问题

给定一个正整数n，其中n < 100

输出菱形图案，如下所示：

输入项 n=1

/\/\
\/\/

输入n=2：

 /\      /\
//\\/\/\//\\
\\//\/\/\\//
 \/      \/

输入n=3：

  /\                /\
 //\\  /\      /\  //\\
///\\\//\\/\/\//\\///\\\
\\\///\\//\/\/\\//\\\///
 \\//  \/      \/  \\//
  \/                \/

输入n=4：

   /\                              /\
  //\\    /\                /\    //\\
 ///\\\  //\\  /\      /\  //\\  ///\\\
////\\\\///\\\//\\/\/\//\\///\\\////\\\\
\\\\////\\\///\\//\/\/\\//\\\///\\\\////
 \\\///  \\//  \/      \/  \\//  \\\///
  \\//    \/                \/    \\//
   \/                              \/

等等。

规则

• 允许的程序和功能。
• 允许尾随空格。
• 没有/或\允许的行上的前导空格。
• 允许尾随换行符。
• 以字节为单位的最短代码获胜

这可能很相关

SOGL V0.12，24个字节

ā.∫ā;∫ \*+}ø┼↔±╬¡;øΚ┼}╬³

在这里尝试！

说明：

ā                         push an empty array (the main canvas)
 .∫                  }    iterate over the input, pushing 1-indexed iteration
   ā;                       push an empty array below the iteration
     ∫    }                 iterate over the iteration counter
       \*                     push current iteration amount of slashes
         +                    append those to the 2nd array
           ø┼               append nothing (so it'd space the array to a square)
             ↔±             reverse horizontally (swapping slashes)
               έ           quad-palindromize with 0 overlap and swapping characters as required
                 ;          get the canvas ontop
                  øΚ        prepend to it an empty line (so the now bigger romb would be one up)
                    ┼       append horizontally the canvas to the current romb
                      ╬³  palindromize horizontally with no overlap and swapping characters

木炭，30 27字节

Ｆ⁺¹Ｎ«Ｆι«Ｆ⁴«↙⁻ικ↑⟲Ｔ»←»Ｍι←»‖Ｍ

在线尝试！链接是详细版本的代码。说明：木炭的绘制图元不能完全绘制钻石，因为对角线移动保持在相同奇偶性的正方形上。编辑：新的解决方案是绘制菱形的一侧，然后旋转整个画布以准备绘制另一侧，从而使菱形可以循环绘制。然后将此循环包含在一个循环中，以绘制每个钻石的所有内部钻石。最外面的循环将所有菱形彼此相邻。最终，图像被镜像。

请注意，此后已扩展了炭笔，可以使用来保存另一个字节Increment。

APL（Dyalog），70 69 66字节

B←{'/\ '['\/'⍳⍺⍺⍵]}
C←⊢,⌽B
C(⊢⍪⊖B)⊃,/{C⊖A↑⊖' /'[⍵≤∘.+⍨⍳⍵+1]}¨⌽⍳A←⎕

在线尝试！

假设⎕IO←0，这在许多系统上都是标准的，因此该程序的索引为0。

这是一个通过STDIN进行输入的交易。

说明

（有点过时）

注意，这⍺是left参数，⍵是right参数，⍺⍺是left运算符。

B是有助于镜像钻石的功能。它使用字符串作为右参数，并将反向函数作为左（B运算符）。

B←{'/\ '['\/'⍳⍺⍺⍵]}
              ⍺⍺⍵            Apply ⍺⍺ on ⍵
         '\/'⍳               Find the index of the reflected string in '\/' (if the character is not found in `'\/'`, then return an index out of the bounds of the string, ie `2` if the character is a space)
   '/\ '[        ]           Use these indexes on '/\ ' to reflect the '/\' characters

现在我们进入程序的主要部分。

A←⎕              Assign the input to variable A
⍳                Create a range 0 .. A-1
⌽                Reverse it so that it becomes A-1 .. 0
¨                For each element do (the right argument is the element):
 ⍳⍵+1             Create a range 0 .. ⍵
 ∘.+⍨             Create an addition table using the range to result in a matrix like so:
                   0+0 0+1 0+2 .. 0+⍵
                   1+0 1+1 1+2 .. 1+⍵
                   2+0 2+1 2+2 .. 2+⍵
                   ...
                   ⍵+0 ⍵+1 ⍵+2 .. ⍵+⍵
 ⍵≤              The elements of the matrix that are greater than or equal to the ⍵,
                 this creates a triangle matrix that looks like this:
                   0 0 .. 0 1
                   0 0 .. 1 1
                   ..
                   1 1 .. 1 1
 ' /'[...]       Index it in ' /' to get a character matrix
                 (ie replace 0s with spaces and 1s with '/'s)
 ⊖               Flip this vertically
 A↑              Pad the top spaces

必须确保为该范围内的每个元素创建的所有三角形都⌽⍳A具有相同的高度，以便以后可以将它们彼此串联。

 ⊖               Flip the matrix vertically again to go back to the original state
 (⊢,  )          Concatenate it with
    ⌽B           itself, but flipped horizontally
,/              Concatenate all triangles formed by the range operator
⊃               The resulting matrix is nested, so this operator "un-nests" it

现在，该模式的左上部分已完成。剩下的就是先垂直翻转然后水平翻转。

(⊢⍪⊖B)          Concatenate the resulting matrix with itself but flipped vertically
                (the vertically flipped matrix is concatenated below of the original matrix)
                Now the left part of the pattern is complete
(⊢,⌽B)         Concatenate the resulting matrix with itself flipped horizontally

就是这样！输出是一个带有/\s并用空格填充的字符矩阵。

05AB1E，47 43 41 35 34 33 32字节

'/×ηηvy∞.C.Bø€∞¹NαGð.ø}})øíJ.B»∞

在线尝试！

（感谢@Emigna提出了3处改进，提供了4个字节）

此说明适用于早期版本，此后进行了几次迭代。

>                                          # [2]
 '/×                                       # ['//']
    η                                      # ['/','//']
     €η                                    # [['/'], ['/', '//']]
       vy                    }             # {For each element...}
         ∞                                 # Mirror horizontally.
          ¶¡                               # Split mirror on newlines.
            N£                             # Shave each diamond down a layer.
              .C                           # Horizontal center.
                .B                         # Pad for the transpose.
                  ø                        # Transpose.
                   €∞                      # Mirror each (vertically).
                     ¹NαFð.ø}              # Pad each list for transpose (verticaly).
                              )            # Wrap back to list...
                               €.B         # Pad each horizontally.
                                  ¦        # Remove the random numbers?
                                   ø       # Back to horizontal...
                                    €R     # Reverse to get correct order.
                                      J    # Join, no spaces.
                                       »   # Join newlines.
                                        ∞  # Final horizontal mirror.

CJam，65 63字节

q~_,:)_W%\+f{_2*S*a@2$-*\_,f{)'/*\Se[_W%'/'\er+}_W%Wf%+1$++}zN*

在线尝试！

说明

在此说明中，我将输入数字称为n。

q~        e# Read and eval the input (push n to the stack).
_,        e# Copy it an get the range [0 .. n-1].
:)        e# Increment each element to get [1 .. n].
_W%       e# Copy it and reverse it.
\+        e# Prepend the reverse to the original range, resulting in [n n-1 .. 1 1 .. n-1 n].
f{        e# Map over each number x in the range using n as an extra parameter:
 _2*S*a   e#  Push a string containing n*2 spaces, and wrap it in an array.
 @2$-     e#  Push n-x.
 *        e#  Repeat the space string from before n-x times.
 \        e#  Bring x back to the top.
 _,       e#  Copy it and get the range [0 .. x-1].
 f{       e#  Map over each number y in this range, using x as an extra parameter:
  )       e#   Increment y.
  '/*     e#   Repeat '/' y times.
  \Se[    e#   Pad the resulting string to length x by adding spaces to the left.
  _W%     e#   Copy the result and reverse it.
  '/'\er  e#   Replace '/' with '\' in that.
  +       e#   Concatenate to the other string. This forms one row of one diamond.
 }        e#  (end map, now we have the top half of a diamond of size x)
 _W%      e#  Copy the half-diamond and reverse it.
 Wf%      e#  Reverse each row.
 +        e#  Concatenate to the top half. Now we have a full diamond of size x.
 1$++     e#  Put the spaces from before at the beginning and end. This is to pad the top
          e#  and bottom of the smaller diamonds.
}         e# (end map)
z         e# Transpose.
N*        e# Join with newlines. Implicit output.

Python 2中，152 147 143 140个字节

^{-1个字节，感谢musicman523}

n=input()
r=range(n)
r+=r[::-1]
for x,i in enumerate(r):a,b='/\\\/'[i<x::2];s=' '*(n+~i);print''.join((s+a*n)[:n-j]+(b*-~i+s)[j:]for j in r)

在线尝试！

这是通过将最大的菱形的内柱切成较小的柱来进行的，[0,..,n,n,..,0]用于控制要去除的柱的数量。

Pyth，35 32字节

j.tsm+Jm.[yQS*"\/"k\ Sd_M_Js__BS

测试套件

了解了我和@LeakyNun的方法会有何不同。

Dyalog APL，46岁

{⊃,/⍵∘{'/ \'[2+((-⍪⊖)⌽,-)(-⍺)↑∘.≥⍨⍳⍵]}¨(⌽,⊢)⍳⍵}

Pyth，49个字节

LC.tb)L+_b.rb"\/"js_,J'MsMCyMy_Mm'Myd._._*\/Q__MJ

在线尝试！

V，38字节

Àá/Àá\ò2ÙlX$xòÍî
òYGpVæHÄÓ¯¨¯*Ü*©Ü/ ± 

在线尝试！