[CF743C] Vladik and fractions

目录

• Description
• Input
• Output
• Sample Input1
• Sample Output1
• Sample Input2
• Sample Output2
• Hint
• 题解

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Description

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 \(\frac {2}{n}\) as a sum of three distinct positive fractions in form \(\frac {1}{m}\). Help Vladik with that, i.e for a given \(n\) find three distinct positive integers \(x\) , \(y\) and \(z\) such that \(\frac {2}{n} = \frac {1}{x} + \frac{1}{y} + \frac {1}{z}\). Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding \(10^{9}\) . If there is no such answer, print -1.

Input

The single line contains single integer \(n\) ( \(1<=n<=10^{4}\) ).

Output

If the answer exists, print \(3\) distinct numbers \(x\) , \(y\) and \(z\) ( \(1<=x,y,z<=10^{9}\) , \(x≠y\) , \(x≠z\) , \(y≠z\) ). Otherwise print -1. If there are multiple answers, print any of them.

Sample Input1

3

Sample Output1

2 7 42

Sample Input2

7

Sample Output2

7 8 56

Hint

对于\(100\)%的数据满足\(n \leq 10^4\)，要求答案中\(x,y,z \leq 2* 10^{9}\)

题解

先贴一下洛谷的题目翻译

题目描述 请找出一组合法的解使得\(\frac {1}{x} + \frac{1}{y} + \frac {1}{z} = \frac {2}{n}\)成立，其中\(x,y,z\)为正整数并且互不相同

输入 一个整数\(n\)

输出 一组合法的解\(x, y ,z\)用空格隔开，若不存在合法的解，输出\(-1\)

数据范围 对于\(100\)%的数据满足\(n \leq 10^4\)，要求答案中\(x,y,z \leq 2* 10^{9}\)

看到\(\frac {2}{n}\)很容易猜想\(x，y\)或者\(z\)中有一个等于\(n\)，那么我们先设\(x=n\)，则有\(\frac{1}{y} + \frac {1}{z} = \frac {1}{n}\)，由样例二我们猜想，\(y=n+1\)，\(z=n\times(n+1)\)，正好有公式\(\frac{1}{n+1} + \frac {1}{n\times(n+1)} = \frac {1}{n}\)，所以我们可以得出一组可行解\(x=n\)，\(y=n+1\)，\(z=n\times(n+1)\)

很显然，这题有多种解，将我们的解代入样例\(1\)，可以发现这是可行的

但是当\(n=1\)时，无解

下面给出两种证法

1. 当\(n=1\)时，\(n+1=2\)，\(n\times(n+1)=2\)，与题目中的\(x≠y\) , \(x≠z\) , \(y≠z\)矛盾

2. 因为\(x≠y\) , \(x≠z\) , \(y≠z\)，且\(x,y,z\)都是整数，所以\(x,y,z\)的最小值分别为\(1,2,3\)，此时\(\frac {1}{x} + \frac{1}{y} + \frac {1}{z} = 1 = \frac {2}{2}\)，很显然\(n≥2\)时才有解

\(My~Code\)

#include<iostream>
#include<cstdio>
using namespace std;

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

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