Codeforces Round #615 (Div. 3)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 nn. Find three distinct integers a,b,ca,b,c such that 2≤a,b,c2≤a,b,c and a⋅b⋅c=na⋅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.
Example
input
Copy
5
64
32
97
2
12345output
Copy
YES
2 4 8
NO
NO
NO
YES
3 5 823 C. Product of Three Numbers
题目链接：https://codeforces.com/contest/1294/problem/C
题意：
给你一个 n ，要求三个整数 a ，b ，c 使得 a * b * c = n 并且 a、b、c >= 2,且a、b、c不等
首先要有解，必要条件是n>=2*3*4=24;
其次不妨设
有
，然后最小的a从2开始枚举，上界是
，然后
,继续根号枚举即可
#include<bits/stdc++.h>
#define ll long long
#define rg register ll
using namespace std;
ll n,t;
int main()
{
cin>>t;
while(t--)
{
cin>>n;
if(n<24)
{
cout<<"NO"<<endl;
}
else
{
rg i,flag=0;
for(i=2;i*i*i<=n;i++);
for(rg i1=2;i1<=i;i1++)
{
ll k=n/i1;
ll i2=sqrt(k);
//cout<<i2<<" "<<i1<<" "<<k<<endl;
//cout<<i1+1<<endl;;
for(rg h=3;h<=i2;h++)
{
ll i3=k/h;
//cout<<i3<<endl;
//cout<<i3<<" "<<i2<<" "<<h<<endl;
if(i3*i1*h==n&&h!=i1&&h!=i3&&i1!=i3)
{
cout<<"YES"<<endl<<i1<<" "<<h<<" "<<i3<<endl;
flag=1;
break;
}
}
if(flag)break;
}
if(!flag)cout<<"NO"<<endl;
}
}
//while(1)getchar();
return 0;
}腾讯云开发者
