在与M恰好K次执行异或运算后检查原始数组是否保留

// C++ implementation for the
// above mentioned problem
#include <bits/stdc++.h>
using namespace std;
// Function to check if original Array
// can be retained by performing XOR
// with M exactly K times
string check(int Arr[], int n,
             int M, int K)
{
    int flag = 0;
    // Check if O is present or not
    for (int i = 0; i < n; i++) {
        if (Arr[i] == 0)
            flag = 1;
    }
    // If K is odd and 0 is not present
    // then the answer will always be No.
    if (K % 2 != 0
        && flag == 0)
        return "No";
    // Else it will be Yes
    else
        return "Yes";
}
// Driver Code
int main()
{
    int Arr[] = { 1, 1, 2, 4, 7, 8 };
    int M = 5;
    int K = 6;
    int n = sizeof(Arr) / sizeof(Arr[0]);
    cout << check(Arr, n, M, K);
    return 0;
}


import java.util.*;
class GFG{
// Function to check if original Array
// can be retained by performing XOR
// with M exactly K times
static String check(int []Arr, int n,
            int M, int K)
{
    int flag = 0;
    // Check if O is present or not
    for (int i = 0; i < n; i++) {
        if (Arr[i] == 0)
            flag = 1;
    }
    // If K is odd and 0 is not present
    // then the answer will always be No.
    if (K % 2 != 0
        && flag == 0)
        return "No";
    // Else it will be Yes
    else
        return "Yes";
}
// Driver Code
public static void main(String args[])
{
    int []Arr = { 1, 1, 2, 4, 7, 8 };
    int M = 5;
    int K = 6;
    int n = Arr.length;
    System.out.println(check(Arr, n, M, K));
}
}
// This code is contributed by Surendra_Gangwar


# Python3 implementation for the
# above mentioned problem
# Function to check if original Array
# can be retained by performing XOR
# with M exactly K times
def check(Arr,  n, M,  K):
    flag = 0
    # Check if O is present or not
    for i in range(n):
        if (Arr[i] == 0):
            flag = 1
    # If K is odd and 0 is not present
    # then the answer will always be No.
    if (K % 2 != 0 and flag == 0):
        return "No"
    # Else it will be Yes
    else:
        return "Yes";
# Driver Code
if __name__=='__main__':
    Arr = [ 1, 1, 2, 4, 7, 8 ]
    M = 5;
    K = 6;
    n = len(Arr);
    print(check(Arr, n, M, K))
# This article contributed by Princi Singh


// C# implementation for the
// above mentioned problem
using System;
class GFG
{
// Function to check if original Array
// can be retained by performing XOR
// with M exactly K times
static String check(int []Arr, int n,int M, int K)
{
    int flag = 0;
    // Check if O is present or not
    for (int i = 0; i < n; i++) {
        if (Arr[i] == 0)
            flag = 1;
    }
    // If K is odd and 0 is not present
    // then the answer will always be No.
    if (K % 2 != 0
        && flag == 0)
        return "No";
    // Else it will be Yes
    else
        return "Yes";
}
// Driver code
public static void Main(String[] args)
{
    int []Arr = { 1, 1, 2, 4, 7, 8 };
    int M = 5;
    int K = 6;
    int n = Arr.Length;
    Console.Write(check(Arr, n, M, K));
}
}
// This code is contributed by shivanisinghss2110


<script>
// Javascript implementation for the
// above mentioned problem
// Function to check if original Array
// can be retained by performing XOR
// with M exactly K times
function check(Arr, n, M, K)
{
    let flag = 0;
    // Check if O is present or not
    for (let i = 0; i < n; i++) {
        if (Arr[i] == 0)
            flag = 1;
    }
    // If K is odd and 0 is not present
    // then the answer will always be No.
    if (K % 2 != 0
        && flag == 0)
        return "No";
    // Else it will be Yes
    else
        return "Yes";
}
// Driver Code
    let Arr = [ 1, 1, 2, 4, 7, 8 ];
    let M = 5;
    let K = 6;
    let n = Arr.length;
    document.write(check(Arr, n, M, K));
script>


Yes