随手练——大量级阶乘 - HDU-2674 N！Again

 N！Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6338    Accepted Submission(s): 3325

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

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类似这种数据级别大的不正常的阶乘，一定有特殊数据，试一下就能找到。

2009=7*7*41=49*41

也就是说，当n≥41时，n！一定是2009的倍数

#include <iostream>
#include <stdio.h>
using namespace std;

int main() {
    long long n;
    while(scanf("%lld",&n)!=EOF){
        ;
        )
        cout<<<<endl;
        else{
        　　;i<=n;i++){
            res=(res*i)%;
        }
        cout<<res<<endl;
        }
    }
    ;
}