对元素对进行计数，以使其AND中的置位位数为B [i]

给定两个数组，每个数组分别包含N个元素的A []和B [] 。任务是找到索引对(i，j)的数量，以使i≤j且F(A [i]＆A [j])= B [j] ，其中F(X)是其中的设置位数X的二进制表示形式。
例子：

Input: A[] = {2, 3, 1, 4, 5}, B[] = {2, 2, 1, 4, 2} 
Output: 4 
All possible pairs are (3, 3), (3, 1), (1, 1) and (5, 5)
Input: A[] = {1, 2, 3, 4, 5}, B[] = {2, 2, 2, 2, 2} 
Output: 2 
 

方法：遍历所有可能的对(i，j)，并检查其AND值中设置位的计数。如果计数等于B [j]，则增加计数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count of pairs
// which satisfy the given condition
int solve(int A[], int B[], int n)
{
    int cnt = 0;
 
    for (int i = 0; i < n; i++)
        for (int j = i; j < n; j++)
 
            // Check if the count of set bits
            // in the AND value is B[j]
            if (__builtin_popcount(A[i] & A[j]) == B[j]) {
                cnt++;
            }
 
    return cnt;
}
 
// Driver code
int main()
{
    int A[] = { 2, 3, 1, 4, 5 };
    int B[] = { 2, 2, 1, 4, 2 };
    int size = sizeof(A) / sizeof(A[0]);
 
    cout << solve(A, B, size);
 
    return 0;
}

Java

// Java implementation of the approach
public class GFG
{
 
    // Function to return the count of pairs
    // which satisfy the given condition
    static int solve(int A[], int B[], int n)
    {
        int cnt = 0;
 
        for (int i = 0; i < n; i++)
        {
            for (int j = i; j < n; j++) // Check if the count of set bits
            // in the AND value is B[j]
            {
                if (Integer.bitCount(A[i] & A[j]) == B[j])
                {
                    cnt++;
                }
            }
        }
 
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int A[] = {2, 3, 1, 4, 5};
        int B[] = {2, 2, 1, 4, 2};
        int size = A.length;
 
        System.out.println(solve(A, B, size));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
 
# Function to return the count of pairs
# which satisfy the given condition
def solve(A, B, n) :
    cnt = 0;
 
    for i in range(n) :
        for j in range(i, n) :
 
            # Check if the count of set bits
            # in the AND value is B[j]
            c = A[i] & A[j]
            if (bin(c).count('1') == B[j]) :
                cnt += 1;
    return cnt;
 
# Driver code
if __name__ == "__main__" :
 
    A = [ 2, 3, 1, 4, 5 ];
    B = [ 2, 2, 1, 4, 2 ];
     
    size = len(A);
 
    print(solve(A, B, size));
 
# This code is contributed
# by AnkitRai01

C#

// C# Implementation of the above approach
using System;
 
class GFG
{
 
    // Function to return the count of pairs
    // which satisfy the given condition
    static int solve(int []A, int []B, int n)
    {
        int cnt = 0;
 
        for (int i = 0; i < n; i++)
        {
            for (int j = i; j < n; j++)
            // Check if the count of set bits
            // in the AND value is B[j]
            {
                if (countSetBits(A[i] & A[j]) == B[j])
                {
                    cnt++;
                }
            }
        }
 
        return cnt;
    }
     
    // Function to get no of set
    // bits in binary representation
    // of positive integer n
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0)
        {
            count += n & 1;
            n >>= 1;
        }
        return count;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int []A = {2, 3, 1, 4, 5};
        int []B = {2, 2, 1, 4, 2};
        int size = A.Length;
 
        Console.WriteLine(solve(A, B, size));
    }
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript Implementation of the above approach
 
// Function to return the count of pairs
// which satisfy the given condition
function solve(A, B, n)
{
    var cnt = 0;
    for (var i = 0; i < n; i++)
    {
        for (var j = i; j < n; j++)
         
        // Check if the count of set bits
        // in the AND value is B[j]
        {
            if (countSetBits(A[i] & A[j]) == B[j])
            {
                cnt++;
            }
        }
    }
    return cnt;
}
 
// Function to get no of set
// bits in binary representation
// of positive integer n
function countSetBits(n)
{
    var count = 0;
    while (n > 0)
    {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
// Driver code
var A = [2, 3, 1, 4, 5];
var B = [2, 2, 1, 4, 2];
var size = A.length;
document.write(solve(A, B, size));
 
// This code is contributed by rutvik_56.

输出：

4