hdoj 2674 N!Again 【好题】
2015-07-25 12:47
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N!Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4008 Accepted Submission(s): 2152
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
因为2009 = 41 * 7 * 7,所以对于大于或者等于41的数,它的阶乘会到达2009,然后2009取余2009 = 0。。。
#include <cstdio> #define LL long long int main() { int n; while(scanf("%d", &n) != EOF) { if(n >= 41) { printf("0\n"); continue; } LL ans = 1; for(int i = 1; i <= n; i++) ans = (ans * i) % 2009; printf("%lld\n", ans); } return 0; }
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