两个数组中所有对的按位与的总和

给定分别为N和M的两个数组A []和B [] ，任务是从这两个数组中查找所有可能的无序对(A [i]，B [j])的按位与之和。

例子：

Input: A[] = {1, 2} , B[] = {3, 4} 
Output: 3 
Explanation: 
Bitwise AND of all possible pairs are 
1 & 3  = 1 
1 & 4 = 0 
2 & 3 = 2 
2 & 4 = 0 
Therefore, the sum of bitwise AND of all possible pairs are = (1 + 0 + 2 + 0) = 3

Input: A[] = {4, 6, 0, 0, 3, 3}, B[] = {0, 5, 6, 5, 0, 3} 
Output: 42

方法：解决问题的方法是遍历两个数组，并从给定的两个数组生成所有可能的对，并继续添加它们各自的按位AND 。最后，打印从两个给定数组中获得的所有可能对(A [i]，B [j])的按位与之和。

请按照以下步骤解决问题：

• 初始化一个变量，例如， pairsAndSum，以存储所有可能对的按位与之和。
• 遍历两个数组并从给定的两个数组生成所有可能的对。
• 最后，从两个数组中计算所有可能的对的按位与之和，然后打印出总和。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find the sum of
// AND of all possible pair
int sumOfAnd(int A[], int B[],
                int N, int M)
{
 
    // Stores sum of bitwise AND
    // of  all possible pair
    int pairsAndSum = 0;
 
    // Traverse the array A[]
    for (int i = 0; i < N; i++) {
         
       // Traverse the array B[]
        for (int j = 0; j < M;
                           j++) {
 
            // Update pairsAndSum
            pairsAndSum +=
                   (A[i] & B[j]);
        }
    }
     
    return pairsAndSum;
}
 
// Driver Code
int main()
{
 
    int A[] = { 4, 6, 0, 0, 3, 3 };
    int B[] = { 0, 5, 6, 5, 0, 3 };
 
    int N = sizeof(A) / sizeof(A[0]);
 
    int M = sizeof(B) / sizeof(B[0]);
     
    cout << sumOfAnd(A, B, N, M);
 
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to find the sum of
// AND of all possible pair
static int sumOfAnd(int A[], int B[],
                    int N, int M)
{
  // Stores sum of bitwise AND
  // of  all possible pair
  int pairsAndSum = 0;
 
  // Traverse the array A[]
  for (int i = 0; i < N; i++)
  {
    // Traverse the array B[]
    for (int j = 0; j < M; j++)
    {
      // Update pairsAndSum
      pairsAndSum += (A[i] & B[j]);
    }
  }
 
  return pairsAndSum;
}
 
// Driver Code
public static void main(String[] args)
{
  int A[] = {4, 6, 0, 0, 3, 3};
  int B[] = {0, 5, 6, 5, 0, 3};
  int N = A.length;
  int M = B.length;
  System.out.print(sumOfAnd(A, B,
                            N, M));
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program to implement
# the above approach
 
# Function to find the sum of
# AND of all possible pair
def sumOfAnd(A, B, N, M):
     
    # Stores sum of bitwise AND
    # of all possible pair
    pairsAndSum = 0
 
    # Traverse the array A
    for i in range(N):
         
        # Traverse the array B
        for j in range(M):
             
            # Update pairsAndSum
            pairsAndSum += (A[i] & B[j])
 
    return pairsAndSum
 
# Driver Code
if __name__ == '__main__':
     
    A = [ 4, 6, 0, 0, 3, 3 ]
    B = [ 0, 5, 6, 5, 0, 3 ]
     
    N = len(A)
    M = len(B)
     
    print(sumOfAnd(A, B, N, M))
 
# This code is contributed by Amit Katiyar

C#

// C# Program to implement
// the above approach
using System;
class GFG{
 
// Function to find the sum of
// AND of all possible pair
static int sumOfAnd(int []A, int []B,
                    int N, int M)
{
  // Stores sum of bitwise AND
  // of  all possible pair
  int pairsAndSum = 0;
 
  // Traverse the array []A
  for (int i = 0; i < N; i++)
  {
    // Traverse the array []B
    for (int j = 0; j < M; j++)
    {
      // Update pairsAndSum
      pairsAndSum += (A[i] & B[j]);
    }
  }
 
  return pairsAndSum;
}
 
// Driver Code
public static void Main(String[] args)
{
  int []A = {4, 6, 0, 0, 3, 3};
  int []B = {0, 5, 6, 5, 0, 3};
  int N = A.Length;
  int M = B.Length;
  Console.Write(sumOfAnd(A, B,
                            N, M));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript Program to implement
// the above approach
 
// Function to find the sum of
// AND of all possible pair
function sumOfAnd(A, B, N, M)
{
 
    // Stores sum of bitwise AND
    // of all possible pair
    var pairsAndSum = 0;
 
    // Traverse the array A[]
    for (var i = 0; i < N; i++) {
         
    // Traverse the array B[]
        for (var j = 0; j < M;
                        j++) {
 
            // Update pairsAndSum
            pairsAndSum +=
                (A[i] & B[j]);
        }
    }
     
    return pairsAndSum;
}
 
// Driver Code
var A = [ 4, 6, 0, 0, 3, 3 ];
var B = [ 0, 5, 6, 5, 0, 3 ];
var N = A.length;
var M = B.length;
 
document.write(sumOfAnd(A, B, N, M));
 

输出：

42

时间复杂度： O(N ² )
辅助空间： O(1)