生成一个最小和数组，其给定数组的相同索引元素的XOR为质数

给定一个N ( 1≤N≤10 ⁵ )个整数的Arr []数组，任务是生成一个由N个非零元素组成的数组B [] ，使得A _i ^ B _i的XOR总是产生质数数字。

注意：应将获得的XOR的总和最小化。

例子：

Input: arr[] = {5, 4, 7, 6} 
Output: {7, 6, 5, 4} 
Explanation: 
2 is the smallest prime number. Therefore, XORing A[i] with (A[i] ^ 2) 
gives us the smallest number which is prime. 
A[i] ^ (A[i] ^ 2) = (A[i] ^ A[i]) ^ 2 = 0 ^ 2 = 2 
because 
1. XOR of 5 ^ 7 = 2, which is prime 
2. XOR of 4 ^ 6 = 2, which is prime. 
3. XOR of 7 ^ 5 = 2, which is prime. 
4. XOR of 6 ^ 4 = 2, which is prime. 
The resultant sum is – 2 + 2 + 2 + 2 = 8, which is the minimum possible

Input: arr[] = {10, 16} 
Output: {8, 18}

方法：可以使用贪婪技术解决此问题。请按照以下步骤解决问题：

1. 由于2是可能的最小质数，因此Arr [i]与B [i] =(Arr [i] ^ 2)的XOR将给我们质数2 。
2. 当任何数组元素本身为Arr [i] = 2时，就会出现矛盾。在这种情况下， B [i] = 2 ^ 2得出0 。
3. 因此，如果Arr [i] = 2 ，则设置B [i] =(2 ^ 3)= 1 ，以使Arr [i] ^ K = 3 (下一个最小质数)。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function to generate an array whose XOR
// with same-indexed elements of the given
// array is always a prime
void minXOR(vector& Arr, int N)
{
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If current array element is 2
        if (Arr[i] == 2) {
 
            // Print its XOR with 3
            cout << (Arr[i] ^ 3) << " ";
        }
        // Otherwise
        else {
 
            // Print its XOR with 2
            cout << (Arr[i] ^ 2) << " ";
        }
    }
}
 
// Driver Code
int main()
{
    // Given array
    vector Arr = { 5, 4, 7, 6 };
 
    // Size of the array
    int N = Arr.size();
 
    // Prints the required array
    minXOR(Arr, N);
    return 0;
}

Java

// Java implementation of the above approach
class GFG{
 
// Function to generate an array whose XOR
// with same-indexed elements of the given
// array is always a prime
private static void minXOR(int Arr[], int N)
{
     
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // If current array element is 2
        if (Arr[i] == 2)
        {
             
            // Print its XOR with 3
            System.out.print((Arr[i] ^ 3) + " ");
        }
         
        // Otherwise
        else
        {
             
            // Print its XOR with 2
            System.out.print((Arr[i] ^ 2) + " ");
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given array
    int Arr[] = { 5, 4, 7, 6 };
     
    // Size of the array
    int N = Arr.length;
     
    // Prints the required array
    minXOR(Arr, N);
}
}
 
// This code is contributed by MuskanKalra1

Python3

# Python3 implementation of the above approach
  
# Function to generate an array whose XOR
# with same-indexed elements of the given
# array is always a prime
def minXOR(Arr, N):
   
    # Traverse the array
    for i in range(N):
  
        # If current array element is 2
        if (Arr[i] == 2):
  
            # Print its XOR with 3
            print(Arr[i] ^ 3,end=" ")
        # Otherwise
        else:
  
            # Print its XOR with 2
            print(Arr[i] ^ 2,end=" ")
  
# Driver Code
if __name__ == '__main__':
   
    # Given array
    Arr = [5, 4, 7, 6 ]
  
    # Size of the array
    N = len(Arr)
  
    # Prints the required array
    minXOR(Arr, N)
  
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach 
using System;
 
class GFG{
 
// Function to generate an array whose XOR
// with same-indexed elements of the given
// array is always a prime
private static void minXOR(int[] Arr, int N)
{
     
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // If current array element is 2
        if (Arr[i] == 2)
        {
             
            // Print its XOR with 3
            Console.Write((Arr[i] ^ 3) + " ");
        }
         
        // Otherwise
        else
        {
             
            // Print its XOR with 2
            Console.Write((Arr[i] ^ 2) + " ");
        }
    }
}
 
// Driver code
public static void Main()
{
     
    // Given array
    int[] Arr = { 5, 4, 7, 6 };
     
    // Size of the array
    int N = Arr.Length;
     
    // Prints the required array
    minXOR(Arr, N);
}
}

Javascript

 
// JavaScript implementation of the above approach
 
// Function to generate an array whose XOR
// with same-indexed elements of the given
// array is always a prime
function minXOR(Arr, N)
{
      
    // Traverse the array
    for(let i = 0; i < N; i++)
    {
          
        // If current array element is 2
        if (Arr[i] == 2)
        {
              
            // Print its XOR with 3
            document.write((Arr[i] ^ 3) + " ");
        }
          
        // Otherwise
        else
        {
              
            // Print its XOR with 2
            document.write((Arr[i] ^ 2) + " ");
        }
    }
}
 
// Driver code
         
    // Given array
    let Arr = [ 5, 4, 7, 6 ];
      
    // Size of the array
    let N = Arr.length;
      
    // Prints the required array
    minXOR(Arr, N);
             

输出：

7 6 5 4

时间复杂度： O(N)
辅助空间： O(1)