Count pairs from 1 to N such that their Sum is divisible by their XOR

Given a number $N$ , the task is to count pairs (x, y) such that their sum (x+y) is divisible by their xor value (x^y) and the condition 1 ? x < y ? N holds true.
Examples:

Input: N = 3
Output: 3
Explanation: 
(1, 2), (1, 3), (2, 3) are the valid pairs

Input: N = 6
Output: 11

Approach:

After taking the array as input, first we need to find out all the possible pairs in that array.
So, find out the pairs from the array
Then for each pair, check whether the sum of the pair is divisible by the xor value of the pair. If it is, then increase the required count by one.
When all the pairs have been checked, return or print the count of such pair.

Below is the implementation of the above approach:

C++

// C++ program to count pairs from 1 to N
// such that their Sum is divisible by their XOR
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count pairs
int countPairs(int n)
{
    // variable to store count
    int count = 0;
 
    // Generate all possible pairs such that
    // 1 <= x < y < n
    for (int x = 1; x < n; x++) {
        for (int y = x + 1; y <= n; y++) {
            if ((y + x) % (y ^ x) == 0)
                count++;
        }
    }
 
    return count;
}
 
// Driver code
int main()
{
    int n = 6;
 
    cout << countPairs(n);
 
    return 0;
}

Java

// Java program to count pairs from 1 to N
// such that their Sum is divisible by their XOR
class GFG
{
     
    // Function to count pairs
    static int countPairs(int n)
    {
        // variable to store count
        int count = 0;
     
        // Generate all possible pairs such that
        // 1 <= x < y < n
        for (int x = 1; x < n; x++)
        {
            for (int y = x + 1; y <= n; y++)
            {
                if ((y + x) % (y ^ x) == 0)
                    count++;
            }
        }
        return count;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 6;
        System.out.println(countPairs(n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 program to count pairs from 1 to N
# such that their Sum is divisible by their XOR
 
# Function to count pairs
def countPairs(n) :
 
    # variable to store count
    count = 0;
 
    # Generate all possible pairs such that
    # 1 <= x < y < n
    for x in range(1, n) :
        for y in range(x + 1, n + 1) :
            if ((y + x) % (y ^ x) == 0) :
                count += 1;
 
    return count;
 
# Driver code
if __name__ == "__main__" :
 
    n = 6;
 
    print(countPairs(n));
 
# This code is contributed by AnkitRai01

C#

// C# program to count pairs from 1 to N
// such that their Sum is divisible by their XOR
using System;
 
public class GFG
{
     
    // Function to count pairs
    static int countPairs(int n)
    {
        // variable to store count
        int count = 0;
     
        // Generate all possible pairs such that
        // 1 <= x < y < n
        for (int x = 1; x < n; x++)
        {
            for (int y = x + 1; y <= n; y++)
            {
                if ((y + x) % (y ^ x) == 0)
                    count++;
            }
        }
        return count;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 6;
        Console.WriteLine(countPairs(n));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
 
// JavaScript program to count pairs from 1 to N
// such that their Sum is divisible by their XOR
 
// Function to count pairs
function countPairs(n)
{
    // variable to store count
    let count = 0;
 
    // Generate all possible pairs such that
    // 1 <= x < y < n
    for (let x = 1; x < n; x++) {
        for (let y = x + 1; y <= n; y++) {
            if ((y + x) % (y ^ x) == 0)
                count++;
        }
    }
 
    return count;
}
 
// Driver code
    let n = 6;
 
    document.write(countPairs(n));
 
// This code is contributed by Surbhi Tyagi.
 
</script>