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

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #define ll long long
 using namespace std;
 ;
 ,b;
 char mp[amn][amn];
 char ans[amn][amn];
 void solve(int x,int y,int tot)
 {
     )
     {
         ans[x][y]='X';
         return;
     }
     int tes=tot/a;
     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....
     {
         for(int j=y; j<tot+y; j+=tes)
         {
             if(mp[(i-x)/tes][(j-y)/tes]=='X')
             {
                 solve(i,j,tes);
             }
         }
     }
 }
 int main()
 {

     )
     {
         ; i<; i++)
         {
             ; j<; j++)
             {
                 ||i==)&&(j==||j==))||(i==&&j==))mp[i][j]='X';
                 else mp[i][j]=' ';
             }
         }
         ;///注意初始化
         ; i<b; i++)c*=a;
         ; i<c; i++)
             ; j<c; j++)
                 ans[i][j]=' ';
         solve(,,c);
         ; i<c; i++)
         {
             ; j>=; j--)///这里是从每一行的最后一个元素开始检测
             {
                 if(ans[i][j]==' ')///将空格标记,遇到第一个"X"时跳出本次循环
                     ans[i][j]='*';
                 else break;
             }
         }
         ; i<c; i++)
         {
             ; j<c; j++)
             {
                 if(ans[i][j]!='*')///如果这个不是被标记的元素则输出
                     printf("%c",ans[i][j]);
                 else break;
             }

             printf("\n");
         }
         printf("-\n");
     }
 }