poj 1941 The Sierpinski Fractal（递归）

最新推荐文章于 2021-02-21 13:23:34 发布

pku_zzy

最新推荐文章于 2021-02-21 13:23:34 发布

阅读量2.7k

点赞数

CC 4.0 BY-SA版权

分类专栏： Algorithm

本文链接：https://blog.youkuaiyun.com/PKU_ZZY/article/details/51441407

该博客介绍了如何使用递归方法来绘制Sierpinski三角形，包括问题描述、输入输出规格以及样例。通过递归实现，解决了在固定分辨率下调整图形大小的问题。文章还提到，对于小规模数据，可以通过设置画布来简化处理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

程序设计实习动态规划作业 poj 1941 The Sierpinski Fractal（递归）
总时间限制: 1000ms 内存限制: 65536kB

描述
Consider a regular triangular area, divide it into four equal triangles of half height and remove the one in the middle. Apply the same operation recursively to each of the three remaining triangles. If we repeated this procedure infinite times, we’d obtain something with an area of zero. The fractal that evolves this way is called the Sierpinski Triangle. Although its topological dimension is 2, its Hausdorff-Besicovitch dimension is log(3)/log(2)~1.58, a fractional value (that’s why it is called a fractal). By the way, the Hausdorff-Besicovitch dimension of the Norwegian coast is approximately 1.52, its topological dimension being 1.

For this problem, you are to outline the Sierpinski Triangle up to a certain recursion depth, using just ASCII characters. Since the drawing resolution is thus fixed, you’ll need to grow the picture appropriately. Draw the smallest triangle (that is not divided any further) with two slashes, to backslashes and two underscores like this:

/\
/_\
To see how to draw larger triangles, take a look at the sample output.

输入
The input contains several testcases. Each is specified by an integer n. Input is terminated by n=0. Otherwise 1<=n<=10 indicates the recursion depth.

输