Pascal's triangle

Motivated by the Hacker News challenge in http://news.ycombinator.com/item?id=3429466 I just made this.


#include <stdio.h>
#include <stdlib.h>

int64_t* next_pascal_row(int64_t* previous_row, int64_t previous_size)
{
  int64_t* retval = (int64_t*)malloc(sizeof(int64_t)*(previous_size+2));
  int64_t i;
  for (i=0;i<previous_size+1;i++) {
    if (i==0 || i == previous_size) {
      retval[i] = 1;
    } else {
      retval[i] = previous_row[i-1] + previous_row[i];
    }
  }
  return retval;
}

void print_row(int64_t* row, int64_t size)
{
  int64_t i;
  for(i = 0; i < size; i++) {
    printf("%lld ",row[i]);
  }
  printf("\n");
}

int main()
{
  int64_t* first_row = (int64_t*)malloc(sizeof(int64_t));
  int64_t* previous_row;
  int64_t* next_row;
  int64_t i,size;
  first_row[0] = 1;
  size = 1;
  print_row(first_row,size);
  previous_row = first_row;
  for(i = 0; i<30;i++) {
    next_row = next_pascal_row(previous_row,size);
    size++;
    print_row(next_row,size);
    free(previous_row);
    previous_row = next_row;
  }
  free(next_row);
}
Opinions? Yeah, int64_t is probably a GNU extension, I don't care as long as it works with gcc. U mad? Also, post you're own solutions. If you can't do the Pascal, you can just go ahead and apply to your're nearest McDonald's, because you ain't worth shit.

>>78
Is it? You can't even multiply (by any variable) in Presburger arithmetic. It's much weaker than Peano Arithmetic, which allows multiplication, but then if you add multiplication, you can encode entire programs within arithmetic, making the deciding the truth value of some sentences undecidable (this result is known in some forms as Godel's incompleteness theorems). No Strong AI will be able to solve the general halting problem, but any strong AI (or even human "AI"), should be able to decide on various specific cases, and they can self-improve to decide on more and more of them, just never 'all'.