N！Again hdu 2674 c++

秒速五厘米 2022-08-10 16:37 3阅读 0赞

# N！Again #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 3434    Accepted Submission(s): 1863**

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

题意：给定n，求n的阶乘对2009 取余的结果；  
分析：水题，给定n的范围看似很大。当n! >2009 时，其阶乘必定含2009 对2009取余必为0，小于2009时 直接算即可。打表可得超过41的阶乘均大于2009.

代码如下：

#include<cstdio>
    #include<cstring>
    
    int main()
    {
        __int64 i,s,n;
        while(scanf("%I64d",&n)!=EOF)
        {
           
           if(n==0)
               printf("1\n");
           else if(n>41)
               printf("0\n");    
           else 
           {
                s=1;
                for(i=n;i>0;i--)
                {
                   s*=i;
                   s=s%2009;
                }   
               printf("%d\n",s);
           }
        }
        return 0;
    }