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hdu2647 N！Again

2022-08-08 18:17:00 【51CTO】

N！Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4781 Accepted Submission(s): 2527

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

Author

WhereIsHeroFrom

Source

HDU女生专场公开赛——谁说女子不如男

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看到n的范围脑子里就知道是规律。。。

其实把结果输出一下就看到了。如果某个值正好整除2009 那么后面的都是0了

       
       import 
       
       java.
       
       math.
       
       BigInteger;
       
       import 
       
       java.
       
       util.
       
       Scanner;
       
       public 
       
       class 
       
       Main {
       
       public 
       
       static 
       
       void 
       
       main(
       
       String[] 
       
       args) {
       
       // TODO Auto-generated method stub
       
       Scanner 
       
       sca
       
       =
       
       new 
       
       Scanner(
       
       System.
       
       in);
       
       while(
       
       sca.
       
       hasNext()){
       
       int 
       
       x
       
       =
       
       sca.
       
       nextInt();
       
       if(
       
       x
       
       >
       
       41){
       
       System.
       
       out.
       
       println(
       
       0);
       
       continue;
       
      }
       
       BigInteger 
       
       res
       
       =
       
       BigInteger.
       
       ONE;
       
       for(
       
       int 
       
       i
       
       =
       
       2;
       
       i
       
       <=
       
       x;
       
       i
       
       ++){
       
       res
       
       =
       
       res.
       
       multiply(
       
       BigInteger.
       
       valueOf(
       
       i));
       
       res
       
       =
       
       res.
       
       mod(
       
       BigInteger.
       
       valueOf(
       
       2009));
       
      }
       
       System.
       
       out.
       
       println(
       
       res);
       
    }
       
  }
       
}
      
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版权声明
本文为[51CTO]所创，转载请带上原文链接，感谢
https://blog.51cto.com/u_15742657/5557043

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