POJ 2083 Fractal (分形&分治)

Fractal

http://poj.org/problem?id=2083

Time Limit:  1000MS

Memory Limit: 30000K

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

小技巧：

用一个数组l存储扩展的长度。

完整代码：

/*94ms,712KB*/

#include<cstdio>
#include<cstring>

char ans[750][750];
int l[8] = {0, 1};

void print(int i, int j, int n)
{
	if (n == 1)
	{
		ans[i][j] = 'X';
		return;
	}
	print(i, j, n - 1);
	print(i, j + (l[n - 1] << 1), n - 1);
	print(i + l[n - 1], j + l[n - 1], n - 1);
	print(i + (l[n - 1] << 1), j, n - 1);
	print(i + (l[n - 1] << 1), j + (l[n - 1] << 1), n - 1);
}

int main(void)
{
	int n, i, j;
	for (int i = 2; i <= 7; ++i)
		l[i] = l[i - 1] * 3;
	while (scanf("%d", &n), ~n)
	{
		memset(ans, ' ', sizeof(ans));
		print(0, 0, n);
		for (i = 0; i < l[n]; ++i)
		{
			for (j = 0; j < l[n]; ++j)
				putchar(ans[i][j]);
			putchar(10);
		}
		puts("-");
	}
	return 0;
}