hdu 2674（N！Again）

刚开始没有注意那个Hint : 0! = 1, N! = N*(N-1)!结果WA了！本题N的数据可以到10的9次方，暴力就算了，冷静分析一下吧！

阶乘数的一个特点就是当n的阶乘数的可以整除一个数m时，那么比n大的阶乘数都含有这样的因子;

因此，本题只需要计算出第一个n的阶乘数f（n），满足f（n）mod2009=0;即可。

后面的数只要大于等于n，直接输出“0”即可。

#include<iostream>
#include<cstdio>
using namespace std;
#define N 1000
long long num[N];
long long lim;
void handle()
{
    long long fac=1;
    long long i;
    num[0]=1;
    for(i=1;i<=1000;i++)
    {
        if(fac%2009==0)  break;
         fac=fac*i;
         fac%=2009;
         num[i]=fac%2009;
    }
    lim=i-1;
   // cout<<lim<<endl;
}
int main()
{
     handle();
     long long n;
     while(cin>>n)
     {
         if(n<=lim)
            cout<<num[n]<<endl;
         else
            cout<<"0"<<endl;

     }
     return 0;

}

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