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)!
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
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("0\n"); else { a=solve(n); printf("%lld\n",a); } } return 0; }
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。