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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define REP(i, m, n) for (int i = (m); i < (int)(n); i++)
#define REPS(i, m, n) for (int i = (m); i <= (int)(n); i++)
#define rep(i, n) REP(i, 0, n)
#define reps(i, n) for (int i = 0; i <= (int)(n); i++)
#define rrep(i, x) for (int i = ((int)(x)-1); i >= 0; i--)
#define rreps(i, x) for (int i = (int)(x); i >= 0; i--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef pair<int, int> P;
const int inf = INT_MAX;
const ll INF = 1LL << 60;
const ll mod = 1e9 + 7;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename A, size_t N, typename T> void Fill(A (&array)[N], const T &val) { fill( (T*)array, (T*)(array+N), val ); }

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

    int n, k; cin >> n >> k;
    vector<ll> num(k);
    // nまででkで割ってi%k個余る数の個数がnum[i%k]個
    REPS(i, 1, n) num[i%k]++;
    ll ans = 0;
    rep(a, k) {
        int b = (k - a) % k;
        int c = (k - a) % k;
        if ((b + c) % k != 0) continue;
        ans += num[a] * num[b] * num[c];
    }
    cout << ans << endl;
    return 0;
}