codeforce 743 C. Vladik and fractions （数学构造）

最新推荐文章于 2022-11-20 12:29:19 发布

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最新推荐文章于 2022-11-20 12:29:19 发布

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本文介绍了一个有趣的数学挑战，即对于给定的正整数n，如何找到三个不同的正整数x、y和z，使得2/n = 1/x + 1/y + 1/z。文中提供了一种构造性的解决方案，并附带了C++代码实现。

Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction $\text{[math]}$ as a sum of three distinct positive fractions in form $\text{[math]}$ .

Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that $\text{[math]}$ . Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 10⁹.

If there is no such answer, print -1.

Input

The single line contains single integer n (1 ≤ n ≤ 10⁴).

Output

If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 10⁹, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.

If there are multiple answers, print any of them.

Examples

Input

Output

2 7 42

Input

Output

7 8 56

题意：给你一个n，找出一组数x，y，z，满足x！=y，y！=z，z！=z，并且2/n=1/x+1/y+1/z

解题思路：构造，因为x，y，z互不相同，可以设x=n，y=n+1，则z=n*（n+1），很明显n=1的时候是不满足的，则n=1的时候输出-1

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==1) printf("-1\n");
        else printf("%d %d %d\n",n,n+1,n*(n+1));
    }
    return 0;
}