poj 2083 分形

Fractal

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6514		Accepted: 3226

Description

A fractal is an object or quantity that displays self-similarity, in a somewhat technical sense, on all scales. The object need not exhibit exactly the same structure at all scales, but the same "type" of structures must appear on all scales.
A box fractal is defined as below :

• A box fractal of degree 1 is simply
  X
  
• A box fractal of degree 2 is
  X X
  X
  X X
  
• If using B(n - 1) to represent the box fractal of degree n - 1, then a box fractal of degree n is defined recursively as following
  

  B(n - 1)        B(n - 1)
  
          B(n - 1)
  
  B(n - 1)        B(n - 1)

Your task is to draw a box fractal of degree n.

Input

The input consists of several test cases. Each line of the input contains a positive integer n which is no greater than 7. The last line of input is a negative integer −1 indicating the end of input.

Output

For each test case, output the box fractal using the 'X' notation. Please notice that 'X' is an uppercase letter. Print a line with only a single dash after each test case.

Sample Input

1
2
3
4
-1

Sample Output

X
-
X X
 X
X X
-
X X   X X
 X     X
X X   X X
   X X
    X
   X X
X X   X X
 X     X
X X   X X
-
X X   X X         X X   X X
 X     X           X     X
X X   X X         X X   X X
   X X               X X
    X                 X
   X X               X X
X X   X X         X X   X X
 X     X           X     X
X X   X X         X X   X X
         X X   X X
          X     X
         X X   X X
            X X
             X
            X X
         X X   X X
          X     X
         X X   X X
X X   X X         X X   X X
 X     X           X     X
X X   X X         X X   X X
   X X               X X
    X                 X
   X X               X X
X X   X X         X X   X X
 X     X           X     X
X X   X X         X X   X X
-

Source

Shanghai 2004 Preliminary

 

（从discuss里copy的代码）

#include <stdio.h>
#include <math.h>
int main()
{
	int n;
	while(scanf("%d",&n)&&n--!=-1)
	{
		for(int i=0;i<pow(3.0,n);i++){
			for(int j=0;j<pow(3.0,n);j++){
				int ii=i;
				int jj=j;
				int k;
				for(k=0;k<n&&(ii%3+jj%3)%2==0;){
					ii/=3;
					jj/=3;
					k++;
				}
				printf("%c",32+56*(k==n));
			}
			printf("\n");
		}
		printf("-\n");
	} 
}