hdu 2674 N！Again（数论，水题）

N！Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4739    Accepted Submission(s): 2500


Problem Description

WhereIsHeroFrom:             Zty, what are you doing ?

Zty:                                     I want to calculate N!......

WhereIsHeroFrom:             So easy! How big N is ?

Zty:                                    1 <=N <=1000000000000000000000000000000000000000000000…

WhereIsHeroFrom:             Oh! You must be crazy! Are you Fa Shao?

Zty:                                     No. I haven's finished my saying. I just said I want to calculate N! mod 2009

Hint : 0! = 1, N! = N*(N-1)!

 

Input

Each line will contain one integer N(0 <= N<=10^9). Process to end of file.

 

Output

For each case, output N! mod 2009

 

Sample Input

4
5

 

Sample Output

24
120

题意：求n的阶乘%2009

思路：首先对与大于等于2009的数，结果必为0应该是显然的，因为阶乘中间必然会出现2009，那么必然可以整除了。

其实这个性质对于2009的最大质因子也是等价的，只要n大于等于2009的最大质因子，阶乘过程中必然会整除2009所有质因子，也就是整除2009本身了。

2009最大质因子只有41，所以当n>=41时输出0，否则O（n）跑一遍阶乘取模即可。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define LL long long
int devide(int n)
{
int maxn=-1;
for(int i=2;i*i<=n;i++)
{
if(n%i==0)
{
maxn=i;
while(n%i==0)
n/=i;
}
}
if(n>1) maxn=n;
return maxn;
}
int main()
{
int n;
int m=devide(2009);
while(~scanf("%d",&n))
{
if(n>=m) printf("0\n");
else
{
LL ans=1;
for(int i=2;i<=n;i++)
ans=ans*i%2009;
printf("%lld\n",ans);
}
}
return 0;
}