N!

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 55283    Accepted Submission(s): 15718

Problem Description
Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!
 

 

Input
One N in one line, process to the end of file.
 

 

Output
For each N, output N! in one line.
 

 

Sample Input
1 2 3
 

 

Sample Output
1 2 6
 

 

Author
JGShining（极光炫影）
 

 

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 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 int main()
 4 {
 5     int N,Len,i,j,sign,Num;
 6     int sum[10086]={0};
 7     while(scanf("%d",&N)!=EOF)  /*输入N，输出N的阶乘*/
 8     {
 9         Len=1;
10         sum[0]=1;
11         for(i=1;i<=N;i++)
12         {
13             for(j=0,sign=0;j<Len;j++)
14             {
15                 Num=sum[j]*i+sign;
16                 if(Num>=10000)
17                 {
18                     sign=Num/10000;
19                     sum[j]=Num%10000;
20                 }
21                 else
22                 {
23                     sum[j]=Num;
24                     sign=0;
25                 }
26                 if(j==Len-1&&sign)
27                 {
28                     Len++;
29                 }
30             }
31         }
32         for(j=Len-1;j>=0;j--)
33         {
34             if(j!=Len-1)
35                printf("%04d",sum[j]);
36             else
37                 printf("%d",sum[j]);
38             sum[j]=0;
39         }
40         putchar(10);
41     }
42     return 0;
43 }
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转载请附上原帖地址：http://www.cnblogs.com/LWF5201314614/articles/4018064.html 

posted @ 2014-10-10 23:05  Wurq  阅读(315)  评论(0)    收藏  举报