我从Codingame那里获得了这一挑战，并且对比我的更好的解决方案感到好奇：

通过标准输入给定宽度，在给定的宽度和长度上绘制一个空心的＃号正方形。

例：

5结果

#####
#   #
#   #
#   #
#####

我用python解决了这个问题，所以我对其他python代码特别感兴趣。但是，请随时以所需的任何语言发布您的解决方案。

Pyke，11个字节

ttDd*n+*.X#

在这里尝试！

ttDd*n+*    - a square of spaces n-2*n-2 big
        .X# - surround in `#`

木炭，6个字节

码：

ＮβＢββ#

说明：

Ｎβ        # Get input from the command line and store into β
   Ｂ      # Draw a hollow box with...
     β     #  Width β
      β    #  Height β
       #   #  Filled with the character '#'
           # Implicitly output the box

在线尝试！

MATL，12字节

:G\1>&*~35*c

在线尝试！

说明

:     % Input n implicitly. Push range [1 2 ... n]
      % STACK: [1 2 3 4 5]
G     % Push n again
      % STACK: [1 2 3 4 5], 5
\     % Modulo
      % STACK: [1 2 3 4 0]
1>    % Does each entry exceed 1?
      % STACK: [0 1 1 1 0]
&*    % Matrix with all pair-wise products
      % STACK: [0 0 0 0 0;
                0 1 1 1 0;
                0 1 1 1 0;
                0 1 1 1 0;
                0 0 0 0 0]
~     % Negate
      % STACK: [1 1 1 1 1;
                1 0 0 0 1;
                1 0 0 0 1;
                1 0 0 0 1;
                1 1 1 1 1]
35*   % Multiply by 35
      % STACK: [35 35 35 35 35;
                35  0  0  0 35;
                35  0  0  0 35;
                35  0  0  0 35;
                35 35 35 35 35]
c     % Convert to char. 0 is interpreted as space. Display implicitly
      % STACK: ['#####';
                '#   #';
                '#   #';
                '#   #';
                '#####']

Jolf，8个字节

,ajj"###
,ajj      draw a box with height (input) and width (input)
    "###  with a hash border

Python 2，62 54字节

f=lambda n:'#'*n+'\n#%s#'%(' '*(n-2))*(n-2)+'\n'+'#'*n

返回#\n#时，输入1

打印的55字节版本

def f(n):a=n-2;print'#'*n,'\n#%s#'%(' '*a)*a,'\n'+'#'*n

62字节版本适用于任何输入：

f=lambda n:'#'*n+'\n#%s#'%(' '*(n-2))*(n-2)+('\n'+'#'*n)*(n>1)

COW，426个 405 348 330字节

MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMMMmoOMMMMoOMoOMoOMoOMoOMoOMoOMoOMoO
MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMMMmoOMMMMoOMoOMoOmoOoomMMM
moOMMMMOOmOomOoMoomoOmoOMOomoomOoMMMmoOMMMMOoMOoMOOmOomOomOomOoMoo
moOmoOMoomoOMMMmoOmoOMMMMOoMOoMOOmOomOomOomOoMoomoOmoOmoOmoOMOomoo
mOomOomOoMoomoOmoOMOomoomOomOomOomOoMoomoOmoOmoOMOOmOoMoomoOMOomoo

在线尝试！ 将第二行中的数字更改为任何数字以更改输出。

我在这里使用的COW解释器是用Perl编写的（比这个挑战还新），但是您仍然可以通过在此处输入代码来获得相同的结果。

说明

; Note: [n] means "value stored in the nth block of memory".
MoOMoOMoOMoOMoOMoOMoOMoOMoOMoO                                                  ;Stores 10 in [0].  10 is the code point for carriage return
MMMmoOMMMMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoO     ;Stores 32 in [1].  32 is the code point for whitespace
MMMmoOMMMMoOMoOMoO                                                              ;Stores 35 in [2].  35 is the code point for #
moOoom                                                                          ;Reads STDIN for an integer, and stores it in [3]
MMMmoOMMM                                                                       ;Copies [3] into [4] 
MOO                                                                             ;Loop as long as [4] is non-zero
    mOomOoMoo                                                                   ;Navigate to [2] and print the character with that code point
    moOmoOMOo                                                                   ;Navigate to [4] and decrement
moo                                                                             ;End loop
mOoMMMmoOMMMMOoMOo                                                              ;Copy [3] into [4] and decrement [4] twice
MOO                                                                             ;Loop as long as [4] is non-zero
    mOomOomOomOoMoo                                                             ;Navigate to [0] and print the character with that code point
    moOmoOMoo                                                                   ;Navigate to [2] and print the character with that code point
    moOMMMmoOmoOMMMMOoMOo                                                       ;Navigate to [3] and copy it into [5], then decrement [5] twice
    MOO                                                                         ;Loop as long as [5] is non-zero
        mOomOomOomOoMoo                                                         ;Navigate to [1] and print the character with that code point
        moOmoOmoOmoOMOo                                                         ;Navigate to [5] and decrement
    moo                                                                         ;End loop
    mOomOomOoMoo                                                                ;Navigate to [2] and print the character with that code point
    moOmoOMOo                                                                   ;Navigate to [4] and decrement
moo                                                                             ;End loop
mOomOomOomOoMoo                                                                 ;Navigate to [0] and print the character with that code point
moOmoOmoO                                                                       ;Navigate to [3]
MOO                                                                             ;Loop as long as [3] is non-zero
    mOoMoo                                                                      ;Navigate to [2] and print the character with that code point
    moOMOo                                                                      ;Navigate to [3] and decrement
moo                                                                             ;End loop

Python 2，59 58字节

n=i=input()
while i:print'#%s#'%((' #'[i%n<2])*(n-2));i-=1

代表

注意：输入1会产生的输出##，但输入小于时，永远不会产生空心正方形3，所以我想这很好。

Java 7中，113个 112 110字节

String c(int n){String r="";for(int i=n,j;i-->0;r+="\n")for(j=0;j<n;r+=i*j<1|n-i<2|n-j++<2?"#":" ");return r;}

@OlivierGrégoire节省了1个字节；@cliffroot
节省了2个字节。

基于我的创建交叉正方形答案的派生解决方案。

在这里尝试。

PowerShell v2 +，48 47字节

param($n)($z='#'*$n--);,("#$(' '*--$n)#")*$n;$z

-1个字节感谢JohnLBevan

将input $n，sets设置$z为$n哈希标记，$n后减。将其封装在括号中以将副本放置在管道上。然后使用逗号运算符来创建预递减的阵列$n的线#，空格#。这些留在管道上。然后$z再次放置在管道上。Write-Output最后通过隐式输出在元素之间引入了换行符，因此我们免费获得换行符。

由于OP的代码不适用于输入n <= 1，因此我认为这也意味着我们也不需要支持输入1。

例子

PS C:\Tools\Scripts\golfing> 2..6|%{"$_";.\draw-a-hollow-square.ps1 $_;""}
2
##
##

3
###
# #
###

4
####
#  #
#  #
####

5
#####
#   #
#   #
#   #
#####

6
######
#    #
#    #
#    #
#    #
######

C，98字节

f(n,i){i=n*(n+1);while(i--){putchar(i%(n+1)==n?10:i<n||i>n*n-1||i%(n+1)==0||i%(n+1)==n-1?35:32);}}

用法：

f(5)

05AB1E，20个字节

'#×D¹Íð×'#.ø¹Í×sJ¹ä»

在线尝试！

或18个字节，如果我们不能忽视1 <= n：

F„ #N¹<%_è¹Í×'#.ø,

在线尝试！

WinDbg中，206个 200 182 170字节

.if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}

从中删除括号.if并使用j而不是使用-6个字节.if

通过使用f而不是a .for来构造字符串的-18个字节。

-12个字节（不是以NULL结尾的字符串），而是将长度传递给 da

输入通过伪寄存器$t0（例如r $t0 = 5; {above-code}）传递。

说明：

.if @$t0                                                *Verify width($t0) at least 1 
{                                                       *(registers have unsigned values) 
    r $t3 = 2000000;                                    *Set $t3 to address where the 
                                                        *string will be constructed
    f @$t3 L@$t0 23;                                    *Put width($t0) '#' at 2000000($t3)
    f 2 * @$t3 L@$t0 20;                                *Put width($t0) ' ' at 4000000(2*$t3)
    eb 2 * @$t3 23;                                     *Put '#' on left of ' ' string
    eb 2 * @$t3 + @$t0 - 1 23;                          *Put '#' on right of ' ' string
    da @$t3 L@$t0;                                      *Print the top of the box
    j 1 < @$t0                                          *If width($t1) at least 2
    '
        .for (r $t1 = @$t0 - 2; @$t1; r $t1 = @$t1 - 1) *Loop width($t0)-2 times to...
        {
            da 2 * @$t3 L@$t0                           *...print the sides of the box
        };
        da @$t3 L@$t0                                   *Print the bottom of the box
    '
}

样本输出：

0:000> r$t0=0
0:000> .if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}

0:000> r$t0=1
0:000> .if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}
Filled 0x1 bytes
Filled 0x1 bytes
02000000  "#"

0:000> r$t0=2
0:000> .if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}
Filled 0x2 bytes
Filled 0x2 bytes
02000000  "##"
02000000  "##"

0:000> r$t0=5
0:000> .if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}
Filled 0x5 bytes
Filled 0x5 bytes
02000000  "#####"
04000000  "#   #"
04000000  "#   #"
04000000  "#   #"
02000000  "#####"

JavaScript，61 58字节

@lmis节省了3个字节！

n=>(b='#'[r='repeat'](n))+`
#${' '[r](n-=2)}#`[r](n)+`
`+b

（不处理0或1）

对于13个额外的字节（为71个字节），您可以！

n=>n?n-1?(b='#'[r='repeat'](n))+`
#${' '[r](n-=2)}#`[r](n)+`
`+b:'#':''

这些解决方案非常简单：它们进行大量存储，以免重复进行以节省一些字节。如果没有变量，它将变为最小：

n => // Anonymous function definition (Param `n` is the size)
    '#'.repeat(n) +      // # `n` times to form the top
    `
#${' '.repeat(n - 2)}#`  // Followed by a newline followed by a hash and `n` - 2 spaces and
                         // another hash to make one of the middle lines
    .repeat(n - 2) +     // The above middle lines repeated `n` - 2 times
    '#'.repeat(n)        // Followed by the top line again

试试吧！

<script type="text/babel">var f=n=>n?n-1?(b='#'[r='repeat'](n))+`\n#${' '[r](n-=2)}#`[r](n)+`\n`+b:'#':'',b,r;function c(){document.getElementById('pre').textContent = f(+document.getElementById('input').value);}</script><input id="input" onkeydown="c();" onkeyup="c();" onchange="c();" onclick="c();" placeholder="Size"><pre id="pre"></pre>

Python，109个字节

n=int(input())
for x in range(n):
 r=list(' '*n);r[0]=r[-1]='#'
 if x%(n-1)==0:r='#'*n
 print("".join(r))

Ruby, 39 bytes

->n{puts a=?#*n,[?#+' '*(n-=2)+?#]*n,a}

Turns out to be shorter this way than all the fancy stuff I was trying. Be advised that this doesn't handle 0 or 1 at all.

Python 2, 50 bytes

m=input()-2
for c in'#'+' '*m+'#':print'#'+m*c+'#'

Works for n>=2. Prints each line with a pound sign, n-2 of the appropriate symbol, then another pound sign.

Aliasing the pound symbol gives same length:

m=input()-2;p='#'
for c in p+' '*m+p:print p+m*c+p

Other attempts:

lambda n:'#'*n+('\n#'+' '*(n-2)+'#')*(n-2)+'\n'+'#'*n

lambda n:'#'*n+'\n#%s#'%((n-2)*' ')*(n-2)+'\n'+'#'*n

lambda n:'\n'.join(['#'*n]+['#'+' '*(n-2)+'#']*(n-2)+['#'*n])

n=input();s='#'+' '*(n-2)+'#'
for c in s:print[s,'#'*n][c>' ']

s='##'+' #'*(input()-2)+'##'
for c in s[::2]:print s[c>' '::2]

s='#'+' '*(input()-2)+'#'
for c in s:print s.replace(' ',c)

Haskell, 49 bytes

n%b='#':(b<$[3..n])++"#\n"
f n=(n%)=<<init(n%' ')

Works for n>=2. Defines the operation of sandwiching a character between # for an n-character newline-terminated string, then applies it twice to make a 2D grid.

Call like:

*Main> putStrLn$ f 5
#####
#   #
#   #
#   #
#####

C, 83 82 80 78 77 Bytes

i,j;f(n){for(i=n;i--;puts(""))for(j=n;j--;putchar(i*j&&i^n-1&&j^n-1?32:35));}

Sneak in a multiply and save a byte...

i,j;f(n){for(i=n;i--;puts(""))for(j=n;j--;putchar(i&&j&&i^n-1&&j^n-1?32:35));}

Also count down j and save a few more...

i,j;f(n){for(i=n;i--;puts(""))for(j=0;j++<n;putchar(i&&j^1&&i^n-1&&j^n?32:35));}

Count down i from n to zero and save a few bytes...

i,j;f(n){for(i=0;i++<n;puts(""))for(j=0;j++<n;putchar(i^1&&j^1&&i^n&&j^n?32:35));}

A bit easier to understand and 1 byte more

i,j;f(n){for(i=0;i++<n;puts(""))for(j=0;j++<n;putchar(i==1|i==n|j==1|j==n?35:32));}

Groovy, 51 50 bytes

{n->a="*"*n+"\n";n-=2;print(a+"*${' '*n}*\n"*n+a)}

PHP, 81 69 bytes

for($n=-1+$i=$argv[1];$i--;)echo str_pad("#",$n," #"[$i%$n<1]),"#\n";

Run with -r; provide input as argument.

Throws a DivisionByZeroError for input=1.

R, 68 70 bytes

Works for n > 1. Thanks to @Billywob for a couple of bytes swapping out the array for a matrix.

cat(rbind(b<-'#',cbind(b,matrix(' ',n<-scan()-2,n),b),b,'
'),sep='')

Uses rbind and cbind to put rows and columns of #'s around an n-2 square matrix of spaces. Newlines are bound to the rows as well. The newline in the source is significant. Input is from STDIN

Common Lisp, 150 130 bytes

-20 thanks to @Cyoce and @AlexL.

(defun s(v)(format t"~v,,,vA~%"v #\# #\#)(dotimes(h(- v 2))(format t"~v,,,vA~A~%"(- v 1)#\  #\# #\#))(format t"~v,,,vA"v #\# #\#))

Usage:

* (s 5)
#####
#   #
#   #
#   #
#####

Basically uses format twice for the top and bottom and a loop for the rows in between. The format call for the top and bottom outputs a line starting with # and padded to the appropriate width with #s. The format call for the rows in between works similarly, except the padding is spaces and a # gets printed at the end of the line.

Note: I'm rather new to Lisp and expect to have a lot of room for improvement on this.

Haskell, 67 bytes

l#n=l<$[1..n]
f n=unlines$'#'#n:('#':' '#(n-2)++"#")#(n-2)++['#'#n]

Usage example:

Prelude> putStrLn $ f 4
####
#  #
#  #
####

How it works:

l#n=l<$[1..n]                      -- helper function that makes n copies of l

   '#'#n                           -- make a string of n copies of #, followed by
                        #(n-2)     -- n-2 copies of
     '#':' '#(n-2)++"#"            -- # followed by n-2 times spaces, followed by #
                           ['#'#n] -- and a final string with n copies of #
unlines                            -- join with newlines in-between

Jelly, 13, bytes

,þ%µỊṀ€€ị⁾# Y

TryItOnline! or try 0 to 15

How?

,þ%µỊṀ€€ị⁾# Y - Main link: n
 þ            - outer product with
,             -    pair:   [[[1,1],[2,1],...,[n,1]],[[1,2],[2,2],...,[n,2]], ... ,[[1,n],[2,n],...,[n,n]]]
  %           - mod n:     [[[1,1],[2,1],...,[0,1]],[[1,2],[2,2],...,[0,2]], ... ,[[1,0],[2,0],...,[0,0]]]
   µ          - monadic chain separation
    Ị         - abs(z)<=1: [[[1,1],[0,1],...,[1,1]],[[1,0],[0,0],...,[1,0]], ... ,[[1,1],[0,1],...,[1,1]]]
      €€      - for each for each
     Ṁ        - maximum:   [[1,    1,    ...,1],    [1,    0,    ..., 1],    ... ,[1,    1,    ..., 1]   ]
        ị     - index into (1 based)
         ⁾#   - "# ":      ["##...#","# ...#", ...,"##...#"]
           Y  - join with line feeds

Pip, 16 bytes

15 bytes of code, +1 for -n flag.

(Y_Xa-2WR'#s)My

Works for input >= 2. Try it online!

Explanation of somewhat ungolfed version

First, we define a function y that takes a string argument, repeats it a-2 times (where a is the first command-line input), and wraps the result in #.

Y _ X a-2 WR '#
  _              Identity function
    X a-2        String-repeated by a-2
          WR '#  Wrapped in #
Y                Yank the resulting function into y

Next, we apply this function twice--once normally, then again with map--to obtain the square as a list of strings:

y M (y s)
    (y s)  Call function y with s (preinitialized to " ") as argument
y M        Map y to each character of the resulting string

For input of 4, (y s) results in "# #" and y M (y s) in ["####"; "# #"; "# #"; "####"]. This latter value is then printed, with the -n flag causing it to be newline-separated.

Golfing tricks

To get from the ungolfed to the golfed version:

• Remove spaces.
• Y is an operator, which means we can use it in an expression. Instead of Y... followed by (ys), we can just do (Y...s).
• The problem is, we have to yank the function before we reference it again as y; so yM(Y_Xa-2WR'#s) won't work. Solution: swap the operands of the Map operator. As long as one of them is a function and the other is an iterable type, it doesn't matter what order they come in.

Racket 113 bytes

(let*((d display)(g(λ()(for((i n))(d"#")))))(g)(d"\n")(for((i(- n 2)))(d"#")(for((i(- n 2)))(d" "))(d"#\n"))(g))

Ungolfed:

(define (f n)
  (let* ((d display)
         (g (λ () 
              (for ((i n))
                (d "#"))
              (d "\n"))))
    (g)
    (for ((i (- n 2)))
      (d "#")
      (for ((i (- n 2)))
        (d " ") )
      (d "#\n"))
    (g)))

Testing:

(f 5)

Output:

#####
#   #
#   #
#   #
#####

SpecBAS - 57 bytes

1 INPUT n: a$="#"*n,n-=2,b$="#"+" "*n+"#"#13: ?a$'b$*n;a$

? is shorthand for PRINT, #13 is carriage return which can be tacked on to the end of a string without needing a + to join them.

The apostrophe moves print cursor down one line.

Stuck, 29 27 Bytes

Pretty darn long for a "golfing" language, but I have forgotten how a lot of it works :P

i_2-_u'#*N+_'#' u*'#N+++u*u

Explanation:

i_2-_u                           # take input and triplicate, subtracting 2 (5 -> [3,3,5])
      '#*N+_                     # create the top and bottom rows
            '#' u*'#N+++u*       # create input - 2 copies of middle rows
                          u      # rotate left 1 to get correct order, implicit output

C#, 154 152 bytes

Golfed:

void F(int n){Console.Write($"{new string('#',n)}\n");for(int i=2;i<n;i++)Console.Write($"#{new string(' ',n-2)}#\n");Console.Write(new string('#',n));}

Ungolfed:

    void F(int n)
    {
        Console.Write($"{new string('#', n)}\n");

        for (int i = 2; i < n; i++)
            Console.Write($"#{new string(' ', n - 2)}#\n");

        Console.Write(new string('#', n));
    }

EDIT1: Loop range optimization.

Lithp, 117 bytes

Line split in two for readability:

#N::((var X (repeat "#" N))(print X)(each (seq 3 N) (scope #X::((print (+ "#" 
     (repeat " " (- N 2)) "#")))))(print X))

Sample usage:

% square.lithp
(
    (import "lists")
    (def s #N::((var X (repeat "#" N))(print X)(each (seq 3 N) (scope #X::((print (+ "#" (repeat " " (- N 2)) "#")))))(print X)))
    (s 10)
)

Output:
$ ./run square.lithp
##########
#        #
#        #
#        #
#        #
#        #
#        #
#        #
#        #
##########