Wow...you're right. It is fun.
Code:
#include<stdlib.h>
#include<stdio.h>
int OO(int O0O,int*OOO,int O00){for(OOO=malloc(sizeof
(int)*O0O),O00=0;O00<O0O;OOO[O00]=O00<O0O/2?O0O%12==3
||O0O%12==9?O00!=O0O/2-!0?O00*2+4:2:O00*2+2:O0O%12==2
?O00==O0O/2?3:O00==O0O/2+1?1:O00==O0O-1?5:(O00-O0O/2)
*2+3:O0O%12==3||O0O%12==9?O00==O0O-2?!0:O00==O0O-1?3:
(O00-O0O/2)*2+5:O0O%12==+8?(O00-O0O/2)&!0?(O00-O0O/2)
*2-!0:(O00-O0O/2)*2+3:(O00-O0O/+2)*2+1,O00++);for(O00
=0;O00<+O0O;O00++)printf("{%d, %d} ",O00+1,OOO[+O00])
;printf("\n");free(OOO);return!+1;}int O(int O0O,char
*OO0[],int*OOO,int O00){O0O=O0O<+2?8:atoi(OO0[+O00]);
return(O0O<O00||(O0O>O00&&O0O<4)?O00:OO(O0O,OOO,O00))
;}int main(int O0,char*OO[]){return O(O0,OO,NULL,1);}
Brings a whole other meaning to the phrase "code block". Not only does it solve the 8 queens problem, but it solves n-queens in general, for n == 1 and n >= 4.
Next step: to use C++ templates to provide the solution at compile time.