N！Again

HDU2674: N！Again

 

 因为2009=7^2*41; so k>=41, k! mod 2009 = 0; k<41时，直接计算就可以了。

 

#include <iostream>
#include <string>
#include <cstring>

using namespace std;

int num[45];

void GetNum(){
    int n;
    memset(num, 0, sizeof(num));
    num[0] = 1;
    for(int i=1; i<41; i++){
        n = 1;
        for(int j=1; j<=i; j++){
            n = n*j % 2009;
        }
        num[i] = n;
    }
}

int main(int argc, char *argv[])
{
    long long n;
    GetNum();
    while(cin>>n){
        if(n<41)
            cout << num[n] <<endl;
        else
            cout << 0 <<endl;
    }

    return 0;
}