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)

/****Author: Barish Namazov****/
#include <bits/stdc++.h>

using namespace std;

/***TEMPLATE***/
#define intt long long
#define pii pair<intt,intt>

#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()

#define F first
#define S second
#define pb push_back

#define IO ios_base::sync_with_stdio(false);cin.tie();
#define endl '\n'

const intt max4 = 10004;
const intt maxx = 100005;
const intt max6 = 1000006;
const intt max7 = 10000007;

const intt lg4 = 13;
const intt lg5 = 17;
const intt lg6 = 20;

const intt INF = 2LL * 1000000000;

/***************/

intt powmod (intt a, intt b, intt mod) {
    intt res = 1;
    a %= mod;
    for (; b; b >>= 1) {
        if (b & 1)
            res = (res * a) % mod;
        a = (a * a) % mod;
    }
    return res;
}

intt gcd (intt a, intt b) {
    while (b > 0) {
        intt t = a % b;
        a = b, b = t;
    }
    return a;
}

intt lcm (intt a, intt b) {
    return (a / gcd (a, b)) * b;
}

intt is_prime (intt n) {
    if (n <= 1 || n > 3 && (n % 2 == 0 || n % 3 == 0))
        return 0;
    for (intt i = 5, t = 2; i * i <= n; i += t, t = 6 - t)
        if (n % i == 0)
            return 0;
    return 1;
}

/**************************/

#define double long double

const double EPS = 1e-12;

int main() {
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    IO;
    intt n;
    cin >> n;
    for (intt x = 1; x <= 3500; x++) {
        for (intt y = 1; y <= 3500; y++) {
            double z = 4.0 / n - 1.0 / x - 1.0 / y;
            if (z < 0) {
                continue;
            }
            if (fabs (floor (1.0 / z) - ceil (1.0 / z)) < EPS) {
                cout << x << " " << y << " " << floor (1.0 / z) << endl;
                return 0;
            }
        }
    }
    return 0;
}