codeforce "c.product of three numbers"

最新推荐文章于 2025-12-06 09:52:47 发布

原创最新推荐文章于 2025-12-06 09:52:47 发布 · 217 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++

本文探讨了一个算法挑战，即给定一个整数n，寻找三个不同的整数a、b、c，使得a*b*c=n，且2≤a,b,c。文章提供了一段C++代码，用于解决这一问题，并通过多个测试案例验证了算法的有效性。

You are given one integer number nn. Find three distinct integers a,b,ca,b,c such that 2≤a,b,c 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 tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.

The next nn lines describe test cases. The ii-th test case is given on a new line as one integer nn (2≤n≤1092≤n≤109).

Output

For each test case, print the answer on it. Print “NO” if it is impossible to represent nn as a⋅b⋅ca⋅b⋅c for some distinct integers a,b,ca,b,c such that 2≤a,b,c2≤a,b,c.

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

#include<iostream>
#include<cmath>
using namespace std;
const long long Max = 1000000000;
int zs(long long x)
{
	int temp = 1;
	for (int i = 2; i < sqrt(x); i++)
	{
		if (x % i == 0)
		{
			return 1;
			temp = 0;
			break;
		}
	}
	if (temp == 1)
		return 0;
}
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		long long n;
		int temp = 0;
		cin >> n;
		if (zs(n))
		{
			for (long long i = 2; i * i * i < n; i++)
			{
				if (n % i == 0 && zs(n / i))
				{
					for (long long j = 3; j * j < n / i; j++)
					{
						if((n/i)%j==0&&i<j&&i<n/i/j&&j<n/i/j )
						{
							
							
								cout << "YES" << endl;
								cout << i << " " << j << " " << n / i / j << endl;
								temp = 1;
								break;
							
						}
					}
				}
				if (temp == 1)
					break;
			}
		}
		else
		{
			cout << "NO" << endl;
			temp = 1;
		}
		if (temp == 0)
			cout << "NO" << endl;
	}
}