POJ 2083 Fractal 递归画分形

Fractal Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6934   Accepted: 3419 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 1 2 3 4 -1 Sample Output 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 X - Source Shanghai 2004 Preliminary

2014，小KD在战斗！！！

虽然学习递归很久了，递归画分形却是第一次遇到。

《数据结构编程实验》P42

set函数的三个参数代表：从坐标x,y开始，画一个n阶的分形。当n不等于1时，分形可以分为5个部分。依次递归。

画分形就是标记二维数组。全部标记完了整体输出。。。

#include <iostream>
#include <cstring>
using namespace std;
const int MAXN=730;
const int N[8]={0,1,3,9,27,81,243,729};
int n,i,j,map[MAXN][MAXN];
void set(int n,int x,int y);
int main()
{
	while (cin>>n && n!=-1){
		memset(map,0,sizeof(map));
		set(n,1,1);
		for (i=1;i<=N[n];++i){
			for(j=1;j<=N[n];++j)
				if (map[i][j]) cout<<'X';
					else cout<<' ';
			cout<<endl;
		}
		cout<<"-\n";
	}
	return 0;
}

void set(int n,int x,int y){
	if (n==1){
		map[x][y]=1;
		return;
	}
	set(n-1,x,y);
	set(n-1,x+2*N[n-1],y);
	set(n-1,x+N[n-1],y+N[n-1]);
	set(n-1,x,y+2*N[n-1]);
	set(n-1,x+2*N[n-1],y+2*N[n-1]);
	return;
}

在POJ的讨论区里面发现一个很牛的代码，短短几行，不明白原理

Posted by jxd at 2006-09-19 13:56:41 on Problem 2083
#include"stdio.h"
#include"math.h"
main()
{
	int i,j,n,ii,jj,k;
	while(scanf("%d",&n)&&n--!=-1)
	{
		for(i=0;i<pow(3,n);i++,printf("\n"))
			for(j=0;j<pow(3,n);j++)
			{
				for(ii=i,jj=j,k=0;k<n&&(ii%3+jj%3)%2==0;ii/=3,jj/=3,k++);
				printf("%c",32+56*(k==n));
			}
		printf("-\n");
	} 
}

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