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这么精彩的程序居然没有人跟贴,诡异.向您致敬~,用最简的(x%3+y%3)%2==0描述了该分形

Posted by tzkq at 2010-07-27 18:16:56 on Problem 2083 and last updated at 2010-07-27 19:30:42
In Reply To:这学期第一道程序，发贴庆祝一下 Posted by:jxd at 2006-09-19 13:56:41 
int i,n,k,x,y,t;
main(){
	for(;scanf("%d",&n)&&0<n;puts("-")){
		k=pow(3,n-1);
		for(i=0;i<k*k;i++){
			for(x=i/k,y=i%k,t=n;--t&&(x%3+y%3)%2==0;x/=3,y/=3);
			printf("%c%s",t?32:88,i%k<k-1?"":"\n");
		}
	}
}
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