POJ 2083 Fractal

链接：http://poj.org/problem?id=2083

Fractal

Time Limit: 1000MS
Memory Limit: 30000KTotal Submissions: 8119
Accepted: 3890

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-

Source

Shanghai 2004 Preliminary

大意——给你一个数n，显示对应的一个分形图形,B(n)由5个B(n-1)的图形组成，如下所示：

B(n-1)          B(n-1)

B(n-1)

B(n-1)          B(n-1)

比如n=1时，图形是
X
n=2时，图形是
X    X
   X
X    X
n=3时，图形是
X    X           X    X
   X                 X
X    X           X    X
         X      X
             X
         X      X
X    X           X    X
   X                 X
X    X           X    X
n最大为7。

思路——因为n比较小，所以我们可以把n=7时的图形打印到2维数组中，然后读到n时，输出对应的图形即可。打印图形到2维数组中，使用递归是最自然的做法。因为这个分形图形本身定义就是递归的。因此我们只需要确定5个n-1规模问题的起始坐标和图形宽度就可以了。首先图形是一个正方形，n=1时，图形宽度是1；n=2时，图形宽度是3；显然，n时，图形宽度是3^(n-1)。接下来考虑起始坐标问题：假设w为n-1规模问题的图形宽度，n规模图形的起始坐标为(x,y)，其被表示成5个n-1规模问题的组合。那么5个n-1规模图形的起始坐标依次是(x,y),(x,y+2w),(x+w,y+w),(x+2w,
y),(x+2w,y+2w)。解决了这两个问题，我们就可以递归打印图形了。

复杂度分析——时间复杂度:O(1),空间复杂度:O(1)

附上AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cmath>#include <iomanip>#include <ctime>#include <climits>#include <cstdlib>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <set>#include <map>#include <stack>//#pragma comment(linker, "/STACK:102400000, 102400000")using namespace std;typedef long long ll;const double pi = acos(-1.0);const double e = exp(1.0);const double eps = 1e-8;const short maxlen = 730;short num[7] = {1, 3, 9, 27, 81, 243, 729}; // 图形宽度char mat[maxlen][maxlen];short n;void print(short n); // 打印图形void init(short n, short x, short y); // 预处理图形，递归实现int main(){ios::sync_with_stdio(false);memset(mat, ' ', sizeof(mat));init(7, 0, 0);while (~scanf("%hd", &n) && n!=-1)print(n-1);return 0;}void print(short n){for (short i=0; i<num[n]; ++i){for (short j=0; j<num[n]; ++j)putchar(mat[i][j]);putchar('\n');}printf("-\n");}void init(short n, short x, short y){if (n == 1)mat[x][y] = 'X';else{short k = num[n-2];init(n-1, x, y); // 五个部分图形起始坐标关系init(n-1, x, y+2*k);init(n-1, x+k, y+k);init(n-1, x+2*k, y);init(n-1, x+2*k, y+2*k);}}