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

Input: N = 6
Output: 11
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to count pairs from 1 to N ` `// such that their Sum is divisible by their XOR ` ` `  `#include ` `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 `

Output:

```11
```

Time Complexity : O(N2)

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Improved By : AnkitRai01