HDU 2674-- N!Again【技巧】
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N!Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4012 Accepted Submission(s): 2154
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
因为2009 = 41 * 7 * 7,所以对于大于41的数,它的阶乘会到达肯定可以用 2009 * 某数表示 ,然后取余2009 = 0。
#include <cstdio>#include <cstring>#define mod 2009int main (){int n;while(scanf("%d", &n)!=EOF){int sum = 1;if(n > 41)printf("0\n");else {for(int i = 1; i <= n; ++i)sum = ( sum * i ) % mod;printf("%d\n", sum);}}return 0;}
0 0
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