recursion - 递归绘图

Question

我了解递归的基础知识，但是当我遇到像this one这样的问题时来自 hackerrank。我很快就对如何处理它感到困惑。

基本上你必须画一个分形ascii树(由字母Y组成)，每一层向下将Y的数量减半。我似乎无法理解基本步骤，因此我可以将其推广到其他层，如下所示:

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__________________1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1_1___________________
___________________1___1___1___1___1___1___1___1___1___1___1___1___1___1___1___1____________________
___________________1___1___1___1___1___1___1___1___1___1___1___1___1___1___1___1____________________
____________________1_1_____1_1_____1_1_____1_1_____1_1_____1_1_____1_1_____1_1_____________________
_____________________1_______1_______1_______1_______1_______1_______1_______1______________________
_____________________1_______1_______1_______1_______1_______1_______1_______1______________________
_____________________1_______1_______1_______1_______1_______1_______1_______1______________________
______________________1_____1_________1_____1_________1_____1_________1_____1_______________________
_______________________1___1___________1___1___________1___1___________1___1________________________
________________________1_1_____________1_1_____________1_1_____________1_1_________________________
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__________________________1_____________1_________________1_____________1___________________________
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____________________________1_________1_____________________1_________1_____________________________
_____________________________1_______1_______________________1_______1______________________________
______________________________1_____1_________________________1_____1_______________________________
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如果有人能在正确的方向上插入我，我将不胜感激。

Answer 1

图像的结构如下:

   Y01
   Y11
---------
   Y02
   Y12
---------
   Y04
   Y14
---------
   Y08
   Y18
---------
   Y016
   Y116

字母Y的Y0x & Y1x:Y0x是分支，Y1x是截断； x 是组成字母所需的行数

例如Y02 & Y12 - 2条线形成分支，2条线形成主干

1---1
 1-1
  1
  1 

您会注意到还有一个“规则”:当您开始绘制新字母时(例如，从 Yx2 到 Yx3)，您复制最后一行。

编辑:假设您将绘图保存在矩阵 char P[63][100] 中我会写 2 个函数(不是最佳方法，但你可以优化它)- drawY - 根据输入绘制 Y 字母- drawPicutre - 绘制 Y 的“层”

drawY(int startPos, int stopPos, int*line, int lvl, int count)
{
   if(lvl>=count)
   {
     P[*line][(startPos+(stopPos-startPos))/2] = '1';
     P[*line + lvl][(startPos+(stopPos-startPos))/2 +count] = '1';
     P[*line + lvl][(startPos+(stopPos-startPos))/2 -count] = '1';
     (*line)--;
     drawY(startPos,stopPos,line,lvl,count+1);
   }

}

int line=63;
drawY(0,100, &line, 16, 1); // draws the first, and largest Y

您只需要实现第二个函数 drawPicture(...) 即可调用 drawY。如果需要，您可以修改 drawY。