hdoj 2674 N！Again
发布时间：2020-12-14 02:19:33 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): 4152????Accepted Submission(s): 2232 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): 4152????Accepted Submission(s): 2232
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
?题意： 2009 =7*7*41，所以输入超过41的数取余后得到0，输入0，时输出1；
???????????? 对于小于41的同余定理求，即可！
??代码：
??????
#include<stdio.h>
int main()
{
int i,j,n;
char s[1010];
while(scanf("%d",&n)!=EOF)
{
if(n>41)
printf("0n");
else if(n==0)
printf("1n");
else if(n<=41)
{
int sum=1;
for(i=1;i<=n;i++)
{
sum=(sum*(i%2009))%2009;
}
printf("%dn",sum);
}
}
return 0;
}
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