N!

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 262144/262144K (Java/Other)

Total Submission(s) : 68 Accepted Submission(s) : 20

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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

这道题要用数组来储存最终的结果。
注意一个很小的细节（俺在那里测了老半天，伤！），就是0的阶乘为1；
下面附上自己写的代码：

 1 #include <stdio.h>
 2 #include <time.h>
 3 #include <string.h>
 4 
 5 
 6 #define mod 10000
 7 
 8 int main() {
 9     short int rem[10000],i,index,j;
10     int n,jw;
11 //    clock_t start,finish;
12 //    start = clock();
13 //    double total = 1;
14     while(~scanf("%d", &n)) {
15         memset(rem, 0, sizeof(rem));
16         rem[1] = 1;
17         index = 2;
18         jw = 0;
19 //        if (n == 0)
20 //            printf("0");
21         for (i = 1; i <= n;i++) {
22             for (j=1;j<index;j++) {
23                 int tem = rem[j]*i+jw;
24                 rem[j] = tem % mod;
25                 jw = tem / mod;
26             }
27             if (jw) {
28                 rem[index] = jw;
29                 jw = 0;
30                 index++;
31             }
32         }
33         for (i=index-1;i>=1;i--) {
34             if (i == index-1)
35                 printf("%d", rem[i]);
36             else
37                 printf("%04d", rem[i]); 
38         }
39         puts("");
40     }
41 //    finish = clock();
42 //    printf("
程序用时为：%lf", (double)(finish-start)/CLOCKS_PER_SEC);
43     return 0;
44 }