HDU2674 N！Again 解题报告【阶乘】

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
解题报告
不难看出，当n>=2009时，n!%2009=0。
那么就很简单了：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=2009;int mul[N+5],n;void init(){    mul[0]=1;    for(int i=1;i<=2008;i++)mul[i]=(mul[i-1]%N*i%N)%N;}int main(){    init();    while(~scanf("%d",&n))    {        if(n>=2009)printf("0\n");        else printf("%d\n",mul[n]%N);    }    return 0;}