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

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 as a sum of three distinct positive fractions in form .

Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that . 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
3
Output
2 7 42
Input
7
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;
}






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