递归绘图

Question

我理解递归的基础知识，但是当我遇到来自hackerrank的this one之类的问题时。我很快就对如何接近它感到困惑。

基本上你必须绘制一个分形ascii树（由字母Y组成），每一层向下一半Ys的数量。我似乎无法绕过基本步骤，所以我可以将它推广到其他层，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_________________________
_________________________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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如果有人能够向我推进正确的方向，我们将不胜感激。

Answer 1

图像的结构如下：

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

Y0x＆amp;字母Y：Y0x的Y1x是分支，Y1x是截断; x是形成字母

所需的行数

E.g。 Y02＆amp; Y12 - 2线形成分支，2线形成树干

1---1
 1-1
  1
  1 

您可以注意到还有一个“规则”：当您开始绘制新字母时（例如，您从Yx2转到Yx3），您将复制最后一行。

编辑： 假设您将图形保存在矩阵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

你只需要实现调用drawY的第二个函数drawPicture（...）。如果需要，可以修改drawY。