检查是否可以从每一行中选择一个数字，以使数字的异或大于零

给定一个有序NXM数组元素的二维数组，任务是检查我们是否可以从每一行中选择一个数字，使所选数字的xor大于0 。

注意：至少有2行。

例子：

Input: a[][] = {{7, 7, 7}, 
                {10, 10, 7}} 
Output: Yes

Input: a[][] = {{1, 1, 1},
                {1, 1, 1}, 
                {1, 1, 1}, 
                {1, 1, 1}} 
Output: No 

方法：最初检查每行的第一列元素的xor是否为0。如果它不为零，则有可能。如果它为零，请检查是否有任何行具有两个或更多不同的元素，那么这也是可能的。如果以上两个条件均不满足，则不可能。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
  
#include 
using namespace std;
#define N 2
#define M 3
  
// Function to check if a number from every row 
// can be selected such that xor of the numbers 
// is greater than zero
bool check(int mat[N][M])
{
    int xorr = 0;
  
    // Find the xor of first
    // column for every row
    for (int i = 0; i < N; i++) {
        xorr ^= mat[i][0];
    }
  
    // If Xorr is 0
    if (xorr != 0)
        return true;
  
    // Traverse in the matrix
    for (int i = 0; i < N; i++) {
        for (int j = 1; j < M; j++) {
  
            // Check is atleast
            // 2 distinct elements
            if (mat[i][j] != mat[i][0])
                return true;
        }
    }
  
    return false;
}
  
// Driver code
int main()
{
    int mat[N][M] = { { 7, 7, 7 },
                      { 10, 10, 7 } };
  
    if (check(mat))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
  
class GFG 
{
    static int N = 2;
    static int M = 3;
      
    // Function to check if a number 
    // from every row can be selected 
    // such that xor of the numbers 
    // is greater than zero
    static boolean check(int mat[][])
    {
        int xorr = 0;
      
        // Find the xor of first
        // column for every row
        for (int i = 0; i < N; i++)
        {
            xorr ^= mat[i] [0];
        }
      
        // If Xorr is 0
        if (xorr != 0)
            return true;
      
        // Traverse in the matrix
        for (int i = 0; i < N; i++) 
        {
            for (int j = 1; j < M; j++) 
            {
      
                // Check is atleast
                // 2 distinct elements
                if (mat[i] [j] != mat[i] [0])
                    return true;
            }
        }
      
        return false;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
          
        int mat[][] = {{ 7, 7, 7 },
                    { 10, 10, 7 }};
      
        if (check(mat))
            System.out.println("Yes");
        else
            System.out.println("No");
  
    }
}
  
// This code is contributed by ajit

Python3

# Python3 program to implement
# the above approach
N = 2
M = 3
  
# Function to check if a number from every row 
# can be selected such that xor of the numbers 
# is greater than zero
def check(mat):
  
    xorr = 0
  
    # Find the xor of first
    # column for every row
    for i in range(N):
        xorr ^= mat[i][0]
  
    # If Xorr is 0
    if (xorr != 0):
        return True
  
    # Traverse in the matrix
    for i in range(N):
        for j in range(1, M):
  
            # Check is atleast
            # 2 distinct elements
            if (mat[i][j] != mat[i][0]):
                return True
          
    return False
  
# Driver code
mat = [[ 7, 7, 7 ],
       [ 10, 10, 7 ]]
  
if (check(mat)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by mohit kumar

C#

// C# program to implement
// the above approach
using System;
  
class GFG
{
    static int N = 2;
    static int M = 3;
      
    // Function to check if a number 
    // from every row can be selected 
    // such that xor of the numbers 
    // is greater than zero
    static bool check(int [,]mat)
    {
        int xorr = 0;
      
        // Find the xor of first
        // column for every row
        for (int i = 0; i < N; i++)
        {
            xorr ^= mat[i, 0];
        }
      
        // If Xorr is 0
        if (xorr != 0)
            return true;
      
        // Traverse in the matrix
        for (int i = 0; i < N; i++) 
        {
            for (int j = 1; j < M; j++) 
            {
      
                // Check is atleast
                // 2 distinct elements
                if (mat[i, j] != mat[i, 0])
                    return true;
            }
        }
      
        return false;
    }
      
    // Driver code
    static void Main()
    {
        int [,]mat = {{ 7, 7, 7 },
                      { 10, 10, 7 }};
      
        if (check(mat))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
  
// This code is contributed by mits

输出：

Yes