Product of Three Numbers

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≤109).

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

题意

实际上就是判断一个数n能否分解为a·b·c

解题思路

这题没什么没用什么算法，就直接暴力枚举。就是注意a，b，c是小于sqrt（n*1.0）的，减少运算次数。还要注意的一点就是sqrt函数（）里的数字必须是浮点型的，在这儿卡了好久…

代码

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int t;
	int n;
	cin >> t;
	while (t--)
	{
		cin >> n;
		int a = 0, b = 0;
		int flag = 0;
		for (int i = 2; i < sqrt(n*1.0); i++)
		{
			if (n % i==0)
			{
				if (a==0)//得出a
				{
					a = i;
					n = n / i;
				}
				else//得出b，c
				{
					b = i;
					n = n / i;
					flag = 1;
					break;
				}
			}
		}
		if (flag)
		{
			cout << "YES" << endl;
			cout << a << ' ' << b << ' ' << n << endl;
		}
		else
			cout << "NO" << endl;
	}
	return 0;
}

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