Gym 100796K Profact（爆搜+剪枝）

K - Profact
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice Gym
100796K

Description

standard input/output

Announcement


Statements

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, the profacts. 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 next n 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".

Sample Input

Input

7
1
2
3
8
12
24
25

Output

YES
YES
NO
YES
YES
YES
NO

Hint

A factorial is any number that can be expressed as 1·2·3·...·k, for some positive integer k,
and is denoted by k!.

题目大意：

输入一个数，问这个数能不能表示为1个或者多个阶乘的乘积。

范围：

a<=10^18。

思路：

可以知道20的阶乘已经超过了10^18，所以我们可以先预处理出来前20的阶乘。然后我们可以对这个数进行爆搜。

但是，这样是TLE！！！

所以考虑加剪枝，举个栗子：我们可以想到，假设如果这个数n能够被19！整除，但是他最后不能被其他阶乘整除了，此时我们没有达到目的。在搜索回溯的过程中，我们重新回到起点，回去判断能否被18！整除。但是！我们根本就不需要再去判了，因为这个数如果能被19！整除，那一定能被18！整除，前面那个不能成功，后者也一定是会失败的。所以此时我们就直接return了。

代码：

#include<stdio.h>
#include<string.h>
#define ll  __int64
ll as[]={
0,1,2,6,24,120,720,5040,  40320,  362880 , 3628800 , 39916800,  479001600 , 6227020800 , 87178291200,
1307674368000 , 20922789888000 , 355687428096000 , 6402373705728000,  121645100408832000 , 2432902008176640000
},xx[8]={2,3,5,7,11,13,17,19};
int flag=0;
void dfs(ll x,ll q)
{
if(flag)return;
if(x==1){
flag=1;
return;
}
if(flag==1)return;
for(int i=q;i>=2;i--)
{
if(x%as[i]==0){dfs(x/as[i],i);
if(!flag){
for(int j=0;j<8;j++)
{
if(xx[j]==i)return;
if(xx[j]>i)break;
}
}
}
if(flag)return;
}
}
int main()
{
//for(int i=1;i<=20;i++)
//    printf("%I64d\n",as[i]);
ll n,a,i,j,k;
scanf("%I64d",&n);
while(n--)
{
scanf("%I64d",&a);
if(a==1){
printf("YES\n");
continue;
}
flag=0;
dfs(a,20);
if(flag)printf("YES\n");
else printf("NO\n");
}
}