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

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 = 3Output: 3Explanation: (1, 2), (1, 3), (2, 3) are the valid pairsInput: N = 6Output: 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>` |

**Output:**

11

**Time Complexity:** O(N^{2}), as we are using nested loops to traverse N*N times.**Auxiliary Space:** O(1), as we are not using any extra space.