能猜结论这件事也很重要

就好像高中时候的压轴题= =

猜出来能过就好啦嘛… 这道题的题意是这样的： = =他说输出任意一个就可以诶 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.

Example Input 3 Output 2 7 42 Input 7 Output 7 8 56

嗯呢就是 输出任意一个嘛 那么就根据高中知识凑啦凑凑凑凑凑凑~ 和 n有关的变量就好= = 这里注意 n=1 的情况

//AC代码
#include <iostream>
using namespace std;
int main (){
    int n;
    cin>>n;
    if(n==1)
        cout<<-1<<endl;
    else 
    cout<<n<<' '<<n+1<<' '<<n*(n+1)<<endl;
}