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: Wrong Answer)

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

// Loops
#define _overload3(_1, _2, _3, name, ...) name
#define _rep(i, n) repi(i, 0, n)
#define repi(i, a, b) for (int i = int(a); i < int(b); ++i)
#define _rrep(i, n) rrepi(i, n, 0)
#define rrepi(i, a, b) for (int i = int(a) - 1; i >= int(b); --i)

#define rep(...) _overload3(__VA_ARGS__, repi, _rep, )(__VA_ARGS__)
#define rrep(...) _overload3(__VA_ARGS__, rrepi, _rrep, )(__VA_ARGS__)
#define each(xi, x) for (auto &&xi : x)
// Note: we can use rep(i,N) or rep(i,from,to)

// typedef
template <class T> using vec = vector<T>;
using ll = long long;
using vi = vec<ll>;
using vvi = vec<vi>;
using vvvi = vec<vvi>;
using pii = pair<ll, ll>;

// Constants

// Shorter repr for frequently used terms
#define pb push_back
#define mp make_pair

// Algorithms
#define all(x) (x).begin(), (x).end()
#define uniq(v) v.erase(unique(all(v)), v.end())
#define perm(c)                                                                \
  sort(all(c));                                                                \
  for (bool c##p = 1; c##p; c##p = next_permutation(all(c)))

template <class T> pair<T, size_t> max(vector<T> &x) {
  auto it = max_element(all(x));
  return mp(*it, it - x.begin());
}
template <class T> pair<T, size_t> min(vector<T> &x) {
  auto it = min_element(all(x));
  return mp(*it, it - x.begin());
}

template <class T> inline bool chmax(T &maxval, const T &newval) {
  if (maxval < newval) {
    maxval = newval;
    return 1;
  }
  return 0;
}
template <class T> inline bool chmin(T &minval, const T &newval) {
  if (minval > newval) {
    minval = newval;
    return 1;
  }
  return 0;
}

// Utilities
// Grid world utilities
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
#define inside(H, W, y, x) 0 <= (x) && (x) < (W) && 0 <= (y) && (y) < (H)

// Debug
#ifdef LOCAL
#include "dump.hpp"
#define debug(x) cerr << #x << ": " << x << '\n'
#else
#define dump(...)
#define debug(x)
#endif

// Yes or No
void yesno(bool ok, string yes = "Yes", string no = "No") {
  cout << (ok ? yes : no) << endl;
}

// Paste snippets here!!

//

void solve(long long N, long long K) {
  ll ret = 0;;

  if(K%2==1){
    // Kが奇数のとき，a, b, c がすべて K の倍数のときのときのみOK
    ret += (N/K)*(N/K)*(N/K);
  }

  if(K%2==0){
    // Kが偶数のときは a, b, c がすべて 奇数 x (K/2) と書けるか，
    // もしくは 偶数 x (K/2)とかける場合に限る
    ll P = N/(K/2);
    ll even = P/2;
    ll odd = P/2 + 1;
    ret += even*even*even;
    ret += odd*odd*odd;
  }
  cout << ret << endl;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false); // Magic for faster cin
  cout.precision(12);
  long long N;
  std::cin >> N;
  long long K;
  std::cin >> K;
  solve(N, K);
  return 0;
}