Codeforces--1294C--Product of Three Numbers

题目描述：
You are given one integer number n. Find three distinct integers a,b,c such that 2≤a,b,c and a⋅b⋅c=n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer t independent test cases.
输入描述:
The first line of the input contains one integer t (1≤t≤100) — the number of test cases.

The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2≤n≤109).
输出描述:
5
64
32
97
2
12345
输入：
YES
2 4 8
NO
NO
NO
YES
3 5 823
输出：
1
4
题意：
给你一个整数n。找出三个不同的整数a、b、c，使得2≤a、b、c和a⋅b⋅c=n，或者说这是不可能的。
题解：
直接在因子里面搜就好了
代码：

#include<cstdio>
#include<vector>
#include<set>
#include<iostream>
#include<algorithm>
using namespace std;int main(){int t,n;scanf("%d",&t);while(t--){vector<int> v;set<int>s;scanf("%d",&n);for(int i = 2; i * i <= n; i ++){if(n % i == 0){v.push_back(i);s.insert(i);s.insert(n / i);}}bool flag = false;for(int i = 0; i < v.size(); i ++){for(int j = i + 1; j < v.size(); j ++){int num = v[i] * v[j];int cnt = n / num;if(n % cnt == 0 && cnt != v[i] && cnt != v[j] && s.find(cnt) != s.end()){printf("YES\n");flag = true;printf("%d %d %d\n", v[i], v[j], cnt);break;}}if(flag) break;}if(!flag) printf("NO\n");}return 0;
}

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