# Number of pairs from the first N natural numbers whose sum is divisible by K in C++

C++Server Side ProgrammingProgramming

Given numbers N and K, we have to count the number of pairs whose sum is divisible by K. Let's see an example.

Input

N = 3
K = 2

Output

1


There is only one pair whose sum is divisible by K. And the pair is (1, 3).

## Algorithm

• Initialise the N and K.
• Generate the natural numbers till N and store them in an array.
• Initialise the count to 0.
• Write two loops to get all pairs from the array.
• Compute the sum of every pair.
• If the pair sum is divisible by K, then increment the count.
• Return the count.

## Implementation

Following is the implementation of the above algorithm in C++

#include <bits/stdc++.h>
using namespace std;
int get2PowersCount(vector<int> arr, int N, int K) {
int count = 0;
for (int i = 0; i < N; i++) {
for (int j = i + 1; j < N; j++) {
int sum = arr[i] + arr[j];
if (sum % K == 0) {
count++;
}
}
}
return count;
}
int main() {
vector<int> arr;
int N = 10, K = 5;
for (int i = 1; i <= N; i++) {
arr.push_back(i);
}
cout << get2PowersCount(arr, N, K) << endl;
return 0;
}

## Output

If you run the above code, then you will get the following result.

9