按位或等于 K 计算给定数组中的对

给定一个由N 个正整数和一个整数K组成的数组arr[] ，任务是计算给定数组中所有可能的对，其中按位或等于K 。

例子：

Input: arr[] = {2, 38, 44, 29, 62}, K = 46
Output: 2
Explanation: Only the following two pairs are present in the array whose Bitwise OR is 46: 

• 2 OR 44 = 46
• 38 OR 44 = 46

Input: arr[] = {1, 5, 20, 15, 14}, K = 20
Output: 5
Explanation:
There are only 5 pairs whose Bitwise OR is 20: 

• 1 OR 15 = 15
• 1 OR 14 = 15
• 5 OR 15 = 15
• 5 OR 14 = 15
• 15 OR 14 = 15

方法：为了解决这个问题，我们的想法是从给定的数组中生成所有可能的对，并计算按位 OR 等于K 的那些对。检查所有对后，打印存储的计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function that counts the pairs from
// the array whose Bitwise OR is K
void countPairs(int arr[], int k, int size)
{
    // Stores the required
    // count of pairs
    int count = 0, x;
 
    // Generate all possible pairs
    for (int i = 0; i < size - 1; i++) {
 
        for (int j = i + 1; j < size; j++) {
 
            // Perform OR operation
            x = arr[i] | arr[j];
 
            // If Bitwise OR is equal
            // to K, increment count
            if (x == k)
                count++;
        }
    }
 
    // Print the total count
    cout << count;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 38, 44, 29, 62 };
    int K = 46;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    countPairs(arr, K, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
  
class GFG{
  
// Function that counts the pairs from
// the array whose Bitwise OR is K
static void countPairs(int[] arr, int k,
                       int size)
{
     
    // Stores the required
    // count of pairs
    int count = 0, x;
  
    // Generate all possible pairs
    for(int i = 0; i < size - 1; i++)
    {
        for(int j = i + 1; j < size; j++)
        {
             
            // Perform OR operation
            x = arr[i] | arr[j];
  
            // If Bitwise OR is equal
            // to K, increment count
            if (x == k)
                count++;
        }
    }
  
    // Print the total count
    System.out.println(count);
}
  
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 2, 38, 44, 29, 62 };
    int K = 46;
    int N = arr.length;
  
    // Function Call
    countPairs(arr, K, N);
}
}
  
// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
  
# Function that counts the pairs from
# the array whose Bitwise OR is K
def countPairs(arr, k, size):
     
    # Stores the required
    # count of pairs
    count = 0
  
    # Generate all possible pairs
    for i in range(size - 1):
        for j in range(i + 1, size):
             
            # Perform OR operation
            x = arr[i] | arr[j]
  
            # If Bitwise OR is equal
            # to K, increment count
            if (x == k):
                count += 1
  
    # Print the total count
    print(count)
 
# Driver Code
arr = [ 2, 38, 44, 29, 62 ]
K = 46
N = len(arr)
  
# Function Call
countPairs(arr, K, N)
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
   
class GFG{
   
// Function that counts the pairs from
// the array whose Bitwise OR is K
static void countPairs(int[] arr, int k,
                       int size)
{
     
    // Stores the required
    // count of pairs
    int count = 0, x;
   
    // Generate all possible pairs
    for(int i = 0; i < size - 1; i++)
    {
        for(int j = i + 1; j < size; j++)
        {
             
            // Perform OR operation
            x = arr[i] | arr[j];
             
            // If Bitwise OR is equal
            // to K, increment count
            if (x == k)
                count++;
        }
    }
     
    // Print the total count
    Console.WriteLine(count);
}
   
// Driver Code
public static void Main()
{
    int[] arr = { 2, 38, 44, 29, 62 };
    int K = 46;
    int N = arr.Length;
     
    // Function Call
    countPairs(arr, K, N);
}
}
   
// This code is contributed by susmitakundugoaldanga

Javascript

// JavaScript program for the above approach
 
// Function that counts the pairs from
// the array whose Bitwise OR is K
function countPairs(arr, k, size)
{
      
    // Stores the required
    // count of pairs
    let count = 0, x;
   
    // Generate all possible pairs
    for(let i = 0; i < size - 1; i++)
    {
        for(let j = i + 1; j < size; j++)
        {
              
            // Perform OR operation
            x = arr[i] | arr[j];
   
            // If Bitwise OR is equal
            // to K, increment count
            if (x == k)
                count++;
        }
    }
   
    // Prlet the total count
    document.write(count);
}
  
// Driver Code
       let arr = [ 2, 38, 44, 29, 62 ];
    let K = 46;
    let N = arr.length;
   
    // Function Call
    countPairs(arr, K, N);
    
   // This code is contributed by avijitmondal998.

输出：

2

时间复杂度： O(N ² )
辅助空间： O(1)

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