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1000 factorial

Name: Anonymous 2011-04-02 22:59

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;
 }

Name: Anonymous 2012-08-03 19:02


in ruby

(0..1000).reduce(:*)

ermagawsh that was difficult
will someone shut these faggots up please

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