- Fractal(3.4.1)

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C - Fractal(3.4.1)

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

SubmitStatus

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

1234-1

Sample Output

X-X X XX X-X X   X X X     XX X   X X   X X    X   X XX X   X X X     XX X   X X-X X   X X         X X   X X X     X           X     XX X   X X         X X   X X   X X               X X    X                 X   X X               X XX X   X X         X X   X X X     X           X     XX X   X X         X X   X X         X X   X X          X     X         X X   X X            X X             X            X X         X X   X X          X     X         X X   X XX X   X X         X X   X X X     X           X     XX X   X X         X X   X X   X X               X X    X                 X   X X               X XX X   X X         X X   X X X     X           X     XX 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] = '\n';     //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;  }

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