我最近以Bookworm Deluxe的形式沉迷于某种怀旧：

如果您以前从未看过它，那么这是一个文字游戏，目标是将相邻的图块连接起来以形成单词。为了确定字符串是否为有效单词，它会根据内部字典检查它，该内部字典以如下所示的压缩格式存储：

aa
2h
3ed
ing
s
2l
3iis
s
2rdvark
8s
4wolf
7ves

解开字典的规则很简单：

1. 阅读该行开头的数字，并从前一个单词的开头复制那么多字符。（如果没有数字，请复制与上次一样多的字符。）

2. 在该词后附加以下字母。

因此，我们的第一个单词是aa，其后是2h，表示“复制的前两个字母aa并附加h”，形成aah。然后3ed变为aahed，并且由于下一行没有数字，因此我们再次复制3个字符以形成aahing。该过程将在字典的其余部分继续进行。小样本输入得到的结果是：

aa
aah
aahed
aahing
aahs
aal
aaliis
aals
aardvark
aardvarks
aardwolf
aardwolves

您面临的挑战是如何以尽可能少的字节执行此拆包。

输入的每一行将包含零个或多个数字，0-9 后跟一个或多个小写字母a-z。您可以接受输入并将输出作为字符串列表，或作为单个字符串，且单词之间用0-9/ 以外的任何字符分隔a-z。

这是另一个小测试用例，示例中未涉及一些边缘情况：

abc cba 1de fg hi 0jkl mno abcdefghijk 10l
=> abc cba cde cfg chi jkl mno abcdefghijk abcdefghijl

您也可以在完整的字典上测试您的代码： input， output。

Vim，57个字节

:%s/\a/ &
:%norm +hkyiwjP
:g/\d/norm diw-@"yl+P
:%s/ //g

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JavaScript（ES6）， 66 62  61字节

a=>a.map(p=s=>a=a.slice([,x,y]=/(\d*)(.*)/.exec(s),p=x||p)+y)

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已评论

a =>                  // a[] = input, re-used to store the previous word
  a.map(p =           // initialize p to a non-numeric value
  s =>                // for each string s in a[]:
    a =               //   update a:
      a.slice(        //     extract the correct prefix from the previous word:
        [, x, y] =    //       load into x and y:
          /(\d*)(.*)/ //         the result of a regular expression which splits the new
          .exec(s),   //         entry into x = leading digits and y = trailing letters
                      //       this array is interpreted as 0 by slice()
        p = x || p    //       update p to x if x is not an empty string; otherwise leave
                      //       it unchanged; use this as the 2nd parameter of slice()
      )               //     end of slice()
      + y             //     append the new suffix
  )                   // end of map()

Perl 6的，50 48个字节

-2字节归功于nwellnhof

{my$l;.map:{$!=S[\d*]=substr $!,0,$l [R||]=~$/}}

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的端口阿尔诺的解决方案。那个R||招是过山车，从“我认为这是可能的”，到“不，这是不可能的”，再到“也许是可能的”，最后是“啊哈！”

说明：

{my$l;.map:{$!=S[\d*]=substr $!,0,$l [R||]=~$/}}
{                                              }  # Anonymous code block
 my$l;    # Declare the variable $l, which is used for the previous number
      .map:{                                  }  # Map the input list to
            $!=              # $! is used to save the previous word
               S[\d*]=       # Substitute the number for
                      substr $!,0    # A substring of the previous word
                                 ,              # With the length of 
                                           ~$0     # The num if it exists
                                  $l [R||]=        # Otherwise the previous num

该$l [R||]=~$/部分大致翻译为，$l= ~$/||+$l但...具有相同数量的字节:(。最初，它使用匿名变量保存字节，因此my$l消失了，但这不起作用，因为作用域现在是替换项，而不是map代码块。那好吧。无论如何，R是反向元运算符，因此它会反转的参数||，因此该$l变量最终会被分配新的数字（~$/存在，），否则将再次自身。

如果Perl 6没有为抛出某种冗余的编译器错误，则可能为47个字节=~。

红宝石，49 45 43字节

$0=$_=$0[/.{0#{p=$_[/\d+/]||p}}/]+$_[/\D+/]

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说明

$0=                                         #Previous word, assign the value of
   $_=                                      #Current word, assign the value of
      $0[/.{0#{              }}/]           #Starting substring of $0 of length p which is
               p=$_[/\d+/]||p               #defined as a number in the start of $_ if any 
                                 +$_[/\D+/] #Plus any remaining non-digits in $_

C，65 57字节

n;f(){char c[99];while(scanf("%d",&n),gets(c+n))puts(c);}

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说明：

n;                     /* n is implicitly int, and initialized to zero. */

f() {                  /* the unpacking function. */

    char c[99];        /* we need a buffer to read into, for the longest line in
                          the full dictionary we need 12 + 1 bytes. */

    while(             /* loop while there is input left. */

        scanf("%d",&n) /* Read into n, if the read fails because this line
                          doesn't have a number n's value does not change.
                          scanf's return value is ignored. */

        ,              /* chain expressions with the comma operator. The loop
                          condition is on the right side of the comma. */

        gets(c+n))     /* we read into c starting from cₙ. c₀, c₁.. up to cₙ is
                          the shared prefix of the word we are reading and the
                          previous word. When gets is successful it returns c+n
                          else it will return NULL. When the loop condition is
                          NULL the loop exits. */

        puts(c);}      /* print the unpacked word. */

脑干，201字节

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

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在输入的末尾需要尾随换行符。没有此要求的版本长6个字节：

笨蛋，207字节

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

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两种版本均假定所有数字均严格小于255。

说明

磁带的布局如下：

tempinputcopy 85 0 inputcopy number 1 a 1 a 1 r 1 d 0 w 0 o 0 l 0 f 0 ...

如果未输入数字，则“数字”单元等于0；如果输入数字n，则“ n + 1”单元等于n + 1。在标有“ 85”的单元格上输入。

,[                     take input and start main loop
 [                     start number input loop
  [-<+>>>+<<]          copy input to tempinputcopy and inputcopy
  >-[---<+>]           put the number 85 in the cell where input was taken
  <[[-<]>>]            test whether input is less than 85; ending position depends on result of comparison
                       (note that digits are 48 through 57 while letters are 97 through 122)
  <[-]>                clean up by zeroing out the cell that didn't already become zero
  >[                   if input was a digit:
   <<,>>               get next input character
   >[-[-<++++++++++>]] multiply current value by 10 and add to current input
   ++++                set number cell to 4 (as part of subtracting 47)
   <[->+<]             add input plus 10*number back to number cell
   -[----->-<]         subtract 51
  <]                   move to cell we would be at if input were a letter
 <]                    move to input cell; this is occupied iff input was a digit

                       part 2: update/output word

 >>>                   move to number cell
 [                     if occupied (number was input):
  [>>]+[-<<]>>         remove existing marker 1s and decrement number cell to true value
  [[>>]+[<<]>>-]       create the correct amount of marker 1s
 ]
 +[>>]<[-]             zero out cell containing next letter from previous word
 <[<<]>                return to inputcopy
 [->[>>]<+<[<<]>]      move input copy to next letter cell
 >[>.>]                output word so far
 +[                    do until newline is read:
  >[-]<                zero out letter cell
  ,.                   input and output next letter or newline
  [->+>+<<]            copy to letter cell and following cell
  >>----------         subtract 10 to compare to newline
 ]
 <[<<]>-               zero out number cell (which was 1 to make copy loop shorter)
 <<<,                  return to input cell and take input
]                      repeat until end of input

Python 3.6以上版本， 172 195 156 123 122 121 104字节

import re
def f(l,n=0,w=""):
 for s in l:t=re.match("\d*",s)[0];n=int(t or n);w=w[:n]+s[len(t):];yield w

• -12个字节归功于Bubbler
• -5个字节感谢Dennis

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说明

我屈服了，并使用了正则表达式。这样可以节省至少17个字节。：

t=re.match("\d*",s)[0]

如果字符串根本不是以数字开头，则该字符串的长度为0。这意味着：

n=int(t or n)

将为nif t为空，int(t)否则为。

w=w[:n]+s[len(t):]

删除从正则表达式中找到s的数字（如果没有找到数字，它将删除0字符，不被s截断），并用n当前单词片段替换前一个单词的除第一个字符以外的所有字符；和：

yield w

输出当前单词。

Haskell，82 81字节

tail.map concat.scanl p["",""]
p[n,l]a|[(i,r)]<-reads a=[take i$n++l,r]|1<2=[n,a]

获取并返回字符串列表。

在线尝试！

        scanl p["",""]        -- fold function 'p' into the input list starting with
                              -- a list of two empty strings and collect the
                              -- intermediate results in a list
  p [n,l] a                   -- 1st string of the list 'n' is the part taken form the last word
                              -- 2nd string of the list 'l' is the part from the current line
                              -- 'a' is the code from the next line
     |[(i,r)]<-reads a        -- if 'a' can be parsed as an integer 'i' and a string 'r'
       =[take i$n++l,r]       -- go on with the first 'i' chars from the last line (-> 'n' and 'l' concatenated) and the new ending 'r'
     |1<2                     -- if parsing is not possible
       =[n,a]                 -- go on with the previous beginning of the word 'n' and the new end 'a'
                              -- e.g. [         "aa",     "2h",      "3ed",       "ing"       ] 
                              -- ->   [["",""],["","aa"],["aa","h"],["aah","ed"],["aah","ing"]]
  map concat                  -- concatenate each sublist
tail                          -- drop first element. 'scanl' saves the initial value in the list of intermediate results. 

编辑：-1字节感谢@Nitrodon。

Japt，19 18 17字节

最初受Arnauld的JS解决方案启发。

;£=¯V=XkB ªV +XoB

试试吧

                      :Implicit input of string array U
 £                    :Map each X
   ¯                  :  Slice U to index
      Xk              :    Remove from X
;       B             :     The lowercase alphabet (leaving only the digits or an empty string, which is falsey)
          ªV          :    Logical OR with V (initially 0)
    V=                :    Assign the result to V for the next iteration
             +        :  Append
              Xo      :  Remove everything from X, except
;               B     :   The lowercase alphabet
  =                   :  Reassign the resulting string to U for the next iteration

果冻，16字节

⁹fØDVo©®⁸ḣ;ḟØDµ\

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这个怎么运作

⁹fØDVo©®⁸ḣ;ḟØDµ\  Main link. Argument: A (array of strings)

              µ\  Cumulatively reduce A by the link to the left.
⁹                     Yield the right argument.
  ØD                  Yield "0123456789".
 f                    Filter; keep only digits.
    V                 Eval the result. An empty string yields 0.
     o©               Perform logical OR and copy the result to the register.
       ®              Yield the value in the register (initially 0).
        ⁸ḣ            Head; keep that many character of the left argument.
          ;           Concatenate the result and the right argument.
            ØD        Yield "0123456789".
           ḟ          Filterfalse; keep only non-digits.

Python 2，118字节

import re
n=0
l=input()
o=l.pop(0)
print o
for i in l:(N,x),=re.findall('(\d*)(.+)',i);n=int(N or n);o=o[:n]+x;print o

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视网膜0.8.2，69字节

+`((\d+).*¶)(\D)
$1$2$3
\d+
$*
+m`^((.)*(.).*¶(?<-2>.)*)(?(2)$)1
$1$3

在线尝试！链接包括较难的测试用例。说明：

+`((\d+).*¶)(\D)
$1$2$3

对于所有以字母开头的行，请复制前一行的数字，并循环播放，直到所有行均以数字开头。

\d+
$*

将数字转换为一元。

+m`^((.)*(.).*¶(?<-2>.)*)(?(2)$)1
$1$3

使用平衡组将所有1s 替换为前一行中的相应字母。（事实证明，这比替换1s的所有游标要稍微打高尔夫球。）

红色，143字节

func[b][a: charset[#"a"-#"z"]u: b/1 n: 0 foreach c b[parse c[copy m to a
p: copy s to end(if p<> c[n: do m]print u: rejoin[copy/part u n s])]]]

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Java（JDK），150字节

a->{String p="",s[];for(int n=0,i=0;i<a.length;a[i]=p=p.substring(0,n=s.length<1?n:new Short(s[0]))+a[i++].replaceAll("\\d",""))s=a[i].split("\\D+");}

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Groovy，74个字节

{w="";d=0;it.replaceAll(/(\d*)(.+)/){d=(it[1]?:d)as int;w=w[0..<d]+it[2]}}

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说明：

{                                                                        }  Closure, sole argument = it
 w="";d=0;                                                                  Initialize variables
          it.replaceAll(/(\d*)(.+)/){                                   }   Replace every line (since this matches every line) and implicitly return. Loop variable is again it
                                     d=(it[1]?:d)as int;                    If a number is matched, set d to the number as an integer, else keep the value
                                                        w=w[0..<d]+it[2]    Set w to the first d characters of w, plus the matched string

果冻，27字节

f€ȯ@\V,ɗḟ€ɗØDZẎḊṖḣ2/Ż;"f€Øa

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Perl 5中 -p，45 41个字节

s:\d*:substr($p,0,$l=$&+$l*/^\D/):e;$p=$_

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说明：

s:\d*:substr($p,0,$l=$&+$l*/^\D/):e;$p=$_ Full program, implicit input
s:   :                           :e;      Replace
  \d*                                       Any number of digits
      substr($p,0,              )           By a prefix of $p (previous result or "")
                  $l=  +                      With a length (assigned to $l) of the sum
                     $&                         of the matched digits
                          *                     and the product
                        $l                        of $l (previous length or 0)
                           /^\D/                  and whether there is no number in the beginning (1 or 0)
                                                (product is $l if no number)
                                    $p=$_ Assign output to $p
                                          Implicit output

Groovy，103 99字节

{w=it[0];d=0;it.collect{m=it=~/(\d+)(.+)/;i=m.find()?{d=m[0][1] as int;m[0][2]}():it;w=w[0..<d]+i}}

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05AB1E，20 19 17 字节

õUvyþDõÊi£U}Xyá«=

在线尝试或验证所有测试用例。

说明：

õ                  # Push an empty string ""
 U                 # Pop and store it in variable `X`
v                  # Loop `y` over the (implicit) input-list
 yþ                #  Push `y`, and leave only the digits (let's call it `n`)
   DõÊi  }         #  If it's NOT equal to an empty string "":
       £           #   Pop and push the first `n` characters of the string
        U          #   Pop and store it in variable `X`
          X        #  Push variable `X`
           yá      #  Push `y`, and leave only the letters
             «     #  Merge them together
              =    #  Print it (without popping)

普通Lisp，181字节

(do(w(p 0))((not(setf g(read-line t()))))(multiple-value-bind(a b)(parse-integer g :junk-allowed t)(setf p(or a p)w(concatenate'string(subseq w 0 p)(subseq g b)))(format t"~a~%"w)))

在线尝试！

取消高尔夫：

(do (w (p 0))   ; w previous word, p previous integer prefix (initialized to 0)
    ((not (setf g (read-line t ()))))   ; read a line into new variable g
                                        ; and if null terminate: 
  (multiple-value-bind (a b)            ; let a, b the current integer prefix
      (parse-integer g :junk-allowed t) ; and the position after the prefix
    (setf p (or a p)                    ; set p to a (if nil (no numeric prefix) to 0)
          w (concatenate 'string        ; set w to the concatenation of prefix
             (subseq w 0 p)             ; characters from the previous word 
             (subseq g b)))             ; and the rest of the current line
    (format t"~a~%"w)))                 ; print the current word

像往常一样，Common Lisp的长标识符使它特别不适用于PPCG。

Python 2中，101个 100 99字节

import re
s=n='0'
for t in input():(m,w),=re.findall('(\d*)(.+)',t);n=m or n;s=s[:int(n)]+w;print s

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C＃（Visual C＃交互式编译器），134字节

a=>{int l=0,m,n;var p="";return a.Select(s=>{for(m=n=0;s[m]<58;n=n*10+s[m++]-48);return p=p.Substring(0,l=m>0?n:l)+s.Substring(m);});}

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-9个字节感谢@ASCIIOnly！

少打高尔夫球...

// a is an input list of strings
a=>{
  // l: last prefix length
  // m: current number of digits
  // n: current prefix length
  int l=0,m,n;
  // previous word
  var p="";
  // run a LINQ select against the input
  // s is the current word
  return a.Select(s=>{
    // nibble digits from start of the
    // current word to build up the
    // current prefix length
    for(m=n=0;
      s[m]<58;
      n=n*10+s[m++]-48);
    // append the prefix from the
    // previous word to the current
    // word and capture values
    // for the next iteration
    return
      p=p.Substring(0,l=m>0?n:l)+
      s.Substring(m);
  });
}

阶，226个 129 102字节

感谢@ ASCII-only在这里的工作（以及Groovy的回答）。

s=>{var(w,d,r)=("",0,"(\\d*)(.+)".r)
s map(_ match{case r(a,b)=>{if(a>"")d=a.toInt
w=w.take(d)+b}
w})}

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