http://www0.us.ioccc.org/years.html

This is a pretty entertaining page... The title says it all. It's an annual contest where people send in the most confusing code they can make. And believe me, there are some really whacky ones.

Here's one I liked. It's the generic "hello world" program.

Code:
#include "stdio.h"
#define	e 3
#define	g (e/e)
#define	h ((g+e)/2)
#define	f (e-g-h)
#define	j (e*e-g)
#define k (j-h)
#define	l(x) tab2[x]/h
#define	m(n,a) ((n&(a))==(a))

long tab1[]={ 989L,5L,26L,0L,88319L,123L,0L,9367L };
int tab2[]={ 4,6,10,14,22,26,34,38,46,58,62,74,82,86 };

main(m1,s) char *s; {
    int a,b,c,d,o[k],n=(int)s;
    if(m1==1){ char b[2*j+f-g]; main(l(h+e)+h+e,b); printf(b); }
    else switch(m1-=h){
	case f:
	    a=(b=(c=(d=g)<<g)<<g)<<g;
	    return(m(n,a|c)|m(n,b)|m(n,a|d)|m(n,c|d));
	case h:
	    for(a=f;a<j;++a)if(tab1[a]&&!(tab1[a]%((long)l(n))))return(a);
	case g:
	    if(n<h)return(g);
	    if(n<j){n-=g;c='D';o[f]=h;o[g]=f;}
	    else{c='\r'-'\b';n-=j-g;o[f]=o[g]=g;}
	    if((b=n)>=e)for(b=g<<g;b<n;++b)o[b]=o[b-h]+o[b-g]+c;
	    return(o[b-g]%n+k-h);
	default:
	    if(m1-=e) main(m1-g+e+h,s+g); else *(s+g)=f;
	    for(*s=a=f;a<e;) *s=(*s<<e)|main(h+a++,(char *)m1);
	}
}
I'm still trying to figure it out...

Some of the codes on the site even look like ASCII art. Nice.

mjk