计算元素对，使其 OR 中的设置位数为 B[i]

给定两个数组A[]和B[] ，每个数组有N个元素。任务是找到索引对(i, j)的数量，使得i ≤ j和F(A[i] | A[j]) = B[j]其中F(X)是设置位的计数X的二进制表示。
例子

Input: A[] = {5, 3, 2, 4, 6, 1}, B[] = {2, 2, 1, 4, 2, 3} 
Output: 7 
All possible pairs are (5, 5), (3, 3), (2, 2), 
(2, 6), (4, 6), (6, 6) and (6, 1).
Input: A[] = {4, 3, 5, 6, 7}, B[] = {1, 3, 2, 4, 5} 
Output: 4 
 

方法：遍历所有可能的对 (i, j) 并检查其 OR 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 OR value is B[j]
            if (__builtin_popcount(A[i] | A[j]) == B[j]) {
                cnt++;
            }
 
    return cnt;
}
 
// Driver code
int main()
{
    int A[] = { 5, 3, 2, 4, 6, 1 };
    int B[] = { 2, 2, 1, 4, 2, 3 };
    int size = sizeof(A) / sizeof(A[0]);
 
    cout << solve(A, B, size);
 
    return 0;
}

Java

// Java implementation of the approach
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 OR 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[] = { 5, 3, 2, 4, 6, 1 };
    int B[] = { 2, 2, 1, 4, 2, 3 };
    int size = A.length;
 
    System.out.println(solve(A, B, size));
}
}
 
// This code is contributed by 29AjayKumar

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 OR value is B[j]
            if (bin(A[i] | A[j]).count('1') == B[j]) :
                cnt += 1;
             
    return cnt
 
 
# Driver code
if __name__ == "__main__" :
 
    A = [ 5, 3, 2, 4, 6, 1 ];
    B = [ 2, 2, 1, 4, 2, 3 ];
    size = len(A);
 
    print(solve(A, B, size));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the 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 OR value is B[j]
            if (bitCount(A[i] | A[j]) == B[j])
            {
                cnt++;
            }
 
    return cnt;
}
 
static int bitCount(long x)
{
    // To store the count
    // of set bits
    int setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}
 
// Driver code
public static void Main(String []args)
{
    int []A = { 5, 3, 2, 4, 6, 1 };
    int []B = { 2, 2, 1, 4, 2, 3 };
    int size = A.Length;
 
    Console.WriteLine(solve(A, B, size));
}
}
 
/* This code is contributed by PrinciRaj1992 */

Javascript

 
// JavaScript implementation of the approach
 
// Function to return the count of pairs
// which satisfy the given condition
function solve(A,B,n)
{
    let cnt = 0;
   
    for (let i = 0; i < n; i++)
        for (let j = i; j < n; j++)
   
            // Check if the count of set bits
            // in the OR value is B[j]
            if (bitCount(A[i] | A[j]) == B[j])
            {
                cnt++;
            }
   
    return cnt;
}
 
function bitCount(x)
{
    // To store the count
    // of set bits
    let setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}
 
 
// Driver code
let A=[5, 3, 2, 4, 6, 1 ];
let B=[2, 2, 1, 4, 2, 3 ];
let size = A.length;
document.write(solve(A, B, size));
 
 
// This code is contributed by rag2127
 

输出：

7

时间复杂度： O(N ² )

辅助空间： O(1)