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From: Ike Naar <ike@sverige.freeshell.org>
Newsgroups: comp.programming
Subject: Re: Project Euler - add all natural numbers below 1000
Date: Wed, 23 May 2012 13:55:11 +0000 (UTC)
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On 2012-05-23, arnuld <sunrise@invalid.address> wrote:
> /* Add all the natural numbers below one thousand that are multiples of 3 
> or 5. 
>  * version 0.1
>  */
>
> #include <stdio.h>
> #include <stdlib.h>
>
> int main(void)
> {
>   int i;
>   unsigned long s = 0;
>   
>   for (i = 3; i <= 1000; i += 3)

Nit: < 1000 (you want to add only numbers *below* 1000).
Here it does not matter because 1000 is not a multiple of 3,
but if you would change the limit to something that *is* a
multiple of 3, you'd get a wrong answer.

>     s += i;
>
>   for (i = 5; i < 1000; i += 5)
>     {
>       if(i % 3) s += i;
>     }
>
>   printf("%lu\n", s);
>
>   return EXIT_SUCCESS;
> }

Just for fun, here's an alternative program, without loops.

#include <stdio.h>

int main(void)
{
  long const n = 1000;
  long const n3 = (n-1) / 3;
  long const n5 = (n-1) / 5;
  long const n15 = (n-1) / 15;
  long const sum = (3*n3*(n3+1) + 5*n5*(n5+1) - 15*n15*(n15+1)) / 2;
  printf("%ld\n", sum);
  return 0;
}