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)


// Problem : C - Triangular Relationship
// Contest : AtCoder - AtCoder Regular Contest 102
// URL : https://atcoder.jp/contests/arc102/tasks/arc102_a
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,b,a) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a,x) for (auto& a : x)
 
#define bug(x) cout<<#x<<"="<<x<<endl; 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define ins insert
 
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
const ld pii=3.14159265359; 
const int MOD = 1000000007;
const char nl = '\n';
const int MX = 200001; //check the limits, dummy
 ll mod_pow(ll x,ll n) {
	if (n == 0) return 1;
	ll res = mod_pow(x*x , n / 2);
	if (n & 1) res = res * x ;
	return res;
}

int main() 
{
	ios_base::sync_with_stdio(0); cin.tie(0);    
	int N,K;
	cin>>N>>K;
	ll ans=0;
	if(K%2==0 && K/2<=N)
		ans++;
	if(K<=N)
	{
		int val=N/K;
		ans+=mod_pow(val,3);
		if(K%2==0)
		{
			if(N%K>=K/2)
				val++;
			ans+=mod_pow(val,3);
			ans--;
		}
	}
	cout<<ans<<nl;
	return 0;
}
 
// read the question correctly (ll vs int)
// template by bqi343