与M进行精确的K次异或后，检查是否保留了原始数组

给定一个数组A和两个整数M和K ，任务是检查并打印“ Yes ”，如果可以通过对数组元素与“ M ”执行精确的“ K ”个按位XOR运算来保留原始数组。否则打印“否”。

注意：可以在数组的任何元素上执行0次或多次XOR操作。

例子：

Input: A[] = {1, 2, 3, 4}, M = 5, K = 6
Output: Yes
Explanation:
If the XOR is performed on 1st element, A[0], for 6 times, we get A[0] back. Therefore, the original array is retained.

Input: A[] = {5, 9, 3, 4, 5}, M = 5, K = 3
Output: No
Explanation:
The original array cant be retained after performing odd number of XOR operations.

方法：可以使用XOR属性解决此问题

A XOR B = C and C XOR B = A

可以看出：

1. 如果对任何正数执行偶数个XOR操作，则可以保留原始数。
2. 但是，0是一个例外。如果对0进行了奇数或偶数的XOR运算，则可以保留原始数。
3. 因此，如果K为偶数且M为0 ，则答案将始终为是。
4. 如果K为奇数且数组中不存在0 ，则答案将始终为否。
5. 如果K为奇数，并且数组中的0至少为1 ，则答案为是。

下面是上述方法的实现：

C++

// C++ implementation for the
// above mentioned problem
  
#include 
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;
}

Java

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

# 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#

// 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

输出：

Yes