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

Javascript

// 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));
 

输出：

Yes

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