HDU 2674 N！Again

N！Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4754    Accepted Submission(s): 2509

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

24120

 

Author

WhereIsHeroFrom

 

Source

HDU女生专场公开赛——谁说女子不如男

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  2672 2668 2673 2669 2671 

 首先要利用同余定理的公式（a*b)mod c=( (a mod c)*(b mod c) )mod c;

但是利用这个直接算会超时，有一点没想到，大于等于2009的阶乘对2009取余结果为0

补充：（a+b）mod c =(a mod c +b mod c) mod c;

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;int main(){int n,i;while(scanf("%d",&n)!=EOF){if(n>=2009)printf("0\n");else{int sum=1;    for(i=1;i<=n;i++)    {  sum=sum*(i%2009);  sum=sum%2009;     }    printf("%d\n",sum);}}return 0; }