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;
const int amn=1e3+10;
int a=3,b;
char mp[amn][amn];
char ans[amn][amn];
void solve(int x,int y,int tot)
{
    if(tot==1)
    {
        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()
{

    while(~scanf("%d",&b)&&b!=-1)
    {
        for(int i=0; i<3; i++)
        {
            for(int j=0; j<3; j++)
            {
                if(((i==0||i==2)&&(j==0||j==2))||(i==1&&j==1))mp[i][j]='X';
                else mp[i][j]=' ';
            }
        }
        int c=1;///注意初始化
        for(int i=1; i<b; i++)c*=a;
        for(int i=0; i<c; i++)
            for(int j=0; j<c; j++)
                ans[i][j]=' ';
        solve(0,0,c);
        for(int i=0; i<c; i++)
        {
            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
            {
                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;
            }
        }
        for(int i=0; i<c; i++)
        {
            for(int j=0; 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");
    }
}