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.

输出

For each test case draw an outline of the Sierpinski Triangle with a side's total length of 2ncharacters. Align your output to the left, that is, print the bottom leftmost slash into the first column. The output must not contain any trailing blanks. Print an empty line after each test case.

样例输入


样例输出

       /\
/__\
/\ /\
/__\/__\
/\ /\
/__\ /__\
/\ /\ /\ /\
/__\/__\/__\/__\ /\
/__\
/\ /\
/__\/__\ /\
/__\

解题思路

一开始总是以小三角形为单位,百思不得其解。看了一眼网上的思路之后恍然大悟要用数组做,递归的整体过程也写的很轻松,然后被各种换行空字符bug折磨,调了一个多小时orz。

AC代码

#include<iostream>
#include<cstring>
using namespace std; char map[][];//x向上延展是行,y向右延展是列 void GetMap(int t, int x, int y)//x,y是每个三角形的起点
{
if (t == )
{
map[x][y] = '/', map[x][y+] = '_', map[x][y+] = '_', map[x][y+] = '\\';
map[x + ][y] = ' ', map[x + ][y + ] = '/', map[x + ][y + ] = '\\';
}
else
{
GetMap(t - , x, y);
GetMap(t - , x, ( << t) + y);
GetMap(t - , ( << (t - )) + x, ( << (t - )) + y);
}
} void Draw(int t)
{
int m = ( << t) + ;
for (int i = <<t; i > ; i--)
{
for (int j = ; j <= m; j++)
{
if (map[i][j])
{
cout << map[i][j];
}
else cout << ' ';
}
cout << endl;
m++;
}
} int main()
{
int t;
int num = ;
while (true)
{
cin >> t;
if (t == )break;
GetMap(t,,);
if(num > ) cout << endl;
num++;
Draw(t);
}
//system("pause");
return ;
}

POJ 1941 The Sierpinski Fractal的更多相关文章

  1. poj 1941 The Sierpinski Fractal 递归

    //poj 1941 //sep9 #include <iostream> using namespace std; const int maxW=2048; const int maxH ...

  2. POJ 1941 The Sierpinski Fractal ——模拟

    只需要开一个数组,记录一下这个图形. 通过一番计算,发现最大的面积大约是2k*2k的 然后递归下去染三角形. 需要计算出左上角的坐标. 然后输出的时候需要记录一下每一行最远延伸的地方,防止行末空格过多 ...

  3. POJ1941 The Sierpinski Fractal

    Description Consider a regular triangular area, divide it into four equal triangles of half height a ...

  4. POJ 题目分类(转载)

    Log 2016-3-21 网上找的POJ分类,来源已经不清楚了.百度能百度到一大把.贴一份在博客上,鞭策自己刷题,不能偷懒!! 初期: 一.基本算法: (1)枚举. (poj1753,poj2965 ...

  5. (转)POJ题目分类

    初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推. ...

  6. poj分类

    初期: 一.基本算法:      (1)枚举. (poj1753,poj2965)      (2)贪心(poj1328,poj2109,poj2586)      (3)递归和分治法.      ( ...

  7. poj 题目分类(1)

    poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01 ...

  8. POJ题目分类(按初级\中级\高级等分类,有助于大家根据个人情况学习)

    本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算 ...

  9. POJ题目分类(转)

    初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推. ...

随机推荐

  1. c#2.0锐利体验《泛型编程》读书笔记

    1.c#泛型及机制 Class Stack<T> { } T 其实为type的缩小,不过也可为其他字符代替T ,被称为“泛型类型”  T为晚绑定的,在编译的时候还不能确定T的确切类型. 2 ...

  2. python练习题(四)

    题目: 根据一个字符串返回一个ip数组,按照ip最后一位排序, 字符串: str = 'ss192.0.0.12?!289.0.0.1!0.0.0.0!192.163.10.28?192.0.0.5' ...

  3. Java 锁(学习笔记)

    关于Java 锁的知识整理与回顾(个人笔记): 锁有哪些,分别用来干嘛? Java实现锁有两种方式,synchronized关键字和Lock (1)Lock(可判断锁状态) Lock是基于JDK层面实 ...

  4. MySql添加字段命令

    使用ALTER TABLE命令来向一个表添加字段,示例如下: -- 向t_user表添加user_age字段 ) DEFAULT NULL COMMENT '年龄' AFTER user_email; ...

  5. 【转】.NET Core 事件总线,分布式事务解决方案:CAP

    [转].NET Core 事件总线,分布式事务解决方案:CAP 背景 相信前面几篇关于微服务的文章也介绍了那么多了,在构建微服务的过程中确实需要这么一个东西,即便不是在构建微服务,那么在构建分布式应用 ...

  6. Python垃圾回收机制?

    Python的GC模块主要运用了“引用计数”(reference counting)来跟踪和回收垃圾.在引用计数的基础上,还可以通过“标记-清除”(mark and sweep)解决容器对象可能产生的 ...

  7. docker容器中oracle数据库导出dmp文件

    Oracle数据库安装在docker容器中 1首先查看容器 docker ps 2进入oracle容器 docker exec -it 7f0f3f2d4f88 /bin/bash 3导出整个库:这个 ...

  8. nexus 3.17.0 简单试用

    老样子,使用docker-compose 运行 环境准备 docker-compose 文件 version: "3" services: nexus: image: sonaty ...

  9. canvas做动画

    一.绘制图片 ①加载图片 <!DOCTYPE html> <html lang="en"> <head> <meta charset=&q ...

  10. Theano安装笔记

    由于实验需要,近三个月来,安装过十几次Theano,基本上每次都是从最基本的nvidia driver装起.总结一些粗浅的安装心得. GPU:Nvidia K40, M40, M60 软件环境:Unb ...