Fractal
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 8341   Accepted: 3965

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
  • 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 OutputX

-
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
-
递归:
思路很巧妙,把每个左上角的点作为起点;开始递归;
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x);
char a[2500][2500];
void dfs(int x,int y,int cur){
if(cur==1){
a[x][y]='X';
return;
}
int n,m;
dfs(x,y,cur-1);
//左上
m=x;n=y+pow(3,cur-2)*2;
dfs(m,n,cur-1);
//右上
m=x+pow(3,cur-2)*2;n=y;
dfs(m,n,cur-1);
//左下
m=x+pow(3,cur-2)*2;n=y+pow(3,cur-2)*2;
dfs(m,n,cur-1);
//右下
m=x+pow(3,cur-2);n=y+pow(3,cur-2);
dfs(m,n,cur-1);
//中
}
int main(){
int N;
while(scanf("%d",&N),N!=-1){
int len=pow(3,N-1);
for(int i=0;i<len;i++){
for(int j=0;j<len;j++)
a[i][j]=' ';
a[i][len]='\0';
}
dfs(0,0,N);
for(int i=0;i<len;i++){
printf("%s\n",a[i]);
}
puts("-");
}
return 0;
}

  

 

Fractal(递归,好题)的更多相关文章

  1. 递归一题总结(OJ P1117倒牛奶)

    题目:                    农民约翰有三个容量分别是A,B,C升的桶,A,B,C分别是三个从1到20的整数,最初,A和B桶都是空的,而C桶是装满牛奶的.有时,约翰把牛奶从一个桶倒到另 ...

  2. 函数递归简单题-hdoj-2044 2018-一只小蜜蜂 母牛的故事

    题目:一只小蜜蜂 递归做法: #include<cstdio> #include<iostream> #include<stdlib.h> #include< ...

  3. poj 2083 Fractal 递归 图形打印

    题目链接: http://poj.org/problem?id=2083 题目描述: n = 1时,图形b[1]是X n = 2时,图形b[2]是X  X        X               ...

  4. 洛谷P1427 小鱼的数字游戏 题解 递归入门题

    题目链接:https://www.luogu.com.cn/problem/P1427 题目大意: 给你一串数(输入到0为止),倒序输出这些数. 解题思路: 首先这道题目可以用数组存数据,然后输出. ...

  5. 十五 链表与递归,leetCode203题

    两种方式: package com.lt.datastructure.LinkedList; /** * leetCode 203题 * /** * Definition for singly-lin ...

  6. 一道Postgresql递归树题

    转载请注明出处: https://www.cnblogs.com/funnyzpc/p/13698249.html 也是偶然的一次,群友出了一道题考考大家,当时正值疫情最最严重的三月(借口...),披 ...

  7. poj 1941 The Sierpinski Fractal 递归

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

  8. HDU 2563 统计问题(递归,思维题)

    统计问题 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submi ...

  9. POJ 2083 Fractal 分形题目

    这两天自学了一线算法导论里分治策略的内容,秉着只有真正投入投入编程,才能更好的理解一种算法的思想的想法,兴致勃勃地找一些入门的题来学习. 搜了一下最后把目光锁定在了Poj fractal这一个题上.以 ...

随机推荐

  1. C++/C#结构体转化-二维数组-bytes To Strings

    C++结构体 typedef struct VidyoClientRequestGetWindowsAndDesktops_ { /*! The number of application windo ...

  2. Qt QToolTip 控件背景的 QSS 设置方法(摘抄)

    Qt/C++/CSS: QTooltip stylesheet background colour Hi there, I've recently come across a problem deve ...

  3. 算法——A*——HDOJ:1813

    Escape from Tetris Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Othe ...

  4. Mybatis使用存储过程(MySql)

    推荐文章:http://www.iteye.com/topic/1132302 http://yhjhappy234.blog.163.com/blog/static/3163283220124557 ...

  5. 常用ajax请求

    JQuery版本的ajax请求:(包括处理WebService中xml字符串) $.ajax({ type: "POST", async: true, url: "&qu ...

  6. codeforces 245H . Queries for Number of Palindromes 区间dp

    题目链接 给一个字符串, q个询问, 每次询问求出[l, r]里有多少个回文串. 区间dp, dp[l][r]表示[l, r]内有多少个回文串. dp[l][r] = dp[l+1][r]+dp[l] ...

  7. asp.net 多站点共享StateServer Session

    <sessionState mode="StateServer" stateConnectionString="tcpip=127.0.0.1:42424" ...

  8. linux下如何查看系统和内核版本

        1. 查看内核版本命令: 1) [root@q1test01 ~]# cat /proc/version Linux version 2.6.9-22.ELsmp (bhcompile@cro ...

  9. JAVA泛型接口

    事例代码: package com.xt.thins_15_3; import java.util.Iterator; /** * 泛型接口 * * @author xue * * @param &l ...

  10. http://www.swoole.com/

    Swoole:重新定义PHP PHP语言的高性能网络通信框架,提供了PHP语言的异步多线程服务器,异步TCP/UDP网络客户端,异步MySQL,数据库连接池,AsyncTask,消息队列,毫秒定时器, ...