N!

题目描述：

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：

#include <stdio.h>
#include <time.h>
#include <string.h>


#define mod 10000

int main() {
    short int rem[10000],i,index,j;
    int n,jw;
//    clock_t start,finish;
//    start = clock();
//    double total = 1;
    while(~scanf("%d", &n)) {
        memset(rem, 0, sizeof(rem));
        rem[1] = 1;
        index = 2;
        jw = 0;
//        if (n == 0)
//            printf("0");
        for (i = 1; i <= n;i++) {
            for (j=1;j<index;j++) {
                int tem = rem[j]*i+jw;
                rem[j] = tem % mod;
                jw = tem / mod;
            }
            if (jw) {
                rem[index] = jw;
                jw = 0;
                index++;
            }
        }
        for (i=index-1;i>=1;i--) {
            if (i == index-1)
                printf("%d", rem[i]);
            else
                printf("%04d", rem[i]); 
        }
        puts("");
    }
//    finish = clock();
//    printf("
程序用时为：%lf", (double)(finish-start)/CLOCKS_PER_SEC);
    return 0;
}