N！Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3243 Accepted Submission(s): 1745
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
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
/*
2634 N！Again
output N! mod 2009
题目很机智啊 n<=10^9算N！肯定超时
2009=41*7*7
所以只要n>=41
n！中必然有41，7,7 所以比是2009的倍数 取余数肯定为0；
懒得打表
*/
#include<iostream>
using namespace std;
int main(){
int i,n;
long sum;
while(cin>>n)
{
if(n<41)
{
sum=1;
for(i=1;i<=n;i++)
{
sum=sum*i;
sum=sum%2009;
}
}
else sum=0;
cout<<sum<<endl;
}
return 0;
}