从给定数组生成的矩阵的子矩阵的按位XOR

给定的阵列ARR []的长度N，尺寸N * N的矩阵是在该阵列ARR定义[]其中，M _I，J =常用3 _I＆ARRĴ。给定四个整数X，Y，S和T ，任务是找到子矩阵从左上角(X，Y)到右下角(S，T)的所有元素的按位XOR。

例子：

Input: N = 3, A[] = {2, 3, 4}, (X, Y)=(0, 1), (S, T)=(2, 2)
Output: 5
Explanation:
Matrix defined on A is
{{(2&2), (2&3), (2&4)}, 
{(3&2), (3&3), (3&4)}, 
{(4&2), (4&3), (4&4)}}

Finally, the matrix will be:
{{2, 2, 0}, 
{2, 3, 0}, 
{0, 0, 4}}
XOR value= (2^0)^(3^0)^(0^4) = 5

Input: N=3, A[]={1, 2, 3}, (X, Y)=(0, 1), (S, T)=(1, 2)
Output: 1

天真的方法：最简单的方法是从给定的数组生成矩阵M ，并计算M的给定子矩阵中存在的所有元素的按位XOR。

时间复杂度： O(N ² )
辅助空间： O(N ² )

高效方法：想法是使用“ XOR”和“ AND”运算的以下分布属性：

(A & B) ^ (A & C) = A & (B ^ C)

因此，可以从下式计算子矩阵从左上角(X，Y)到右下角(S，T)的最终XOR：

Final XOR
= (XOR of row X)^(XOR of row X+1)^. . . . ^(XOR of row S)
= (A_X & (A_Y ^. . . .^ A_T)) ^ ….
. . . ^(A_S & (A_Y^. . . .^A_T))
= (A_Y^. . . .^A_T)&(A_X^. . .^A_S)

• 从索引Y到T遍历数组，然后计算元素的XOR 。
• 从索引X到S遍历数组，并计算元素的XOR。
• 最后，计算所计算的XOR的按位与，等于从(X，Y)到(S，T)的子矩阵的按位XOR。

下面是上述方法的实现：

C++

// C++ program of the
// above approach
 
#include 
using namespace std;
 
// Function to find the submatrix
// XOR of the given matrix
int submatrix_xor(int* A, int N,
                  int X, int Y,
                  int S, int T)
{
    int left_xor = 0, i, right_xor = 0;
 
    // Calculating left xor
    // i.e A[Y]^A[Y+1]^. . .^A[T]
    for (i = Y; i <= T; i++) {
        left_xor ^= A[i];
    }
 
    // Calculating right xor
    // i.e A[X]^A[X+1]^. . .^A[S]
    for (i = X; i <= S; i++) {
        right_xor ^= A[i];
    }
 
    // Bitwise AND of left_xor and
    // right_xor gives required result
    return left_xor & right_xor;
}
 
// Driver Code
int main()
{
    int A[3] = { 2, 3, 4 }, X = 0,
        Y = 1, S = 2, T = 2, N = 3;
 
    // Printing xor of submatrix
    cout << submatrix_xor(A, N, X,
                          Y, S, T);
    return 0;
}

Java

// Java program of the
// above approach
import java.io.*;
 
class GFG{
  
// Function to find the submatrix
// XOR of the given matrix
static int submatrix_xor(int[] A, int N,
                         int X, int Y,
                         int S, int T)
{
    int left_xor = 0, i, right_xor = 0;
  
    // Calculating left xor
    // i.e A[Y]^A[Y+1]^. . .^A[T]
    for(i = Y; i <= T; i++)
    {
        left_xor ^= A[i];
    }
  
    // Calculating right xor
    // i.e A[X]^A[X+1]^. . .^A[S]
    for(i = X; i <= S; i++)
    {
        right_xor ^= A[i];
    }
  
    // Bitwise AND of left_xor and
    // right_xor gives required result
    return left_xor & right_xor;
}
  
// Driver Code
public static void main (String[] args)
{
    int[] A = { 2, 3, 4 };
    int X = 0, Y = 1, S = 2,
        T = 2, N = 3;
  
    // Printing xor of submatrix
    System.out.print(submatrix_xor(A, N, X,
                                   Y, S, T));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program of the
# above approach
 
# Function to find the submatrix
# XOR of the given matrix
def submatrix_xor(A, N, X, Y, S, T):
     
    left_xor = 0
    i = 0
    right_xor = 0
 
    # Calculating left xor
    # i.e A[Y]^A[Y+1]^. . .^A[T]
    for i in range(Y, T + 1):
        left_xor ^= A[i]
 
    # Calculating right xor
    # i.e A[X]^A[X+1]^. . .^A[S]
    for i in range(X, S + 1):
        right_xor ^= A[i]
 
    # Bitwise AND of left_xor and
    # right_xor gives required result
    return left_xor & right_xor
 
# Driver Code
if __name__ == '__main__':
     
    A = [ 2, 3, 4 ]
    X = 0
    Y = 1
    S = 2
    T = 2
    N = 3
 
    # Printing xor of submatrix
    print(submatrix_xor(A, N, X, Y, S, T))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach 
using System;
 
class GFG{
  
// Function to find the submatrix
// XOR of the given matrix
static int submatrix_xor(int[] A, int N,
                         int X, int Y,
                         int S, int T)
{
    int left_xor = 0, i, right_xor = 0;
  
    // Calculating left xor
    // i.e A[Y]^A[Y+1]^. . .^A[T]
    for(i = Y; i <= T; i++)
    {
        left_xor ^= A[i];
    }
  
    // Calculating right xor
    // i.e A[X]^A[X+1]^. . .^A[S]
    for(i = X; i <= S; i++)
    {
        right_xor ^= A[i];
    }
  
    // Bitwise AND of left_xor and
    // right_xor gives required result
    return left_xor & right_xor;
}
  
// Driver Code
public static void Main ()
{
    int[] A = { 2, 3, 4 };
    int X = 0, Y = 1, S = 2,
        T = 2, N = 3;
  
    // Printing xor of submatrix
    Console.Write(submatrix_xor(A, N, X,
                                Y, S, T));
}
}
 
// This code is contributed by code_hunt

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to find the submatrix
// XOR of the given matrix
function submatrix_xor(A, N, X, Y, S, T)
{
    let left_xor = 0, i, right_xor = 0;
   
    // Calculating left xor
    // i.e A[Y]^A[Y+1]^. . .^A[T]
    for(i = Y; i <= T; i++)
    {
        left_xor ^= A[i];
    }
   
    // Calculating right xor
    // i.e A[X]^A[X+1]^. . .^A[S]
    for(i = X; i <= S; i++)
    {
        right_xor ^= A[i];
    }
   
    // Bitwise AND of left_xor and
    // right_xor gives required result
    return left_xor & right_xor;
}
 
// Driver code
    let A = [ 2, 3, 4 ];
    let X = 0, Y = 1, S = 2,
        T = 2, N = 3;
   
    // Printing xor of submatrix
    document.write(submatrix_xor(A, N, X,
                                   Y, S, T));
 
// This code is contributed by target_2.

输出：

5

时间复杂度： O(N)，其中N是数组的大小
辅助空间： O(1)