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:

Below is the implementation of the above approach:

C++



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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

11

Time Complexity : O(N2)

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Improved By : AnkitRai01