这是警察的话题。强盗的线索在这里。

编写仅由可打印ASCII字符（不包括制表符和换行符）组成的程序或函数，该程序或函数至少按升序（从空格到代字号/ 32到126）输出5个可打印ASCII字符。字符可以输出多次，只要它们彼此相邻即可。接受一条尾随的换行符。您必须提供完整的输出，并且对于代码中的每6个字符，您必须在代码中显示的位置提供一个字符。如果您的代码包含5个或更少的字符，那么您只需要显示长度即可。您的代码中每6个字符不得超过1个字符。

因此，如果您的代码alphaprinter从az输出字母，那么您需要显示代码的2个字符（对于其他字符，请使用下划线），例如：

al__________  // or
__p__p______  // or
__________er

规则/规格：

• 您不能追加没有功能的尾随空格。
• 您不能使用注释（但强盗可以在破解代码时使用注释）
• 不允许使用内置的加密原语（包括任何rng，加密，解密和哈希）。
• 在默认输出类似于MATLAB：的语言中ans =，则可以接受，只要明确说明并显示ans =输出即可。还应明确说明这是否属于“递增输出”的一部分。
• 输出必须是确定性的
• 不接受使用非自由语言
• 输出不必是连续字符。这两个abcde和abcdg有效输出。
• 只要输出包含至少5个字符，就不必区分字符。aaaaa并且aaacc有效。
• 假定具有REPL环境的答案不被视为有效的程序或功能，因此是不允许的。
• STDERR不被视为有效输出，但是可以将数据写入STDERR。

如果显示的字符带有下划线，则必须按如下所示进行识别。在这种情况下，第二个和第六个字符显示为下划线，而其他字符未知。

_____________
 |   |

注意：强盗仅需要找到产生相同输出的代码。语言，强盗解决方案的长度以及所显示字符的位置也必须匹配。

您的分数就是代码中的字符数。获胜者将是7天内未破解的最低分数。只有2015年（UTC）发布的作品才有资格获奖。欢迎晚于此发布的提交，但不能获胜。

为了获得胜利，您需要显示完整代码（7天后）。

您的帖子应采用以下格式（nn是字符数）：

语言，nn个字符

输出：

abcdefghijklmnopqrstuvwxyz

代码（12个字符）：

al__________

如果代码被破解，则在标题中插入[Cracked]（链接到cracker）。如果提交是安全的，请在标题中插入“安全”，并在答案中显示完整代码。只有显示完整代码的答案才有资格获胜。

六角形，8个字符，安全

_______.

版画

123456

您可以在线尝试Hexagony

解

`&)!."@.

的`仅仅是误导使代码看上去像是需要边长3.除非解释与执行-d确定所述布局前倒引号被简单地从源代码剥离。之后，代码适合侧面长度2：

 & )
! . "
 @ .

这会将内存指针（MP）绕内存网格的一个六边形移动，同时将值复制到下一个边缘，将其递增并打印。

这是内存网格的样子，MP开始显示标记的位置：

最初，所有边都是零。第一个&是no-op，但会)增加边并!打印1。然后"移回左侧（标记为2的边缘）。在那里，&复制一个值。由于当前边缘为零，因此将复制左邻居（在MP方向上），这就是1我们刚刚打印的结果。)增加它，!打印2。只要我们要访问新的边沿并且将所有数字一直打印到，此过程就会继续6。当我们击中边缘时，我们&将复制右边的邻居（因为边缘值是正值），因此边缘0再次变为，控制流跳至最后一行，@ 终止程序。

在线尝试。

Perl，46个字符（安全）

输出：

((()))*++,---../00011123445556667779::;;<<==??@@AAACEFGHHHIKKMMNOOOPPQQQRRSSSVWXXYY[[[\]]]^^^```aaeeffggghjjjnpppqqrttuwxxyzz{{{

码：

print _______________________________________;

我设法在一周的时间内删除了原件，但是我认为这是对的：

print chr()x(($i=($i*41+7)%97)&3)for(40..123);

Brainfuck，48个字符，由Mitch Schwartz 破解

我为强盗做了这个。绝对不会是获奖作品：）

____[_____[_______]__]___[___________]___[_____]

输出：

ABCDEFGHIJKLMNOPQRSTUVWXYZ

解：

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

我在这里测试过。

提示：不要试图查找由在线生成器生成的程序。这是手写的，只能通过逻辑思维解决：)

CJam，13个字符

_____________
  ||

版画

7LLMYahnsuv

您可以在线尝试CJam。

解

{`__Jf^er$}_~

我认为以警察为基础的笼子是聪明而新颖的。发布此刻的那一刻，我意识到__并且er完全没有用，这就是为什么我发布8字节CJam警察以获得更高的竞争力。Pietu很快破解了那个，所以我担心他也会弄清楚这个。我认为不必要的繁琐字符音译可以将其保存下来。

无论如何，代码会使用自己的字符（除外_~），将每个字符与19进行XOR进行混淆，然后对其进行排序。

这个警察使我想到了xorting的“发现” ，尽管我实际上并没有在这里使用它（并且可能几乎不可能用简短的广义quine使用它）。

嘈杂的3SP，89个字符（安全）

0 _____________________________ _ _ _ _ _ _ _ _ _ __2_2__________________________________

原始程序：

0 0 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 3 2 2 2 2 0 3 0 1

该程序打印

 !"#$$

（注意前导空格），仅此而已！

说明（* n表示内存中的第n个单元）：

0 0 0                                Increment *0 to 3 (via the 0s in *0, *1, and *2)
2 (32 times)                         Increment *3 to 32 (via the 0 in *2 and the 3 in *0)
                                      (Note that *3 mod 256 is the character that prints)
0                                    Increment *32 (via the 3 in *0 and the 32 in *3)
3                                    Increment *1 (via the 32 in *3 and the 1 in *32)
                                      (Note that printing is on when *1 mod 2 is 1,
                                       which means " " prints right here)
2 (4 times)                          Increment *3 to 36 (printing '!"#$')
0                                    Increment *36 (via *3) (printing "$" again)
3                                    Increment *1 to 2 (via *36) (printing turns off)
0                                    Increment *36 to 2 (via *3)
1                                    Increment *0 to 4 (via *2)

最后两个命令足以防止在以后通过程序时再次将1递增，这意味着不再打印任何内容，尽管它将继续将所有内存地址从* 36开始永久设置为值36 。

JavaScript（ES6），60个字符，由user81655破解

不希望赢得比赛，但希望能乐在其中：

e_e_______a__a____e___e________e__o___________o______o______

此函数返回：

    :ERacddeeeeeeffiinnnoorrrrst
^^^^ -> 4 whitespaces

编辑

user81655逐个字符地将其破解：

e=e=>{try{a}catch(e){return[...e.toString()].sort().join``}}

Matlab Octave，27个字符，安全

挑战

关于新的语言限制：它也适用于Octave。

____________3_3___4_3______

输出（以八度为单位）：（ans =不是输出的一部分，输出的第一个字符是"）

ans = """""""""""""""""""""""""""""""""""""""""""########$$$$$$%%%%&&&&'''(()))**++,,-../001223456789:;<=>?@BCDFGHJKMNPRTUWY[]_acehjloq

输出（在Matlab中）：（ans = \n\n不是输出的一部分，输出只是最后一行）

ans =

"""""""""""""""""""""""""""""""""""""""""""########$$$$$$%%%%&&&&'''(()))**++,,-../001223456789:;<=>?@BCDFGHJKMNPRTUWY[]_acehjloq

解

[fix((0:.07/3:3).^4+34),'']

我们基本上是在看转换为ASCII值的函数。

R，60字节，由plannapus 破解

输出：

0123456789ABCDEFGHTUVWXYZdefghijklmnopqrstuvw

码：

___(_(_:_____________]_____________6]_______s_4_____,___=___

我认为这不会那么难。我们会看到。

Jolf，27个字符，由Adnan破解

________W___________--_____
                         |

版画

abcdefghijklmnopqrst

口译员。

原始代码：

on*2toHEWn)oH+H `pl$--$n}_H
on                          set n
  *2t                        to twenty
     oHE                    set H to the empty string
        Wn)                 while(n){
           oH+H             append to H
                `pl$--$n     get the nth character of the alphabet, decrement n before
                        }   }
                         _H print H reversed

恭喜阿德南！你吓到我了。

在这里，您可以测试提交的内容，以查看必须删除的字符数。

setInterval(function(){r=document.querySelector("textarea").value;document.getElementById("output").innerHTML=r.length/6-r.split("").filter(function(x){return x=="_"}).length|0},20)

*{font-family:Consolas,monospace}textarea{width:100%;resize:none;height:auto;min-height:1.2em;}b{color:red;}

<textarea>Example Program</textarea><div>You must reveal <b id="output">n</b> characters.</div>

Fortran，45个字符，安全

完整的输出：

 !"$)0<Ka|

猫试图在这里破解它

代码（45个字符）：

pr_gr_m____________________/___3____________

程序代码：

program t;print*,(char(i**3/8+33),i=1,9);end

PHP，46个字符，安全

资源：

for(_______________________=________________);

输出一个84个字符长的字符串：

!!"#$%&'()*+,-./0123456789::;<=>?@AABCDEEFGHIIJKLLMNNOPPQRRSTTUUVWWXXYYZZ[[\\]]^^__`

暗示

你说什么 我的表亲参与其中？

公开代码

此方法相对简单，因为它最多可以打印64个字符chr(32)。令人困惑的是，迭代器i不是线性增加的。它以自己的余弦值递增，这将导致重复和丢​​失字符。

for(;$i<64;)echo chr(32+$i+=cos(deg2rad($i)));
for(_______________________=________________);

CJam，8个字节，由Pietu1998破解

__:_____

版画

%*;a|~

您可以在线尝试CJam。

05AB1E，13个字符，被quintopia 破解

我是个白痴。我忘了执行电源功能（-_-｡）。这是混淆的代码：

__D____<_____

我的原始代码是：

99DF2}r<F*}bR

说明：

99DF2}r<F*}bR

99            # Pushes 99
  D           # Duplicates the top item
   F }        # Creates a for loop: For N in range(0, 99)
    2         # Pushes 2 (this will be done 99 times)
      r       # Reverses the stack
       <      # Decrement on the last item
        F }   # Creates a for loop: For N in range(0, 98)
         *    # Multiplies the last two item of the stack, which now looks like
              # [2, 2, 2, 2, 2, 2, 2, 2, ...]. Basically we're calculating 2^99
           b  # Converts the last item to binary
            R # Reverses the last item of the stack

这将输出：

0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

是的，那是99个零，后跟一个1。

注意，我现在实现了幂函数。因此，这可以用六个字节完成：

Y99mbR

Python 3.4，127个字符

这是我的第一个铜柱。我认为/希望它不太难或不太明显。

混淆的代码：

_______inspect__________________getsource_____________________________________________print__________c*s_______________________

创建此输出（开头有5个空格；总长度为7740个字符）：

     """"""""""""(((((((((((((((((((((((((((((((((((((((((((((((((((((((()))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))**********************...................................................................................................................;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;==============================[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccdddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiijjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooopppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppprrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrsssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Pyth，17个字符，被Pietu1998破解

输出：

""''''''''''''''''''''0000000000111111111122222222223333333333XXXXXXXXXX[[[[[[]]]]]]

码：

___________#____1

我的解决方案：

Sjkm^dT%"'%#X"291

Pietu1998的解决方案：

S`u]G6*T`t"#X0231

Malbolge，254个字符，由Adnan破解

码：

_____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ __

输出：

`AB\cd`

我为每个第6个角色都留了一个空格而有点作弊。那好吧...

Adnan的破解代码：

('&%: ^"!65 4Xzyx w4-Qs rpo'K mlk"' ~Dfdc ba}v< ::8[Z XnVlq }0/mO ,*MbK JH^6# 4`C|0 ?U=Sw uPsaq L4on2 Mjjh, ged)c '<$$p ]!};Y WiVxS eRQ>= ).K]l Ij"Fh gfedc b,=;z L\q%H X3E2C /ng-k N*KJI 8%6#D ~2^Ai >g<<d tUr`6 oJ"!1 /|{CU f)d>b <A_^! \};Yj yUxTS dtsr` ML

原始代码：

bCBA@ ?>=<; :9876 54321 0/.-, +*)(' &%$#" !~}|{ zyxwv utsrq ponml kjihg fedcb a`_^] \[ZYX WVUTS RQPON MLKJI HGFED CBA@? >=<;: 9y165 43210 /.-,+ *)('& }C#"! ~wv<z \xwvu tsrqp onmlk jihgf e^$ba `_^W\ UyYXW VUTSR QPONM FKJIH GFEDC BA@9] =<;:9 27654 3210) M:

我再次使用此代码的文本转Malbage生成器作弊。

Mathematica 10.1，98个字符，安全

输出在这里。整个输出有838,316个字符和尾随换行符。

码：

Print[""<>Chara__________________________________________________________________________________]

是的，必须在10.1中运行。_（提示）原始程序：

Print[""<>Characters@ToCamelCase@ExampleData@{"Text","DonQuixoteIEnglish"}~SortBy~ToCharacterCode]

MATL，8个字符，安全

希望能在Octave中运行的稳定版本的编译器早日面世。同时，此GitHub commit可以正常运行，尤其是运行产生指定输出的原始代码。

此提交计数为2015年12月28日世界标准时间15:30发布的时间

输出：

$**0016688??@HHQ

代码（8个字符）：

_:______

解

6:9tX*Sc

它能做什么：

6:9          % numeric vector [6 7 8 9]
t            % duplicate
X*           % Kronecker tensor product
S            % sort in ascending order 
c            % convert to char. Implicit display

迷宫，5个字节，由Adnan 破解

_____

版画

000000

我的原始代码是：

<!!@ 

注意尾随空格。

我想我应该去找更多的字符并显示其中一个!，因为当添加更多时!，阿德南的解决方案只会线性增长，而我的解则是二次增长。

PHP，28个字符

<?=________________________;

输出字符串

abcdefghijklmnopqrstuvwxyz

ES6，17个字符，破获由CᴏɴᴏʀO'Bʀɪᴇɴ

__=____=_________

这是一个将返回字符串的函数

11233

JavaScript，83个字符，由MartinBüttner破解

输出量

Hi~~~

码

_=____r____=u____=__s__=_________=__________________u_______A__________e______"_~_"

输出在Chrome / Firefox等浏览器的JS控制台中。
原始代码（可能无效）：

q=w=e=r=t=y=u=i=o=p=s=d=f=g=h=j=k=l=z=x=c=v=b=n=m="Huh"[0]+"Alarms in here"[7]+"~~~"

Pyth，6个字符，被Pietu1998 破解

码：

S_____

输出：

.0113345678888999

你可以在这里尝试Pyth

JavaScript ES6，134个字符

boy，这很有趣。祝好运！

_=_=_____l____.______.________(___}__;___=______ru____"_____+_[________]______!____]__[_________e________.__U_________)__________\_____

输出：

((((((()))))))+...000000000000000000002233=>AAAAAAAAAAABBBBBCCCCCCCCCCCCCCCCCCDDDDDDDEEEEEEEEEEEEEEEEEEEFFFFFFFGHIIIIIIIIIIIIIIIIJJJJLLMMNNNNNNNNNNNNNNNNNNNNNNNNNOOOOOOOOOOOOOOOOOOOOOOPRRRRRRRSSSSTTTTTTTTTTTTTTTTTTTTTTTUUUUUUUUUVVVVVVXXY[[[[[[[[]]]]]]]]````ddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeffffffffffffffffffffffffffffffffffffffffffffffffffffffffiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiinnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu{{{{{{}}}}}}

CJam，12个字符，由jimmy23013 破解

输出量

#$%&'()*+,-./:;<=>?@[\]^_`

码

_:____#_____

严重的是，7个字节，由MartinBüttner破解

______S

输出：

   deeknnoow

即使结果并不完全符合您的期望，也应该不太困难。

Python 3，共58个字符，由Mitch Schwartz破解

码：

______________print(_______(_______________________)_____)

导入字符串;打印（''.join（sorted（string.printable））[5：]）

输出：

 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~

Befunge，11个字符，由MegaTom 破解（在funge-98中）

输出：

1223788

混淆代码：

________.__

因为规则禁止换行，所以只一行。使用http://www.quirkster.com/iano/js/befunge.html进行了测试。

编辑：由于我没有指定版本，因此这在法律上是可以破解的，但是请注意，那里仍然有Befunge-93解决方案。

Perl 5，30个字节，由Adnan 破解

码：

print_________________________

输出：

9:;@AFGHLMNRSTYZ_`a

破解：

阿德南发现了一个微不足道的 裂缝：很明显，我没想到这个难题，通过不够好。我改写了另一个。

Python 2，62个字符，由Adnan 破解

______________________________________________________________
         | |   ||                  |  |  ||

有点像一张桌子。

这将输出字符串abcde。

Adnan的代码：

print "ab_c_de___".replace("_","")#___________________________

我的代码：

f=lamda _,__,___:"abcde"+'f'*sum(_,__,___)*000;print f(1,2,3);