Divisible Sum Pairs - HackerRank || HackerRank Divisible Sum Pairs Solution

#include <stdio.h>

int main()
{
    int n, k, i, j, sum = 0, result = 0;
    scanf("%d %d", &n, &k);
    int array[n];
    for (i = 0; i < n; i++) scanf("%d", &array[i]);
    for (i = 0; i < n; i++)
    {
        for (j = i + 1; j < n; j++)
        {
            sum = array[i] + array[j];
            if (sum % k == 0) result++;
        }
    }
    printf("%d\n", result);

    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int n, k, i, j, sum = 0, result = 0;
    cin >> n >> k;
    int array[n];
    for (i = 0; i < n; i++) cin >> array[i];
    for (i = 0; i < n; i++)
    {
        for (j = i + 1; j < n; j++)
        {
            sum = array[i] + array[j];
            if (sum % k == 0) result++;
        }
    }
    cout << result << endl;

    return 0;
}

import java.util.Scanner;

public class Divisible_Sum_Pairs {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int k = input.nextInt();
        int[] array = new int[n];
        int count = 0;
        for (int i = 0; i < n; i++) array[i] = input.nextInt();
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) if ((array[i] + array[j]) % k == 0) count++;
        }
        System.out.println(count);
    }
}