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
-1Sample 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

一递归：
先看下较为常规的递归解决：

由于图形是重复的，小的图形只是把大图形的左上角一部分输出，，只计算最大的图形，打表可加快速度。

#include 
#include 
int p[8] = {1,3,9,27,81,243,729};
char map[730][730];
//n当前的图形大小，x，y图形所在的坐标
void print(int n,int x,int y){
	if(n == 0){
		map[x][y] = 'X'; //
		return;
	}
	print(n-1, x, y); //左上
	print(n-1, x+2*p[n-1], y); //右上
	print(n-1, x+p[n-1], y+p[n-1]); //中间
	print(n-1, x, y+2*p[n-1]);
	print(n-1, x+2*p[n-1],  y+2*p[n-1]);
}
int n;
int main(){
	for(int i=0; i= 0){
		for(int i=0; i 
 
二 数学方法
 
 
在discuss里面看到这个短小精悍的程序。
 
 
 
 
 
#include"stdio.h"
#include"math.h"
main()
{
	int i,j,n,ii,jj,k;
	while(scanf("%d",&n)&&n--!=-1)
	{
		for(i=0;i 
 
关于这个图形，还可以用来证明
 
 
寻找1/5 + 1/25 + 1/125 + .. = 1/4的图形证明