仕事

空白以外の印刷可能な文字を1つ指定して、その入力の3x3の正方形表現を作成します。たとえば、入力がの場合#、出力は次のとおりです。

###
# #
###

ルール

• 出力形式は厳密ですが、末尾の改行を使用できます。つまり、中央にスペースが必要であり、3行を区切る2つの改行文字も必要です。

テストケース

入力： #

出力：

###
# #
###

入力： A

出力：

AAA
A A
AAA

入力： 0

出力：

000
0 0
000

得点

これはcode-golfです。バイト単位の最短回答が優先されます。

炭、5 3バイト

Ｂ³Ｓ

オンラインでお試しください！編集：@carusocomputingのおかげで40％節約できました。説明：

Ｂ   Draw a box
³   3×3 (second dimension is implicit if omitted)
Ｓ   Using the input character

ニンジン、11バイト

###
# #
###

オンラインでお試しください！

プログラムはキャレットモードで、#sは入力に置き換えられます。

Python 2、32バイト

lambda s:s+s.join(s+'\n \n'+s)+s

オンラインでお試しください！
For s='a'：中間文字列がs+'\n \n'+s生成されa\n \na、s.joinそれを有効にします（太字は追加するものです）。文字列を反復可能として受け入れるため、欠落している2つの文字で囲まれているためです。aa\na a\naaa.join.join

MATL、5バイト

3Y6*c

オンラインでお試しください！

説明

3Y6   % Push predefined literal: [true true true; true false true; true true true]
*     % Implicitly input a character. Multiply element-wise by its code point
c     % Convert to char. Implicitly display. Char 0 is displayed as space

05AB1E、8バイト

4×ð«û3ô»

オンラインでお試しください！

INPUT    # ['R']                 | Implicit Input: 'R'
---------#-----------------------+-------------------------------
4×       # ['RRRR']              | Repeat string 4 times.     
  ð      # ['RRRR',' ']          | Push space onto top of stack.
   «     # ['RRRR ']             | Concatenate last 2 items.
    û    # ['RRRR RRRR']         | Palindromize.
     3ô  # [['RRR','R R','RRR']] | Split into 3 pieces.
       » # ['RRR\nR R\nRRR']     | Join with newlines
---------#-----------------------+-------------------------------
OUTPUT   # RRR                   | Implicitly print the top
         # R R                   | of the stack on exit.
         # RRR                   |

30を2進数として使用する元のアイデア（未完成、他の言語で誰かがこれを試してみます）：

05AB1E、12バイト

30bûTIð«‡3ô»

オンラインでお試しください！

Python 3.6, 33 bytes

lambda c:f'{3*c}\n{c} {c}\n{3*c}'

Try it online!

RPL (Reverse Polish Lisp), 60 characters

→STR 1 4 START DUP NEXT " " + SWAP + 4 ROLLD + + SWAP 2 PICK

(Note that "→" is a single character on the HP48 and compatible calculators)

Would visually represent what you want by having three items on the stack:

3.: "###"
2.: "# #"
1.: "###"

If you insist to return it as one string, one has to add the newline characters as well and combine the strings, left as exercise to the next tester.

Explanation:

• →STR: Make the last object in the stack into a string. (So the input can be anything, e.g. a number.)
• 1 4: Push the number 1 and 4 to the stack.
• START [...] NEXT：forループに似ていますが、カウンター変数にアクセスできません。スタックから2つの数値を取得し（ここではプッシュ1と4）[...]、対応する回数（ここでは4回）コードを実行します。
• DUP：スタック内の最後のエントリを複製します。
• " "：文字列（つまり、空白が1つある文字列）をスタックにプッシュします。
• +：スタックから2つのオブジェクトを取得し、それらを一緒に追加して返します。
• 4：番号4をスタックにプッシュします。
• ROLLD：4スタックから最後の要素（ここではプッシュしたばかり）を取得し、スタックから取得した数値が指定するスタックの次の要素をロールダウンします。
• SWAP：最後の2つのスタック要素を交換します。
• 2：2スタックにプッシュします。
• PICK: Takes an element (here: The 2 we just pushed to the stack), interprets it as a number n, and copies the nth element from the stack.

JavaScript, 28 bytes

c=>c+c+c+`
${c} ${c}
`+c+c+c

Try it

f=
c=>c+c+c+`
${c} ${c}
`+c+c+c
o.innerText=f(i.value="#")
i.oninput=_=>o.innerText=f(i.value)

<input id=i maxlength=1><pre id=o>

Jelly, 8 bytes

1 byte thanks to Erik the Outgolfer.

x4,`Ks3Y

Try it online!

Java 7, 56 55 bytes

-1 Thanks to Leaky Nun for pointing out the space I missed

String a(char s){return"...\n. .\n...".replace('.',s);}

Simply replaces the periods with the given character, for input #:

...       ###
. .  =>   # #
...       ###

Try it online!

PHP, 32 Bytes

<?=strtr("000
0 0
000",0,$argn);

Try it online!

sed, 28 18 bytes

s:.:&&&\n& &\n&&&:

Try it online!

Brain-Flak, 76 70 + 1 = 71 bytes

Requires the -c flag

(((((((({})))<((([])[]{}<>)<>)>)<(<>({})({}){}()()<>)>)<([]()()())>)))

Try it online!

Pyth, 7 bytes

jc3.[9d

Try this online.

Explanation:

jc3.[9d Expects quoted input.
  3     3
     9  9
      d ' '
        Q (eval'd input) as implicit argument
   .[   Pad B on both sides with C until its length is a multiple of A
 c      Split B to chunks of length A, last chunk may be shorter
j       Join A on newlines

Brain-Flak, 61, 59 bytes

(((((((({})))<([][][]())>)<(([][][]()){})>)<([]()()())>)))

Try it online!

This is 58 bytes of code +1 byte for the -c flag which enables ASCII input and output.

Explanation:

(((
   (
    (
     (

      #Duplicate the input 3 times
      ((({})))

#Push 10 (newline)
<([][][]())>

     #Push the input again
     )

#Push 32 (space)
<(([][][]()){})>

    #Push the input again
    )

#Push 10 (newline)
<([]()()())>)

#Push input 3 times
)))

C (gcc), 49 47 bytes

Saved 2 bytes thanks to 2501!

j;f(i){for(j=12;j;)putchar(--j%4?j-6?i:32:10);}

Try it online! has a trailing newline

Python 3, 34 bytes

lambda s:s*3+"\n"+s+" "+s+"\n"+s*3

Try it online!

Octave, 36 bytes

x=repmat(input(0),3);x(5)=32;disp(x)

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Explanation

This creates a 3x3 char matrix with the input char repeated, and sets its 5th entry in column-major order (i.e. its center) to 32 (ASCII for space).

Pyke, 5 bytes

dQA.X

Try it online!

Ruby, 27 25 bytes

Saved 2 bytes thanks to Level River St

->x{[s=x*3,x+" "+x,s]*$/}

Try it online!

Brainfuck, 40 bytes

+++++[->++<<++++++>],...>.<.<++.>.>.<...

Try it online! Requires an implementation that can access left of the starting position.

Also see: Graviton's brainfuck answer which takes a different approach (but is longer).

Explanation:

Brainfuck can do a lot of cool tricks with its limited instruction set. Unfortunately, this answer doesn't use any of them, because it's cheaper (in terms of bytes) to just hardcode everything.

+++++[->++<<++++++>]                         Sets the cells to |5*6|>0<|5*2|
,                   Takes input character into the middle cell | 30|>#<| 10|
...                                Print the top of the square | 30|>#<| 10| ###
>.                                   Print a newline character | 30| # |>10|    \n
<.                               Print another input character | 30|>#<| 10| #
<++.                  Add 30+2 for a space character and print |>32| # | 10|  _
>.                   And just print the 5 remaining characters | 32|>#<| 10|   #
>.                                                             | 32| # |>10|    \n
<...                                                           | 32|>#<| 10| ###

# = input character, _ = space (ASCII 32), \n = newline (ASCII 10)

Results in this beautiful box (for input '+'):

+++
+ +
+++

05AB1E, 7 6 bytes

-1 byte thanks to carusocomputing.

ж¹ðJû

Explanation:

         # Implicit input                  # ['R']
 Ð       # Repeat string three times       # ['R', 'R', 'R']
  ¶      # Push newline character          # ['R', 'R', 'R', '\n']
   ¹     # Push first input                # ['R', 'R', 'R', '\n', 'R']
    ð    # Push space                      # ['R', 'R', 'R', '\n', 'R', ' ']
     J   # Join stack                      # ['RRR\nR ']
      û  # Palindromize ("abc" -> "abcba") # ['RRR\nR R\nRRR']
         # Implicit output                 # []

Uses the CP-1252 encoding. Try it online!

Pyth, 11 bytes

jc++K*z4dK3

Try it online!

Explanation:

jc++K*z4dK3    expects a single char as input

j              joins on new line
 c        3    chops array into 3 sized pieces
  +            joins +K*z4d and K
   +           joins K*z4 and d
    K          initialize variable K as *z4
     *z4       duplicate the input 4 times
        d      variable initialized to string " "
         K     calls variable K, in this case *z4

><>, 24 23 bytes

i:o:o:oao:o84*o:a0!.<o$

Try it online!

Swift3, 50 bytes

[1,2,3].map{$0==2 ? print(c+" "+c) : print(c+c+c)}

This uses the ternary operator to print different strings, depending on the row.

Try it online

V, 8 bytes

x3p2Ùlr 

Try it online!

C#,50 bytes

a=>Console.Write(a+a+a+"\n"+a+" "+a+"\n"+a+a+a);

Test Case:

var f = new Action<string>(
a=>Console.Write(a+a+a+"\n"+a+" "+a+"\n"+a+a+a);
);
f("#");

Vim, 9 keystrokes

Assuming the input char is present in a buffer, vim makes this straightforward

x3pY2plr<space>

There is probably some magic vim commands of use here (there always seem to be some) so improvement suggestions are welcome. Only one keystroke behind V!

Z80 or 8080 Assembly, 21 bytes machine code

Assume a memory mapped I/O device:

              Z80                  8080
3A xx xx    ld  a, (input)      lda  input       ; get input character
11 0A 20    ld  de, 200ah       lxi  d, 200ah   ; space & newline
21 yy yy    ld  hl, output      lxi  h, output  ; get output address
77          ld  (hl), a         mov  m, a       ; output character * 3
77          ld  (hl), a         mov  m, a
77          ld  (hl), a         mov  m, a
73          ld  (hl), e         mov  m, e       ; output newline
77          ld  (hl), a         mov  m, a       ; output character
72          ld  (hl), d         mov  m, d       ; output space
77          ld  (hl), a         mov  m, a       ; output character
73          ld  (hl), e         mov  m, e       ; output newline
77          ld  (hl), a         mov  m, a       ; output character * 3
77          ld  (hl), a         mov  m, a
77          ld  (hl), a         mov  m, a
76          halt                hlt             ; or C9 ret

No interpreter needed!

Hexdump:

0000: 3A 00 FF 11 0A 20 21 01 FF 77 77 77 73 77 72 77
0010: 73 77 77 77 76

where the input address is at FF00h and the output address is mapped at FF01h. The actual addresses will depend on the actual hardware. Of course this assumes the I/O is memory mapped. If it is I/O mapped, it would take several extra bytes because Z80 & 8080 I/O instructions are two bytes each. This also assumes the output device interprets 0Ah as a newline and doesn't require a CR (0Dh) which would add an extra 4 bytes to the program.

J-uby, 22 20 bytes

-2 bytes thanks to @Jordan

:tr&"...
. .
..."&?.

Explanation

String#tr is Ruby's character-wise replace method. The first & binds :tr to "...\n. .\n...", and the second partially applies '.' to it. Effectively, this is ->s{"...\n. .\n...".tr('.',s)}