Problem:
Score : 300 points

Problem Statement

You are given an integer N.

Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w.

If there are multiple solutions, any of them will be accepted.

Constraints

It is guaranteed that, for the given integer N, there exists a solution such that h,n,w \leq 3500.

Inputs

Input is given from Standard Input in the following format:

N

Outputs

Print a triple of positive integers h, n and w that satisfies the condition, in the following format:

h n w

Sample Input 1

2

Sample Output 1

1 2 2

4/2 = 1/1 + 1/2 + 1/2.

Sample Input 2

3485

Sample Output 2

872 1012974 1539173474040

It is allowed to use an integer exceeding 3500 in a solution.

Sample Input 3

4664

Sample Output 3

3498 3498 3498

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <bitset>
#include <complex>
#include <functional>

// output
#define SPBR(w, n) std::cout<<(w + 1 == n ? '\n' : ' ');
#define YES cout << "YES" << endl
#define Yes cout << "Yes" << endl
#define NO cout << "NO" << endl
#define No cout << "No" << endl
// utility
#define ALL(i) (i).begin(), (i).end()
#define FOR(i, a, n) for(int i=(a);i<(n);++i)
#define RFOR(i, a, n) for(int i=(n)-1;i>=(a);--i)
#define REP(i, n) for(int i=0;i<int(n);++i)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define IN(a, x, b) (a<=x && x<b)
#define OUT(a, x, b) (x<a || b<=x)
template<class T> inline T chmax(T & a, const T b) { return a = (a < b) ? b : a; }
template<class T> inline T chmin(T& a, const T b) { return a = (a > b) ? b : a; }

// type/const
#define int ll
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const int MOD = 1000000007;
/* const int MOD = 998244353; */
const int INF = 1e18;
const double PI = acos(-1);

using namespace std;


signed main() {
  int N;
  cin >> N;

  int M = 3500;
  FOR(a, 1, M+1) FOR(b, 1, M+1){
    if(a*N+b*N-4*a*b == 0) continue;
    double c = 4.0/N-1.0/a-1.0/b;
    c = 1/c;
    c = (int)c;
    if(c < 0) continue;
    if(c > M) continue;
    if((4*a*b*c) == ((a*b+b*c+c*a)*N)){
      cout << a << " " << b << " " << c << endl;
      return 0;
    }
  }

  return 0;
}