hdoj 2674 N!

N！Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5078    Accepted Submission(s): 2707

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

水题

思路：将1-2009的阶乘先存在f[2010]，之后用a，b来存N／2009和N％ 2009的结果，最后结果为ans=f[2009]^a*f[b]%2009.

#include "cstdio"
#include "algorithm"
#include "cstring"
];
int main()
{
    int n,ans;
    int a,b;
    f[]=;
    ;i<=;i++){
        f[i]=f[i-]*i%;
    }
    ){
        ans=;
        a=n/;
        b=n%;
        ;i<a;i++){
            ans=ans*f[]%;
        }
        ) {ans=ans*f[b]%;}
        printf("%d\n",ans);

    }
    ;
}