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
5
Sample Output
24
120
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);
}
;
}