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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;

/*
 * kが奇数の場合：条件を満たす３数はすべてkの倍数である
 * kが偶数の場合：条件を満たす３数はすべてkの倍数かすべてkの剰余がk/2となる数である
 */

int n;
int k;

long oddK() {
	int num = n / k;
	return (long)pow((double)num, 3);
}

long evenK() {
	int num1 = n / k;
	int num2;
	if (n % k < k / 2) num2 = n / k;
	else num2 = (n / k) + 1;

	return (long)pow((double)num1, 3) + (int)pow((double)num2, 3);
}

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n >> k;

	long result;
	if (k % 2 == 1) result = oddK();
	else result = evenK();

	cout << result << "\n";
	return 0;
}