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 C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <cstdio>
#include <deque>
#include <iomanip>
#include <iostream>
#include <queue>
#include <stack>
#include <utility>
#include <vector>
using std::cin;
using std::cout;
using std::endl;

int main(void) {
    cout << std::fixed << std::setprecision(10);
    cin.tie(0);
    std::ios::sync_with_stdio(false);
    int n, k;
    int64_t result = 0;
    cin >> n >> k;
    int secnum, length;
    for (int i = 1; i <= n; i++) {
        if (i % k == 0) {
            secnum = k;
            length = n - secnum;
            result += (length / k + 1) * (length / k + 1);
        } else if (k != 2 && k % 2 == 0 && i % k == k / 2) {
            secnum = k / 2;
            length = n - secnum;
            result += (length / k + 1) * (length / k + 1);
        } else {
            secnum  = k - (i % k);
            int mod = secnum % k;
            if ((mod * 2) % k == 0) {
                length = n - secnum;
                result += (length / k + 1) * (length / k + 1);
            }
        }
        // cout << result << endl;
    }

    cout << result << endl;

    return 0;
}