hdu2674N！Again（水题）

题意：
$求N! mod\ \ 2009$
$N<=1e9$
思路：水题。
当N >= 2009是全为0。所以预处理好1 ~ 2009的阶乘，然后大于等于2009的全为0。
其实大于40的就全为0了，因为2009 = $7^2*41$。
$AC \ Code:$

#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#include<cstdio>
#include<sstream>
#include<vector>
#include<bitset>
#include<algorithm>
#include<unordered_map>
using namespace std;
#define read(x) scanf("%d",&x)
#define Read(x,y) scanf("%d%d",&x,&y)
#define gc(x)  scanf(" %c",&x);
#define mmt(x,y)  memset(x,y,sizeof x)
#define write(x) printf("%d\n",x)
#define pii pair<int,int>
#define INF 0x3f3f3f3f
#define ll long long
const int N = 100000 + 100;
const int M = 3e6 + 1005;
typedef long long LL;
int prim[N],tot = 0;
int cnt[N] ;
ll f[10000];
int main(){
    ll x;
    ll mod = 2009;
    f[0] = 1;
    for(ll i = 1;i <= 2009;++i){
        f[i] = f[i-1]*i % mod;
    }
    while(scanf("%lld",&x) == 1){
       x = min(x,2009LL);
       cout<<f[x]<<endl;
    }
}