杭电-2674N！Again（大数阶乘）
发布时间：2020-12-14 02:11:00 所属栏目：大数据 来源：网络整理
导读：N！Again Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4267????Accepted Submission(s): 2279 Problem Description WhereIsHeroFrom: ????????????Zty,what are you doing ? Zty: ?????????
N！Again
Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4267????Accepted Submission(s): 2279
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
?
Sample Output
?
大数阶乘，但是规律是41及其以后取余2009都为0，一定要优化这一句，否则超时
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long solve(long long n)
{
long long t=1;
for(long long i=n;i>=2;i--)
t=i*t%2009;
return t;
}
int main()
{
long long a,n,i;
while(scanf("%lld",&n)!=EOF)
{
if(n>=41)
printf("0n");
else
{
a=solve(n);
printf("%lldn",a);
}
}
return 0;
}
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