zoj 2423 Fractal

Fractal

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 XX 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. Don't output any trailing spaces at the end of each line, or you may get an 'Presentation Error'!

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-

分析：

此题用递归实现。

因为输出时都是从左到右，所以不能分块打印。

故考虑数组保存这个二维图形。找出规律，对于每个n，都可以递归处理5个大X，对于每个大X，又可以分成5个小X。

最后注意处理时X在数组中的位置。

代码：

以前做的，直接粘了。

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;const int maxn = 1000;char b[ maxn ][ maxn ];int flag[ maxn ];void f( int n, int x, int y ){if ( n == 1 ){b[ x ][ y ] = 'X';return;}int t = ( int )pow( 3.0, n - 2 );f( n - 1, x, y );f( n - 1, x + t, y + t );f( n - 1, x + 2 * t, y );f( n - 1, x, y + 2 * t );f( n - 1, x + 2 * t, y + 2 * t );}int main(){int n;while ( cin >> n && n != -1 ){memset( b, '\0', sizeof( b ) );memset( flag, 0, sizeof( flag ) );int i, j;int temp = ( int )pow( 3.0, n - 1 );for ( i = 0; i < temp; i++ ){for ( j = 0; j < temp; j++ )b[ i ][ j ] = ' ';// b[ i ][ j ] = '\n';}f( n, 0, 0 );for ( i = 0; i < temp; i++ ) {            for ( j = temp - 1; j >= 0; j-- )                if ( b[ i ][ j ] == 'X' ) {                    flag[ i ] = j;                    break;                }}for ( i = 0; i < temp; i++ ){for ( j = 0; j <= flag[ i ]; j++ )printf("%c", b[ i ][ j ]);printf("\n");}printf("-\n");}return 0;}