ZOJ - 2423-Fractal

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. Don't output any trailing spaces at the end of each line, or you may get an 'Presentation Error'!

题意:输出n层"X"型分型,遇到-1时结束输入,可以看成一个3X3的正方形X为原型,输出后还要在另一行输出一个"-"

注意:图形最右方"X"后的空格不能输出,这里给出的一个解决方法是单独把他们标记出来,输出时检测这些标记来控制输出格式

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #define ll long long
 5 using namespace std;
 6 const int amn=1e3+10;
 7 int a=3,b;
 8 char mp[amn][amn];
 9 char ans[amn][amn];
10 void solve(int x,int y,int tot)
11 {
12     if(tot==1)
13     {
14         ans[x][y]='X';
15         return;
16     }
17     int tes=tot/a;
18     for(int i=x; i<tot+x; i+=tes)///注意这里i=x,j=y,找了好久才发现错误....
19     {
20         for(int j=y; j<tot+y; j+=tes)
21         {
22             if(mp[(i-x)/tes][(j-y)/tes]=='X')
23             {
24                 solve(i,j,tes);
25             }
26         }
27     }
28 }
29 int main()
30 {
31 
32     while(~scanf("%d",&b)&&b!=-1)
33     {
34         for(int i=0; i<3; i++)
35         {
36             for(int j=0; j<3; j++)
37             {
38                 if(((i==0||i==2)&&(j==0||j==2))||(i==1&&j==1))mp[i][j]='X';
39                 else mp[i][j]=' ';
40             }
41         }
42         int c=1;///注意初始化
43         for(int i=1; i<b; i++)c*=a;
44         for(int i=0; i<c; i++)
45             for(int j=0; j<c; j++)
46                 ans[i][j]=' ';
47         solve(0,0,c);
48         for(int i=0; i<c; i++)
49         {
50             for(int j=c-1; j>=0; j--)///这里是从每一行的最后一个元素开始检测
51             {
52                 if(ans[i][j]==' ')///将空格标记,遇到第一个"X"时跳出本次循环
53                     ans[i][j]='*';
54                 else break;
55             }
56         }
57         for(int i=0; i<c; i++)
58         {
59             for(int j=0; j<c; j++)
60             {
61                 if(ans[i][j]!='*')///如果这个不是被标记的元素则输出
62                     printf("%c",ans[i][j]);
63                 else break;
64             }
65 
66             printf("\n");
67         }
68         printf("-\n");
69     }
70 }