poj Fractal 2083 （模拟&&dfs）

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Fractal

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 8339 Accepted: 3963

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

大神的思路及分析，，。。。

/*n度的盒分形的规模为3^(n-1),即n度的盒分形图为一个长宽为3^(n-1)的正方形。
设置递归函数printBox(n,x,y)生成以坐标（x,y）为左上角的n度盒分形。

1）递归边界：若n=1,则在（x,y）输出‘X’
2）若n>1,则计算n-1度的盒子的规模 m = 3^(n-2),分别在左上方，右上方，中间，左下方和右下方画出5个n-1度的盒子。
对于左上方的n-1度的盒子，左上角的坐标为（x,y）,递归printBox(n-1,x,y)生成；
对于右上方的n-1度的盒子，左上角的坐标为（x+2m,y）,递归printBox(n-1,x+2m,y)生成；
对于中间的n-1度的盒子，左上角的坐标为（x+m,y+m）,递归printBox(n-1,x+m,y+m)生成；
对于左下方的n-1度的盒子，左上角的坐标为（x,y+2m）,递归printBox(n-1,x,y+2m)生成；
对于右上方n-1度的盒子，左上角的坐标为（x+2m,y+2m）,递归printBox(n-1,x+2m,y+2m)生成；

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#define N 1010using namespace std;char map[N][N];void cop(int n,int x,int y){if(n==1)map[x][y]='X';else{int m=pow(3,n-2);cop(n-1,x,y);cop(n-1,x+2*m,y);cop(n-1,x,y+2*m);cop(n-1,x+m,y+m);cop(n-1,x+2*m,y+2*m);}}int main(){int n,i,j;while(scanf("%d",&n)&&n!=-1){int m=pow(3,n-1);for(i=0;i<m;i++){for(j=0;j<m;j++){map[i][j]=' ';map[i][m]='\0';}}cop(n,0,0);for(i=0;i<m;i++)printf("%s\n",map[i]);printf("-\n");}return 0;}

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