Problem:
Score : 300 points

Problem Statement

You are given integers N and K. Find the number of triples (a,b,c) of positive integers not greater than N such that a+b,b+c and c+a are all multiples of K.
The order of a,b,c does matter, and some of them can be the same.

Constraints

1 \leq N,K \leq 2\times 10^5

N and K are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of triples (a,b,c) of positive integers not greater than N such that a+b,b+c and c+a are all multiples of K.

Sample Input 1

3 2

Sample Output 1

9

(1,1,1),(1,1,3),(1,3,1),(1,3,3),(2,2,2),(3,1,1),(3,1,3),(3,3,1) and (3,3,3) satisfy the condition.

Sample Input 2

5 3

Sample Output 2

1

Sample Input 3

31415 9265

Sample Output 3

27

Sample Input 4

35897 932

Sample Output 4

114191

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int N = sc.nextInt();
        int K = sc.nextInt();


        int count = 0;
        for (int a = 1; a < N + 1; a++) {
            int modA = (a % K);  // aの余り
            int modB = K - modA; // a + bはKで割り切れる必要があるので bの余りはK-aの余り
            int modC = K - modA; // b + cはKで割り切れる必要があるので cの余りはK-bの余り

            if ((modB + modC) % K == 0) {
                // N以下でKで割るとmodBになるBの個数 x N以下でKで割るとmodCになるCの個数
                int b = ((N + K - modB) / K);
                int c = ((N + K - modC) / K);
                b = (modB == 0) ? b - 1 : b;
                c = (modC == 0) ? c - 1 : c;
                int num = b * c;
                count += num;
            }
        }
        System.out.println(count);
    }
}