Problem Description
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.
Examples
Input
5
64
32
97
2
12345Output
YES
2 4 8
NO
NO
NO
YES
3 5 823
题意：t 组数据，每组给出一个数 n，问 n 是否能分解成 a*b*c 的形式，要求 a、b、c 大于等于 2 并且不相等，若存在，输出任意一个方案
思路：
将 n 分解成 a*b*c 的形式，那么 a、b、c 这三个数一定是 n 的因子，因此首先将 n 进行因子分解，并统计其素因子个数，然后判断是否有三个不一样的因子
当 n 存在一个素因子 x 时：最简单的方案是 x * (x*x) * (x*x*x)，也即 n 中至少存在 6 个以上的 x 时，才存在合法方案
当 n 存在两个素因子 x1、x2 时：最简单的方案是 (x1*x2)*x1*x2、(x1*x1)*x1*x2、(x2*x2)*x1*x2，也即 n 中的 x1、x2 个数和至少达到 4 个时，才存在合法方案
当 n 存在三个及以上的素因子时，一定存在合法方案
而输出合法方案是一个问题，可以发现，如果有了 a、b，那么 c 就不需要进行暴力枚举，直接利用 c=n/(a*b) 即可计算得出
Source Program
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
#define Pair pair<int,int>
LL quickPow(LL a,LL b){ LL res=1; while(b){if(b&1)res*=a; a*=a; b>>=1;} return res; }
LL multMod(LL a,LL b,LL mod){ a%=mod; b%=mod; LL res=0; while(b){if(b&1)res=(res+a)%mod; a=(a<<=1)%mod; b>>=1; } return res%mod;}
LL quickMultPowMod(LL a, LL b,LL mod){ LL res=1,k=a; while(b){if((b&1))res=multMod(res,k,mod)%mod; k=multMod(k,k,mod)%mod; b>>=1;} return res%mod;}
LL quickPowMod(LL a,LL b,LL mod){ LL res=1; while(b){if(b&1)res=(a*res)%mod; a=(a*a)%mod; b>>=1; } return res; }
LL getInv(LL a,LL mod){ return quickPowMod(a,mod-2,mod); }
LL GCD(LL x,LL y){ return !y?x:GCD(y,x%y); }
LL LCM(LL x,LL y){ return x/GCD(x,y)*y; }
const double EPS = 1E-6;
const int MOD = 1000000000+7;
const int N = 1000+5;
const int dx[] = {0,0,-1,1,1,-1,1,1};
const int dy[] = {1,-1,0,0,-1,1,-1,1};
using namespace std;map<LL, int> num; //记录素因子个数
LL factor[N], cntF; //记录素因子
void getFactor(LL n) {cntF = 0;num.clear();for (LL i = 2; i <= n / i; i++) {if (n % i == 0) //记录素因子factor[cntF++] = i;while (n % i == 0) {n /= i;num[i]++; //记录素因子个数}}if (n > 1) {factor[cntF++] = n;num[n]++;}
}
int main() {int t;scanf("%d", &t);while (t--) {LL n;scanf("%lld", &n);getFactor(n);if (cntF == 1) {if (num[factor[0]] >= 6) {LL a = factor[0];LL b = factor[0] * factor[0];LL c = n / (a * b);printf("YES\n");printf("%lld %lld %lld\n", a, b, c);} elseprintf("NO\n");} else if (cntF == 2) {if (num[factor[0]] + num[factor[1]] >= 4) {LL a = factor[0];LL b = factor[1];LL c = n / (a * b);printf("YES\n");printf("%lld %lld %lld\n", a, b, c);} elseprintf("NO\n");} else {LL a = factor[0];LL b = factor[1];LL c = n / (a * b);printf("YES\n");printf("%lld %lld %lld\n", a, b, c);}}return 0;
}
