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 <algorithm>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <bitset>
# include <complex>
#include<limits.h>
#include<random>
#include<deque>
#include<cstdio>
#include<fstream>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 2000100;
const int MOD = (int)pow(10, 9) + 7;
const ll INF = 1LL << 60;
#define rep(i,n) for(ll i=(ll)0;i<(ll)n;++i)
#define ALL(x) x.begin(),x.end()
#define pp pair<ll,ll>
#define ppp pair<ll,pp>
#define pps pair<string ,ll>
#define fi first
#define se second
#define pb push_back
#define fix(n) fixed<<setprecision(n)
#define itr vector<ll>::iterator
#define print(s) cout<<s<<endl
#define inp(s) cin>>s
string yn(bool b) { return (b ? "Yes" : "No"); }
string YN(bool b) { return (b ? "YES" : "NO"); }
ll ans,n,v[100010],k;
int main() {
	cin >> n>>k;
	rep(a, k) {
		ll b, c,cta,ctb,ctc;
		b = c= (k - a)%k;
		if ((b + c) % k != 0)continue;
		//cout << b << " " << c << endl;
		cta = n / k + (a == 0 ? 0 : 1);
		ctb = n / k + (b == 0 ? 0 : 1);
		ctc = n / k + (c == 0 ? 0 : 1);
		if (a == b == c)ans += cta*ctb*ctc;
		else ans += cta * ctb*ctc;
		//cout << a << " " << ans << endl;
	}
	cout << ans << endl;
	return 0;
}