- Fractal(3.4.1)

C - Fractal(3.4.1)

Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u

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

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
-
#include <iostream>  
#include <string>  
#include <cstring>  
#include <cmath>  
using namespace std;   
#define M 1000  
char s[M][M];   
void print( int n, int x, int y)  
{  
    int m;//m 表示方盒子规模 即边长 
	double t=3;
    if( n == 1 )  
    {  
        s[x][y] = 'X';   //回溯，当n=1时，为（x，y）坐标赋值'X'  
        return ;  
    }  
    else if( n > 1 )  
    {  
        m = pow( t, n - 2 );  //计算盒子规模  
        print( n - 1, x, y );   //左上角第一个图形进行递归打印，第一个图形左上角坐标为（x，y）  
        print( n - 1, x, y + 2 * m ); //右上角第二个图形进行递归打印，第二个图形左上角坐标为（x，y+2*m）  
        print( n - 1, x + m, y + m );  //中间第三个图形进行递归打印，第三个图形左上角坐标为（x+m,y+m）  
        print( n - 1, x + 2 * m, y );  //左下角第四个图形进行递归打印，第四个图形左下角坐标为（x+2*m,y）  
        print( n - 1, x + 2 * m, y + 2 * m);//右下角第五个图形进行递归打印，第五个图形右下角坐标为（x+2*m,y+2*m）  
    }  
}  
  
int main()  
{  
    int n, i, p; double t=3;
    while(cin >> n && n != -1 )  
    {  
        p = pow( t, n - 1 );  //计算方盒子规模  
        for( i = 0; i < p; i++ )   //对数组进行初始化  
        {  
            memset( s + i, ' ', p );   //每一行均初始化为空格  
            s[i][p] = '
';     //每一行最后一列初始化为换行,如果不这样做，会出错，图形无法输出  
        }  
        print( n, 0, 0 );   //从左上角（x，y）坐标开始打印  
        for( i = 0; i < p; i++)  
            cout << s[i] ;     //输出图形  
        cout << '-' << endl;  
    }  
    return 0;  
}