Gym - 100796K Profact 阶乘的乘积 DFS+剪枝

2015-2016 ACM ICPC Baltic Selection Contest

K. Profact

time limit per test:1 second

memory limit per test:256 megabytes

Alice is bored out of her mind by her math classes. She craves for something much more exciting. That is why she invented a new type of numbers, theprofacts. Alice calls a positive integer number a profact if it can
be expressed as a product of one or several factorials.

Just today Alice received n bills. She wonders whether the costs on the bills are profact numbers. But the numbers are too large, help Alice check this!

Input
The first line contains a single integer n, the number of bills (1 ≤ n ≤ 105). Each of the nextn lines contains
a single integer ai, the cost on the
i-th bill (1 ≤ ai ≤ 1018).

Output
Output n lines, on the
i-th line output the answer for the number
ai. If the number
ai is a profact, output "YES", otherwise output "NO".

Examples

Input

7
1
2
3
8
12
24
25

Output

YES
YES
NO
YES
YES
YES
NO

Note
A factorial is any number that can be expressed as1·2·3·...·k, for some positive integer
k, and is denoted by
k!.

    先输入总共有几组数据，然后每组数据一个数字，判断这个数字是否可以通过一个或多个阶乘数的乘积得到，比如8=2！*2！*2！。可以就输出YES，不可以输出NO，然后问题就在于如何判断了，因为题目范围是10^18，所以提前打了个表看了一下，19！就已经大于10^18了，所以先打表打到19的阶乘都得出来然后再进行处理。

    之后用dfs不断对这个数字取余判断就可以了，如果能顺利除到1就是可以的，如果不能就不可以。当然直接暴力的话肯定是会超时的，简单的剪枝就不说了，比较关键的是对一些素数阶乘的处理，如果在这个数中存在素数，就一定要用素数！来进行处理，否则一定不能满足题目条件。

    下面AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long fac[25];
int t;
int pri[10]={2,3,5,7,11,13,17,19};

int know()
{
int i;
fac[0]=1;
for(i=1;i<20;i++)
{
fac[i]=fac[i-1]*(i+1);
//cout<<fac[i]<<endl;
}
return 0;
}

int dfs(long long a,int b)
{
int i,j;
if(t==1)
return 0;
if(a==1)
{
t=1;
return 0;
}
for(i=b;i>=2;i--)
{
if(a%fac[i-1]==0)
{
dfs(a/fac[i-1],i);
if(t==0)
{
for(j=0;j<8;j++)
{
if(pri[j]==i)
return 0;
if(pri[j]>i)
break;
}
}
}
if(t==1)
return 0;
}
return 0;
}

int main()
{
int T;
long long n;
int i;
know();
scanf("%d",&T);
while(T--)
{
t=0;
scanf("%lld",&n);
dfs(n,20);
if(t==1)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}