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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <deque>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <list>
#include <unordered_set>
#include <tuple>
#include <cmath>
#include <limits>
#include <type_traits>
#include <iomanip>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <set>
#include <bitset>
#include <regex>
#include <random>
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
#define rep(i,n)for(ll i=0;i<n;++i)
#define exout(x) printf("%.10f\n", x)
const double pi = acos(-1.0);
const ll MOD = 1000000007;
const ll INF = 1e18;
const ll MAX_N = 201010;

// 組み合わせの余りを求める
ll fac[MAX_N], finv[MAX_N], inv[MAX_N];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX_N; i++) {
		fac[i] = fac[i - 1] * i;
	}
}

// 二項係数計算
long long COM(ll n, ll k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] / (fac[k] * fac[n - k]);
}


ll dx[4] = { 0,0,-1,1 };
ll dy[4] = { -1,1,0,0 };
char c[310][310];
ll dp[101];

//long longしか使わない
//素数は1より大きい
//lower_boundは指定したkey以上の要素の一番左のイテレータをかえす
//upper_boundは指定したkeyより大きい要素の一番左のイテレータをかえす
int main() {
	ll n, k;
	cin >> n >> k;
	ll res = 0;
	ll amari = 0;
	for (ll i = 1;i <= n;++i) {
		if (i % k == 0)res++;
		if (k%2==0) {
			if (i % k == k / 2)amari++;
		}
	}
	ll ans = 0;
	ans += res * res * res;
	if (k % 2 == 0) {
		ans += amari * amari * amari;
	}
	cout << ans << endl;
	return 0;
}