CF 615 C Product of Three Numbers

C. Product of Three Numbers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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
Copy

5
64
32
97
2
12345

Output
Copy

YES
2 4 8
NO
NO
NO
YES
3 5 823

判断三个数互不相等
ix != iy && iy != iz && ix != iz
刚开始最后一个判断没加上， wa了一发。吃一堑长一智。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

int a[10000];

void solve() {
	memset(a, 0, sizeof(a));
	int n, tot = 0;
	cin >> n;
	int mx = sqrt(n);
	for(int i = 2; i <= mx; i++) {
		while(n % i == 0) {
			n /= i;
			a[tot++] = i;
		}
	}
	if(n > 1) 
		a[tot++] = n;
//	for(int i = 0; i < tot; i++)
//		cout << a[i] <<" ";
//	cout << endl;
	int ix = 1, iy = 1, iz = 1;
	if(tot < 3) {
		cout << "NO" << endl;
		return;
	}
	if(tot == 3) {
		if(a[0] != a[1] && a[1] != a[2] && a[0] != a[2]) {
			cout << "YES" << endl;
			cout << a[0] << " " << a[1] << " " << a[2] << endl;
			return;
		}
		else {
			cout << "NO" << endl;
			return;
		}
	}
	ix = a[0], iy = a[1];
	int i = 1;
	if(ix == iy) {
		i = 3;
		iy *= a[2];
	}
	else
		i = 2;
	for(; i < tot; i++) {
		iz *= a[i];
	}
	if(ix != iy && iy != iz && ix != iz) {
		cout << "YES" << endl;
		cout << ix << " " << iy << " " << iz << endl;
		return; 
	}
	else {
		cout << "NO" << endl;
		return;
	}
}
	
int main() {
	int t;
	cin >> t;
	while(t--)
		solve();
	return 0;
}