TOJ1011 阶乘末尾非零数求和

/*
阶乘末尾非零数求和
http://acm.tongji.edu.cn/people/ps/showproblem.php?problem_id=1011
Time Limit:1s Memory Limit:1000k
Total Submit:4671 Accepted:1317

Problem

对于小于25000的自然数n,求阶乘n!,(n-1)!,(n-2)!...3!,2!,1!右边的非零数之和。
例如：
当n=5时，
5!=120,右边非零数为2;
4!=24,右边非零数为4;
3!=6,右边非零数为6;
2!=2,右边非零数为2;
1!=1,右边非零数为1。
其右边的非零数之和为15。

Input

本题有多组数据，每组数据包含一个正整数N(N不大于25000)占一行。

Output

对给定的每组输入数据，输出一个整数。每个结果占一行。不要输出额外的空行。

Sample Input

5
10
1

Sample Output

15
39
1

T_T 超郁闷的题目……
Run ID User Problem Result Memory Time Language Date
263600 kingwei 1011 Accepted 28 k 659 ms C 2005-06-01 16:59:22
263554 kingwei 1011 Time Limit Exceeded C 2005-06-01 16:03:42
263477 kingwei 1011 Wrong Answer 36 k 10 ms C 2005-06-01 14:37:16
263466 kingwei 1011 Wrong Answer 36 k 6 ms C 2005-06-01 14:28:36
263463 kingwei 1011 Wrong Answer 44 k 6 ms C 2005-06-01 14:27:20
*/
#include <stdio.h>
#define MAX_NUM 25000
#define MAX_LEN 1600
int start, end;
int workarr[MAX_LEN] = {1, 0};
int res[MAX_NUM] = {0, 1};
int main()
{
int n, i, j, carry, temp;

start = 0;
end = 0;
for (i=2; i<MAX_NUM; i++)
{
carry = 0;
for (j=start; j<=end; j++)
{
workarr[j] = workarr[j]*i+carry;
carry = workarr[j]/10000;
workarr[j] %= 10000;
}
while (carry > 0 && end<MAX_LEN)
{
end++;
workarr[end] = carry%10000;
carry /= 10000;
}
while (workarr[start] == 0)
start++;
temp = workarr[start];
while (temp%10 == 0)
temp /= 10;
res[i] = res[i-1]+temp%10;
}

while (scanf("%d", &n) != EOF)
{
printf("%d/n", res
);
}
return 0;
}