(int, input().split()))
flag = True
tempa = []
tempb = []
for i in range(n + k):
tempa.clear()
tempb.clear()
if flag == False:
break
for j in range(i + 1):
if i - j > n - 1 or j > k - 1:
continue
tempa.append(a[i - j][j])
tempb.append(b[i - j][j])
tempa.sort()
tempb.sort()
sz = len(tempa)
for j in range(sz):
if flag and tempa[j] != tempb[j]:
flag = False
break
if flag:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST BIN_OP LIST VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST BIN_OP LIST VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = [[] for i in range(n)]
b = [[] for i in range(n)]
va = [[] for i in range(n + m)]
vb = [[] for i in range(n + m)]
for i in range(n):
a[i] = list(map(int, input().split()))
for j in range(m):
va[i + j].append(a[i][j])
for i in range(n):
b[i] = list(map(int, input().split()))
for j in range(m):
vb[i + j].append(b[i][j])
for i in range(n + m - 1):
va[i].sort()
vb[i].sort()
for j in range(len(va[i])):
if va[i][j] != vb[i][j]:
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
arr = {}
j = 0
for i in range(n):
s = list(map(int, input().split()))
for k in range(m):
y = i + k
x = s[k]
ky = str(x) + "a" + str(y)
if ky in arr:
arr[ky] += 1
else:
arr[ky] = 1
j = 0
flag = 1
for i in range(n):
s = list(map(int, input().split()))
for k in range(m):
y = i + k
x = s[k]
ky = str(x) + "a" + str(y)
if ky in arr:
arr[ky] -= 1
else:
flag = 0
if flag == 0:
print("NO")
exit()
for k in arr:
if arr[k] != 0:
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR STRING FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR STRING FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
size = [int(number) for number in input().split()]
matrixa = []
for i in range(size[0]):
matrixa.append([int(number) for number in input().split()])
matrixb = []
for i in range(size[0]):
matrixb.append([int(number) for number in input().split()])
def checkDiagonal(cords):
if cords[0] > 0 and cords[1] < size[1] - 1:
return checkDiagonal([cords[0] - 1, cords[1] + 1])
else:
done = False
alist = []
blist = []
while not done:
if cords[0] < size[0] and cords[1] >= 0:
alist.append(matrixa[cords[0]][cords[1]])
blist.append(matrixb[cords[0]][cords[1]])
cords[0] += 1
cords[1] -= 1
else:
done = True
alist.sort()
blist.sort()
if alist == blist:
return True
else:
return False
truth = "yes"
for i in range(size[0]):
if checkDiagonal([i, size[1] - 1]) == False:
truth = "no"
for i in range(size[1]):
if checkDiagonal([0, i]) == False:
truth = "no"
print(truth)
|
ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF IF VAR NUMBER NUMBER VAR NUMBER BIN_OP VAR NUMBER NUMBER RETURN FUNC_CALL VAR LIST BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR IF VAR NUMBER VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR LIST NUMBER VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
input = sys.stdin.readline
MOD = 1000000007
MOD2 = 998244353
ii = lambda: int(input().strip("\n"))
si = lambda: input().strip("\n")
dgl = lambda: list(map(int, input().strip("\n")))
f = lambda: map(int, input().strip("\n").split())
il = lambda: list(map(int, input().strip("\n").split()))
ls = lambda: list(input().strip("\n"))
let = "abcdefghijklmnopqrstuvwxyz"
n, m = f()
m1 = [[] for i in range(n + m)]
m2 = [[] for i in range(n + m)]
for i in range(n):
l = il()
for j in range(m):
m1[i + j].append(l[j])
for i in range(n):
l = il()
for j in range(m):
m2[i + j].append(l[j])
for i in range(n + m):
m1[i].sort()
m2[i].sort()
if m1[i] != m2[i]:
print("NO")
exit()
print("YES")
|
IMPORT ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR STRING ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
M1 = [list(map(int, input().split())) for i in range(n)]
M2 = [list(map(int, input().split())) for i in range(n)]
h = [0] * n
for i in range(n):
M1[i].extend(h)
for i in range(n):
M2[i].extend(h)
fl = 0
for i in range(1, m + n - 1):
s1 = dict()
s2 = dict()
x = i
y = i
while m + n > x >= 0 and n > i - x >= 0:
if M1[i - x][x] not in s1:
s1[M1[i - x][x]] = 0
s1[M1[i - x][x]] += 1
x -= 1
while m + n > y >= 0 and n > i - y >= 0:
if M2[i - y][y] not in s2:
s2[M2[i - y][y]] = 0
s2[M2[i - y][y]] += 1
y -= 1
if s1 != s2:
fl = 1
if M1[0][0] != M2[0][0] or M1[-1][-1] != M2[-1][-1]:
fl = 1
if fl:
print("NO")
else:
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR VAR WHILE BIN_OP VAR VAR VAR NUMBER VAR BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER WHILE BIN_OP VAR VAR VAR NUMBER VAR BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
def collect_tr(mat, n, m, i, j):
tr = []
while i >= 0 and j < m:
tr.append(mat[i][j])
i -= 1
j += 1
return tr
def main():
n, m = map(int, input().split(" "))
mat1 = [list(map(int, input().split(" "))) for _ in range(n)]
mat2 = [list(map(int, input().split(" "))) for _ in range(n)]
for i in range(n):
tr1 = sorted(collect_tr(mat1, n, m, i, 0))
tr2 = sorted(collect_tr(mat2, n, m, i, 0))
if tr1 != tr2:
print("NO")
return 0
for j in range(1, m):
tr1 = sorted(collect_tr(mat1, n, m, n - 1, j))
tr2 = sorted(collect_tr(mat2, n, m, n - 1, j))
if tr1 != tr2:
print("NO")
return 0
print("YES")
return 0
exit(main())
|
FUNC_DEF ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING RETURN NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR IF VAR VAR EXPR FUNC_CALL VAR STRING RETURN NUMBER EXPR FUNC_CALL VAR STRING RETURN NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
i = 0
k = []
while i < n:
s = list(map(int, input().split()))
i += 1
k += [s]
i = 0
x = []
while i < n:
g = []
j = 0
p = i
while p > -1 and j > -1 and p < n and j < m:
g += [k[p][j]]
j += 1
p -= 1
g.sort()
x += [g]
i += 1
i = n - 1
j = 1
while j < m:
p = i
q = j
g = []
while p >= 0 and p < n and q >= 0 and q < m:
g += [k[p][q]]
q += 1
p -= 1
g.sort()
x += [g]
j += 1
y = []
i = 0
k = []
while i < n:
s = list(map(int, input().split()))
i += 1
k += [s]
i = 0
while i < n:
g = []
j = 0
p = i
while p > -1 and j > -1 and p < n and j < m:
g += [k[p][j]]
j += 1
p -= 1
g.sort()
y += [g]
i += 1
i = n - 1
j = 1
while j < m:
p = i
q = j
g = []
while p >= 0 and p < n and q >= 0 and q < m:
g += [k[p][q]]
q += 1
p -= 1
g.sort()
y += [g]
j += 1
if x == y:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR LIST VAR ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR LIST VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR LIST VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR LIST VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR LIST VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR LIST VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR LIST VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR LIST VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR LIST VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR LIST VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = [list(map(int, input().split())) for i in range(n)]
b = [list(map(int, input().split())) for i in range(n)]
a1 = [[] for i in range(n + m + 100)]
b1 = [[] for i in range(n + m + 100)]
for i in range(n):
for j in range(m):
a1[i + j].append(a[i][j])
b1[i + j].append(b[i][j])
for i in range(n + m + 5):
a1[i].sort()
b1[i].sort()
if len(a1[i]) != len(b1[i]):
print("NO")
exit(0)
for j in range(len(a1[i])):
if a1[i][j] != b1[i][j]:
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
tokens = input().split()
rows = int(tokens[0])
columns = int(tokens[1])
A = []
B = []
for i in range(rows):
tokens = input().split()
tokens = [int(w) for w in tokens]
A.append(tokens)
for i in range(rows):
tokens = input().split()
tokens = [int(w) for w in tokens]
B.append(tokens)
status = True
for count in range(rows):
i = count
j = 0
a = []
b = []
while i >= 0 and j < columns:
a.append(A[i][j])
b.append(B[i][j])
i -= 1
j += 1
a.sort()
b.sort()
for i in range(len(a)):
if a[i] != b[i]:
status = False
break
for count in range(columns):
i = rows - 1
j = count
a = []
b = []
while i >= 0 and j < columns:
a.append(A[i][j])
b.append(B[i][j])
i -= 1
j += 1
a.sort()
b.sort()
for i in range(len(a)):
if a[i] != b[i]:
status = False
break
if status:
print("YES")
else:
print("NO")
|
ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = [int(x) for x in input().split()]
A = []
B = []
for _ in range(n):
A.append([int(x) for x in input().split()])
for _ in range(n):
B.append([int(x) for x in input().split()])
AD = []
BD = []
for s in range(2, m + n + 1):
at = []
bt = []
for r in range(1, n + 1):
c = s - r
if 1 <= c and c <= m:
at.append(A[r - 1][c - 1])
bt.append(B[r - 1][c - 1])
at.sort()
bt.sort()
AD.append(at)
BD.append(bt)
if AD == BD:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split(" "))
ma = []
mb = []
for i in range(n):
li = list(map(int, input().split(" ", m)[:m]))
ma.append(li)
for i in range(n):
li = list(map(int, input().split(" ", m)[:m]))
mb.append(li)
lia = []
lib = []
i, j = 0, 0
kl = 0
ad = 0
for l in range(n + m - 1):
li = []
while i >= 0 and j < m:
li.append(ma[i][j])
i -= 1
j += 1
lia.append(li)
if kl == n - 1:
i = n - 1
ad += 1
j = ad
else:
kl += 1
i = kl
j = 0
i, j = 0, 0
kl = 0
ad = 0
for l in range(n + m - 1):
li = []
while i >= 0 and j < m:
li.append(mb[i][j])
i -= 1
j += 1
lib.append(li)
if kl == n - 1:
i = n - 1
ad += 1
j = ad
else:
kl += 1
i = kl
j = 0
fl = 0
for i in range(len(lia)):
lia[i].sort()
lib[i].sort()
if lia[i] != lib[i]:
fl = 1
break
if fl == 0:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
a = []
b = []
test = []
test_ = []
for i in range(n + m):
test.append([0])
test_.append([0])
for i in range(n):
cache = list(map(int, input().split()))
a.append(cache)
for i in range(n):
cache = list(map(int, input().split()))
b.append(cache)
for i in range(n):
for j in range(m):
test[i + j].append(a[i][j])
test_[i + j].append(b[i][j])
k = 0
for i in range(n + m - 1):
test[i].sort()
test_[i].sort()
for i in range(n + m - 1):
if test[i] == test_[i]:
k += 1
if k == n + m - 1:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR LIST NUMBER EXPR FUNC_CALL VAR LIST NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = list(map(int, input().split()))
a = []
for i in range(n):
a.append(list(map(int, input().split())))
b = []
for i in range(n):
b.append(list(map(int, input().split())))
for i in range(m + n - 1):
d1 = {}
d2 = {}
for j in range(n):
l = j
r = i - j
if l >= 0 and l < n and r >= 0 and r < m:
e = a[l][r]
if e not in d1:
d1[e] = 0
d1[e] += 1
for j in range(n):
l = j
r = i - j
if l >= 0 and l < n and r >= 0 and r < m:
e = b[l][r]
if e not in d2:
d2[e] = 0
d2[e] += 1
for k in d1:
if k not in d2:
print("NO")
return
if d1[k] != d2[k]:
print("NO")
return
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER VAR VAR NUMBER FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING RETURN IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
A = []
B = []
for i in range(n):
A.append([int(x) for x in input().split()])
for i in range(n):
B.append([int(x) for x in input().split()])
flag = 0
for i in range(n + m - 1):
a = []
b = []
if i < m:
r, c = 0, i
else:
r, c = i - m + 1, m - 1
while r < n and c >= 0:
a.append(A[r][c])
b.append(B[r][c])
r = r + 1
c = c - 1
a = sorted(a)
b = sorted(b)
if a != b:
print("NO")
flag = 1
break
if flag == 0:
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST IF VAR VAR ASSIGN VAR VAR NUMBER VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
the_string = input()
rows, coloumns = the_string.split()
row = int(rows)
coloumn = int(coloumns)
biggestdiag = min(row, coloumn)
A = []
B = []
for i in range(row):
string = input()
string = string.split()
A.append(string)
for i in range(row):
string = input()
string = string.split()
B.append(string)
for i in range(biggestdiag):
for j in range(biggestdiag):
A[i].append(None)
for i in range(max(row, coloumn)):
temp = []
for j in range(max(row, coloumn)):
temp.append(None)
A.append(temp)
DiagA = []
for i in range(coloumn):
diagtemp = []
for j in range(biggestdiag):
if A[j][i - j] != None:
diagtemp.append(A[j][i - j])
A[j][i - j] = None
DiagA.append(diagtemp)
diagtemp = []
for i in range(row - 1):
diagtemp = []
for j in range(biggestdiag):
if A[i + j + 1][coloumn - j - 1] != None:
diagtemp.append(A[i + j + 1][coloumn - j - 1])
A[i + j + 1][coloumn - j - 1] = None
DiagA.append(diagtemp)
diagtemp = []
DiagB = []
for i in range(biggestdiag):
for j in range(biggestdiag):
B[i].append(None)
for i in range(max(row, coloumn)):
temp = []
for j in range(max(row, coloumn)):
temp.append(None)
B.append(temp)
DiagB = []
for i in range(coloumn):
diagtemp = []
for j in range(biggestdiag):
if B[j][i - j] != None:
diagtemp.append(B[j][i - j])
B[j][i - j] = None
DiagB.append(diagtemp)
diagtemp = []
for i in range(row - 1):
diagtemp = []
for j in range(biggestdiag):
if B[i + j + 1][coloumn - j - 1] != None:
diagtemp.append(B[i + j + 1][coloumn - j - 1])
B[i + j + 1][coloumn - j - 1] = None
DiagB.append(diagtemp)
diagtemp = []
output = True
for i in range(row + coloumn - 1):
DiagA[i].sort()
DiagB[i].sort()
if DiagA[i] != DiagB[i]:
output = False
break
if output == True:
print("YES")
else:
print("NO")
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NONE FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NONE EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR VAR NONE EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NONE EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NONE EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NONE EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NONE FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NONE EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR VAR NONE EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NONE EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NONE EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NONE EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = [int(x) for x in input().split()]
A = [[int(x) for x in input().split()] for _ in range(n)]
B = [[int(x) for x in input().split()] for _ in range(n)]
def valid(i, j):
return i >= 0 and j >= 0 and i < n and j < m
can = True
for i in range(n):
i1, j1 = i, 0
a = []
b = []
while valid(i1, j1):
a.append(A[i1][j1])
b.append(B[i1][j1])
i1 -= 1
j1 += 1
if sorted(a) != sorted(b):
can = False
break
if not can:
print("NO")
exit(0)
for j in range(m):
i1, j1 = n - 1, j
a = []
b = []
while valid(i1, j1):
a.append(A[i1][j1])
b.append(B[i1][j1])
i1 -= 1
j1 += 1
if sorted(a) != sorted(b):
can = False
break
if can:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN VAR NUMBER VAR NUMBER VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR LIST ASSIGN VAR LIST WHILE FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split(" "))
a = []
b = []
for i in range(n):
a.append(list(map(int, input().split(" "))))
for i in range(n):
b.append(list(map(int, input().split(" "))))
for i in range(n):
j = i
k = 0
c = []
d = []
while j >= 0 and k < m:
c.append(a[j][k])
d.append(b[j][k])
j -= 1
k += 1
c.sort()
d.sort()
for j in range(len(c)):
if c[j] != d[j]:
print("NO")
exit(0)
for i in range(m):
j = n - 1
k = i
c = []
d = []
while j >= 0 and k < m:
c.append(a[j][k])
d.append(b[j][k])
j -= 1
k += 1
c.sort()
d.sort()
for j in range(len(c)):
if c[j] != d[j]:
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
from sys import stdin
n, m = map(int, stdin.readline().strip().split())
mt1 = []
for i in range(n):
mt1.append(list(map(int, stdin.readline().strip().split())))
mt2 = []
for i in range(n):
mt2.append(list(map(int, stdin.readline().strip().split())))
t = True
for i in range(n):
x = i
y = 0
s = []
s1 = []
while x >= 0 and y < m:
s.append(mt1[x][y])
s1.append(mt2[x][y])
x -= 1
y += 1
s.sort()
s1.sort()
if s != s1:
t = False
for i in range(1, m):
y = i
x = n - 1
s = []
s1 = []
while y < m and x >= 0:
s.append(mt1[x][y])
s1.append(mt2[x][y])
x -= 1
y += 1
s.sort()
s1.sort()
if s != s1:
t = False
if t:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
diag_A = []
diag_B = []
for i in range(n + m):
diag_A.append([])
diag_B.append([])
A = []
for i in range(n):
row = input().split()
A.append(row)
B = []
for i in range(n):
row = input().split()
B.append(row)
for row in range(n):
for col in range(m):
diag_A[row + col].append(A[row][col])
diag_B[row + col].append(B[row][col])
for i in range(n + m):
if sorted(diag_A[i]) != sorted(diag_B[i]):
print("NO")
break
else:
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR LIST EXPR FUNC_CALL VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
R = lambda: map(int, input().split())
n, m = R()
def f():
a = [()] * (n + m)
for i in range(n):
for x in R():
a[i] += (x,)
i += 1
return [*map(sorted, a)]
print("YNEOS"[f() != f() :: 2])
|
ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR BIN_OP LIST BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER RETURN LIST FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING FUNC_CALL VAR FUNC_CALL VAR NUMBER
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
EMPTY = 0
WHITE = 1
BLACK = 8
BOARDWIDTH = 7
BOARDHEIGHT = 6
def isWinner(board, tile):
for x in range(BOARDWIDTH - 3):
for y in range(3, BOARDHEIGHT):
if (
board[x][y] == tile
and board[x + 1][y - 1] == tile
and board[x + 2][y - 2] == tile
and board[x + 3][y - 3] == tile
):
return True
def printDiagonal(num):
aret = []
bret = []
y = d if d < M else M - 1
x = 0 if d < M else d - M
while x < N and y >= 0:
aret.append(A[x][y])
bret.append(B[x][y])
y -= 1
x += 1
aret.sort()
bret.sort()
if aret == bret:
return True
else:
return False
N, M = map(int, input().split())
A = []
for n in range(N):
A.append(list(map(int, input().split())))
B = []
for n in range(N):
B.append(list(map(int, input().split())))
for d in range(N + M):
if not printDiagonal(d):
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR RETURN NUMBER FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER BIN_OP VAR VAR WHILE VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
ii = lambda: int(input())
mi = lambda: map(int, input().split())
li = lambda: list(mi())
n, m = mi()
a = [li() for i in range(n)]
b = [li() for i in range(n)]
p = [[] for i in range(n + m)]
q = [[] for i in range(n + m)]
for i in range(n):
for j in range(m):
p[i + j].append(a[i][j])
q[i + j].append(b[i][j])
print("YES" if all(sorted(x) == sorted(y) for x, y in zip(p, q)) else "NO")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR STRING STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
import sys
def equal(arr, arr_1):
for i in range(len(arr)):
if arr[i] != arr_1[i]:
return False
return True
n, m = [int(s) for s in input().split()]
array = []
array_2 = []
for i in range(n):
array.append([int(s) for s in input().split()])
for i in range(n):
array_2.append([int(s) for s in input().split()])
diags = []
diags_2 = []
for i in range(n):
tmp = []
tmp_2 = []
j = 0
while i - j > -1 and j < m:
tmp.append(array[i - j][j])
tmp_2.append(array_2[i - j][j])
j += 1
diags.append(tmp)
diags_2.append(tmp_2)
for i in range(1, m):
tmp = []
tmp_2 = []
j = 0
while i + j < m and n - j - 1 > -1:
tmp.append(array[n - j - 1][i + j])
tmp_2.append(array_2[n - j - 1][i + j])
j += 1
diags.append(tmp)
diags_2.append(tmp_2)
for i in range(len(diags)):
diags[i] = sorted(diags[i])
diags_2[i] = sorted(diags_2[i])
for i in range(len(diags)):
if not equal(diags[i], diags_2[i]):
print("NO")
sys.exit(0)
print("YES")
|
IMPORT FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
ans = "YES"
M = [list(map(int, input().split())) for j in range(n)]
M2 = [list(map(int, input().split())) for j in range(n)]
for hehe in range(1, n + m):
start_m = max(0, hehe - n)
count = min(hehe, m - start_m, n)
dic1, dic2 = {}, {}
for j in range(0, count):
if M[min(n, hehe) - j - 1][start_m + j] in dic1:
dic1[M[min(n, hehe) - j - 1][start_m + j]] += 1
else:
dic1[M[min(n, hehe) - j - 1][start_m + j]] = 1
if M2[min(n, hehe) - j - 1][start_m + j] in dic2:
dic2[M2[min(n, hehe) - j - 1][start_m + j]] += 1
else:
dic2[M2[min(n, hehe) - j - 1][start_m + j]] = 1
if dic1 != dic2:
ans = "NO"
print(ans)
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR DICT DICT FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR VAR VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER IF VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR VAR VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = [int(x) for x in input().split()]
A = []
B = []
for i in range(n):
a = [int(x) for x in input().split()]
A.append(a)
for i in range(n):
a = [int(x) for x in input().split()]
B.append(a)
a_list = []
for i in range(n):
a = []
j = 0
a.append(A[i][j])
while 1:
if i - 1 >= 0 and j + 1 <= m - 1:
i -= 1
j += 1
a.append(A[i][j])
else:
break
a_list.append(a)
i = n - 1
for j in range(1, m):
i = n - 1
a = []
a.append(A[i][j])
while 1:
if i - 1 >= 0 and j + 1 <= m - 1:
i -= 1
j += 1
a.append(A[i][j])
else:
break
a_list.append(a)
b_list = []
for i in range(n):
a = []
j = 0
a.append(B[i][j])
while 1:
if i - 1 >= 0 and j + 1 <= m - 1:
i -= 1
j += 1
a.append(B[i][j])
else:
break
b_list.append(a)
i = n - 1
for j in range(1, m):
a = []
i = n - 1
a.append(B[i][j])
while 1:
if i - 1 >= 0 and j + 1 <= m - 1:
i -= 1
j += 1
a.append(B[i][j])
else:
break
b_list.append(a)
for i in range(len(a_list)):
a = sorted(a_list[i])
b = sorted(b_list[i])
if a != b:
print("NO")
break
else:
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR WHILE NUMBER IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR VAR VAR WHILE NUMBER IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR WHILE NUMBER IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR WHILE NUMBER IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
R = lambda: map(int, input().split())
n, m = R()
def f():
a = [[] for _ in [0] * (n + m)]
for i in range(n):
j = 0
for x in R():
a[i + j] += (x,)
j += 1
return a
print("YNEOS"[any(sorted(x) != sorted(y) for x, y in zip(f(), f())) :: 2])
|
ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR LIST VAR BIN_OP LIST NUMBER BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR STRING FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
A = [[] for i in range(n)]
B = [[] for i in range(n)]
for i in range(n):
a = list(map(int, input().split()))
A[i] = a
for i in range(n):
a = list(map(int, input().split()))
B[i] = a
A1 = [[] for i in range(n + m - 1)]
B1 = [[] for i in range(n + m - 1)]
for k in range(n + m - 1):
for i in range(n):
for j in range(m):
if i + j == k:
A1[k].append(A[i][j])
B1[k].append(B[i][j])
A1[k] = sorted(A1[k])
B1[k] = sorted(B1[k])
for l in range(len(A1[k])):
if A1[k][l] != B1[k][l]:
print("NO")
exit()
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
a = []
n, m = map(int, input().split())
for i in range(n):
a.append(input().split())
b = []
for i in range(n):
b.append(input().split())
for k in range(m):
i = k
j = 0
c = []
d = []
while i >= 0 and j <= n - 1:
c.append(a[j][i])
d.append(b[j][i])
i = i - 1
j = j + 1
c = sorted(c)
d = sorted(d)
if c != d:
print("No")
exit(0)
for k in range(1, n):
j = k
i = m - 1
c = []
d = []
while i >= 0 and j <= n - 1:
c.append(a[j][i])
d.append(b[j][i])
i = i - 1
j = j + 1
c = sorted(c)
d = sorted(d)
if c != d:
print("No")
exit(0)
print("Yes")
|
ASSIGN VAR LIST ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m, *_ = list(map(int, input().split()))
b1 = []
for i in range(n):
b1.append(input().split())
b2 = []
for i in range(n):
b2.append(input().split())
def works(n, m, b1, b2):
for diag in range(n + m - 1):
l1 = []
l2 = []
for col in range(diag + 1):
point = [col, diag - col]
if 0 <= point[0] < n and 0 <= point[1] < m:
l1.append(b1[point[0]][point[1]])
l2.append(b2[point[0]][point[1]])
if sorted(l1) != sorted(l2):
return False
return True
if works(n, m, b1, b2):
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR LIST VAR BIN_OP VAR VAR IF NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
a, b = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(a)]
B = [list(map(int, input().split())) for _ in range(a)]
for i in range(a):
A[i] += [0] * (b + a)
B[i] += [0] * (b + a)
for i in range(a - 1):
a1 = {}
b1 = {}
for j in range(i + 1):
a1[A[j][i - j]] = a1.get(A[j][i - j], 0) + 1
b1[B[j][i - j]] = b1.get(B[j][i - j], 0) + 1
if a1 != b1:
print("NO")
exit(0)
for i in range(b):
a1 = {}
b1 = {}
for j in range(a):
a1[A[j][a + i - j - 1]] = a1.get(A[j][a + i - j - 1], 0) + 1
b1[B[j][a + i - j - 1]] = b1.get(B[j][a + i - j - 1], 0) + 1
if a1 != b1:
print("NO")
exit(0)
print("YES")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP LIST NUMBER BIN_OP VAR VAR VAR VAR BIN_OP LIST NUMBER BIN_OP VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
A = []
B = []
for i in range(n):
A.append(list(map(int, input().split())))
for i in range(n):
B.append(list(map(int, input().split())))
lenkey = n - 1 + (m - 1) + 1
checkA = []
checkB = []
for i in range(lenkey):
checkA.append([])
checkB.append([])
for i in range(n):
for j in range(m):
checkA[i + j].append(A[i][j])
checkB[i + j].append(B[i][j])
checkMat = True
for i in range(lenkey):
if sorted(checkA[i]) != sorted(checkB[i]):
checkMat = False
break
if checkMat:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR LIST EXPR FUNC_CALL VAR LIST FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().split())
A = []
B = []
flag = 1
for i in range(n):
A.append(list(map(int, input().split())))
for i in range(n):
B.append(list(map(int, input().split())))
for i in range(n + m - 1):
j = 0
Alist = []
Blist = []
while i >= 0:
try:
Alist.append(A[i][j])
Blist.append(B[i][j])
except IndexError:
pass
j += 1
i -= 1
Alist.sort()
Blist.sort()
if Alist != Blist:
flag = 0
if flag == 1:
print("YES")
else:
print("NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
inp = lambda: map(int, input().split())
def ans():
global n, m
a = [[] for _ in [0] * (n + m)]
for i in range(n):
for j in inp():
a[i].append(j)
i += 1
return [*map(sorted, a)]
n, m = inp()
if ans() != ans():
print("NO")
else:
print("YES")
|
ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR LIST VAR BIN_OP LIST NUMBER BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER RETURN LIST FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR IF FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $A$ and $B$ are given, each of them has size $n \times m$. Nastya can perform the following operation to matrix $A$ unlimited number of times: take any square square submatrix of $A$ and transpose it (i.e. the element of the submatrix which was in the $i$-th row and $j$-th column of the submatrix will be in the $j$-th row and $i$-th column after transposing, and the transposed submatrix itself will keep its place in the matrix $A$).
Nastya's task is to check whether it is possible to transform the matrix $A$ to the matrix $B$.
$\left. \begin{array}{|c|c|c|c|c|c|c|c|} \hline 6 & {3} & {2} & {11} \\ \hline 5 & {9} & {4} & {2} \\ \hline 3 & {3} & {3} & {3} \\ \hline 4 & {8} & {2} & {2} \\ \hline 7 & {8} & {6} & {4} \\ \hline \end{array} \right.$ Example of the operation
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $M$ is a matrix which consist of all elements which comes from one of the rows with indeces $x, x+1, \dots, x+k-1$ of matrix $M$ and comes from one of the columns with indeces $y, y+1, \dots, y+k-1$ of matrix $M$. $k$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
-----Input-----
The first line contains two integers $n$ and $m$ separated by space ($1 \leq n, m \leq 500$) — the numbers of rows and columns in $A$ and $B$ respectively.
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $A$ ($1 \leq A_{ij} \leq 10^{9}$).
Each of the next $n$ lines contains $m$ integers, the $j$-th number in the $i$-th of these lines denotes the $j$-th element of the $i$-th row of the matrix $B$ ($1 \leq B_{ij} \leq 10^{9}$).
-----Output-----
Print "YES" (without quotes) if it is possible to transform $A$ to $B$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2 2
1 1
6 1
1 6
1 1
Output
YES
Input
2 2
4 4
4 5
5 4
4 4
Output
NO
Input
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 6
3 8 9
Output
YES
-----Note-----
Consider the third example. The matrix $A$ initially looks as follows.
$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$
Then we choose the whole matrix as transposed submatrix and it becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$
Then we transpose the submatrix with corners in cells $(2, 2)$ and $(3, 3)$.
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$
So matrix becomes
$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$
and it is $B$.
|
n, m = map(int, input().strip().split())
def transpose(matrix):
to_return = [[] for i in range(m)]
for i in range(m):
for j in range(n):
to_return[i].append(matrix[j][i])
return to_return
A, B = [], []
for i in range(n):
A.append([int(i) for i in input().strip().split()])
for i in range(n):
B.append([int(i) for i in input().strip().split()])
yes = True
pivot = 0
not_exceed = 0
if m < n:
A = transpose(A)
B = transpose(B)
m, n = n, m
index = 0
while index < m:
col_index = index
row_index = 0
numbers = dict()
while row_index < n and col_index >= 0:
if A[row_index][col_index] not in numbers:
numbers[A[row_index][col_index]] = 1
else:
numbers[A[row_index][col_index]] += 1
row_index += 1
col_index -= 1
col_index = index
row_index = 0
while row_index < n and col_index >= 0:
if B[row_index][col_index] in numbers and numbers[B[row_index][col_index]] > 0:
numbers[B[row_index][col_index]] -= 1
else:
yes = False
break
row_index += 1
col_index -= 1
index += 1
if not yes:
break
index = 1
while index < n:
col_index = m - 1
row_index = index
numbers = dict()
while row_index < n and col_index >= 0:
if A[row_index][col_index] not in numbers:
numbers[A[row_index][col_index]] = 1
else:
numbers[A[row_index][col_index]] += 1
row_index += 1
col_index -= 1
col_index = m - 1
row_index = index
while row_index < n and col_index >= 0:
if B[row_index][col_index] in numbers and numbers[B[row_index][col_index]] > 0:
numbers[B[row_index][col_index]] -= 1
else:
yes = False
break
row_index += 1
col_index -= 1
index += 1
if not yes:
break
print("YES" if yes else "NO")
|
ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR RETURN VAR ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR WHILE VAR VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR WHILE VAR VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR STRING STRING
|
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