Recursion 递归绘图

我了解递归的基本原理，但当我遇到类似于hackerrank的问题时。我很快就搞不清楚该怎么做了

基本上，你必须画一个分形ascii树（由字母Y组成），每一层向下，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_________________________
_________________________1_______________1_______________1_______________1__________________________
_________________________1_______________1_______________1_______________1__________________________
_________________________1_______________1_______________1_______________1__________________________
_________________________1_______________1_______________1_______________1__________________________
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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________________________________
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如果有人能给我一个正确的方向，将不胜感激

图像的结构如下所示：

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

字母Y的Y0x&Y1x:Y0x是分支，Y1x是trunc；x是构成字母所需的行数

例如，Y02和Y12-2条线形成分支，2条线形成主干

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（例如，从Yx2到Yx3）时，您复制最后一行

编辑： 假设图形保存在矩阵

charp[63][100]

我将编写2个函数（不是最佳方法，但您可以对其进行优化） -drawY-根据输入绘制Y字母 -drawPicutre-绘制Y的“层”

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

您的图片与您的描述不匹配。您可以重申这个问题，因为后面的每一行都有一半的Y字符。第一行有32个Y字符，用空格分隔。下一行将有16个Y字符，位于第一行Y字符之间。你可以算出剩下的。向上是Ys的两倍，但我重新表述了这个问题，因为它将向下生成。我想你需要生成准确的图像，你能说出图像的大小吗：行x cols？很抱歉，延迟了，错过了更新。共有63行100列。我为您发布了一个答案，请让我知道这是否有帮助，或者您需要其他建议-我已经解决了这个问题：）就个人而言，我会从底部开始，一路向上；但这只是偏好的问题。我想，即使知道图像的结构，我也很难将其分解为递归步骤和基本情况。我有点明白你的意思，只是我无法通过挑战声明变量，所以矩阵不在表中。我要画出整个结构好吧，直到这一点，你才谈到变量：）但这并不重要，你只需打印到控制台，而不是将数据保存在矩阵中。您必须调整逻辑以适应直线只能增加的约束。基本的想法是一样的。谢谢你的帮助，我应该可以从这里找到答案。

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