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

  • Last Updated : 04 Mar, 2021

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:  

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  • 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>
Output: 
11

 

Time Complexity: O(N2)
Auxiliary Space: O(1)




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