Day 48 C. Product of Three Numbers-优快云博客

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给定一个整数n，找出三个不同的整数a、b、c，使得2≤a,b,c且a×b×c=n。如果可以找到这样的三个数，输出'YES'和一种可能的组合；否则，输出'NO'。" 106267059,8535079,Oracle数据库管理实验：用户权限、对象操作与事务管理,"['数据库', 'Oracle', 'SQL']

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Problem
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.

Input
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≤10^9).

Output
For each test case, print the answer on it. Print “NO” if it is impossible to represent n as a⋅b⋅c for some distinct integers a,b,c such that 2≤a,b,c.

Otherwise, print “YES” and any possible such representation.

Example
input
5
64
32
97
2
12345
output
YES
2 4 8
NO
NO
NO
YES
3 5 823

#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<stdio.h>
#include<limits.h>
#include<cmath>
#include<set>
#include<map>
#define ll long long
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 2e5 + 5;
using namespace std;
//题目意思简洁明了 这里采取的是暴力拆解质因数 
int main() {
    int t;
    cin >> t;
    while (t--) 
    {
        int n;
        int ans = 0;
        bool f = false;
        cin >> n;
        int m = sqrt(n);
        for (int i = 2; i <= m; i++)
        {
            if (n % i == 0)
            {
                ans = n / i;
                int k = sqrt(ans);
                for (int j = 2; j <= k; j++)
                {
                    if (ans % j == 0 && j != i && ans / j != j && ans / j != i)
                    {
                        cout << "YES" << endl;
                        cout << i << " " << j << " " << ans / j << endl;
                        f = true;
                        break;
                    }
                }
                if (f)
                {
                    break;
                }
            }
        }
        if (!f)
        {
            cout << "NO" << endl;
        }
    }
    return 0;
}