Given a number 
, 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 pairsint 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 codeint 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 pairsfunction 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), as we are using nested loops to traverse N*N times.
Auxiliary Space: O(1), as we are not using any extra space.
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