1000 factorial

Well, someone wrote that they could write a program to do 1000! in one line of Haskell. I said that it would take less than 100 lines of C to do the same. I wrote the program, and it is less than 100 lines of C:

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

#define LENGTH 10000

int max (int a, int b) { return (a > b) ? a : b; }
int min (int a, int b) { return (b > a) ? a : b; }

int length (unsigned char * a)
 {
   int i;
   i = LENGTH - 1;
   while ((a[i] == 0) && (i >= 0)) i--;
   return i + 1;
 }

void add (unsigned char * a, unsigned char * b)
 {
   int i, c, C;
   c = min(max(length(a), length(b)) + 1, LENGTH);
   for (i = 0, C = 0; i < c; i++)
    {
      a[i] += b[i] + C;
      C = a[i] > 99;
      if (C) a[i] -= 100;
    }
 }

   unsigned char x [LENGTH], y [LENGTH];
void mul (unsigned char * a, unsigned char * b)
 {
   int i, j, c, d, C;
   memset(y, '\0', LENGTH);
   c = min(length(a) + length(b) + 1, LENGTH);
   d = length(b);
   for (i = 0; i < d; i++)
    {
      memset(x, '\0', LENGTH);
      for (j = 0, C = 0; (i + j) < c; j++)
       {
         x[i + j] = (a[j] * b[i] + C) % 100;
         C = (a[j] * b[i] + C) / 100;
       }
      add(y, x);
    }
   memcpy(a, y, LENGTH);
 }

void numset (unsigned char * a, unsigned long b)
 {
   int i;
   for (i = 0; i < LENGTH; i++)
    {
      a[i] = b % 100;
      b /= 100;
    }
 }

int tz (unsigned char * a)
 {
   int i;
   for (i = 0; i < LENGTH; i++) if (a[i] != 0) return 0;
   return 1;
 }

   unsigned char minus1 [LENGTH], prod [LENGTH];
void fact (unsigned char * a)
 {
   numset(prod, 1);
   while (!tz(a))
    {
      mul(prod, a);
      add(a, minus1);
    }
   memcpy(a, prod, LENGTH);
 }

   unsigned char arg_res [LENGTH];
int main (void)
 {
   unsigned long a;
   int i;
   memset (minus1, 99, LENGTH);
   printf("Input the number to take the factorial of: ");
   scanf("%lu", &a);
   numset(arg_res, a);
   fact(arg_res);
   i = LENGTH - 1;
   while (arg_res[i] == 0) i--;
   while (i >= 0)
    {
      printf("%02d", (int) arg_res[i]);
      i--;
    }
   putchar('\n');
   return 0;
 }


factorialere sic
  da meo numerum parprimum.
  unum meo resulto da.
  si numerum tum nullum aequalitam fac unum redde.
  dum numerum fac sic
   resulto damentum numerum tum numerum unum deme multiplica.
   numerum damentum numerum unum deme.
  cis
  numerum redde.
cis