给定以ASCII艺术形式输入的“旅馆”并标有单个房间，请根据某些规则输出房间号。

这是ASCII酒店的示例：

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

以下是有关ASCII酒店的更多信息：

• 每个“建筑物”都由一个矩形的#字符#表示，其中每个字符 代表一个“房间”。

• 以上酒店由三栋建筑组成。每个建筑物都由两列空格隔开，最低的“楼层”将始终位于最后一行。

• 每个建筑物在每个楼层上始终具有1-9个“地板”（行）和1-9个“房间”的任何位置。也将永远有1-9座建筑物。

• 客房编号如下：[building #][floor #][room on floor #]。例如，让我们在上图中标记一些房间：

                 ##
                 ##
  #####          ##
  #####  ####$#  ##
  ##%##  ######  ##
  #####  ######  #@
  

  标有的%房间是123房间（1、2楼，3楼的房间）。同样，标有的房间$是房间235，@是的房间312。

• 建筑物，楼层和“第n个房间在楼层”始终为1索引。

输入将包括一个ASCII酒店，其中一个房间由星号（*）代替。这是您必须为其输出房间号的房间。输入必须作为单个字符串，但是您可以使用逗号作为换行符，而不是换行符（以防您选择的语言不能采用多行输入，或者采用单行输入时较短）。您可以选择要求以逗号/换行符结尾。您也可以用尾随空格填充线条，以使输入成为完整的矩形。

由于这是code-golf，因此以字节为单位的最短代码为准。

测试用例（包含在单个代码块中，以节省垂直空间）：

Input:
*

Output: 111

Input:
#  #  *  #  #

Output: 311

Input:
#####
#####
####*
#####
#####

Output: 135

Input:
         #####
         #####           ######
         #####           ######
#  #  #  #####  #  #  #  ######  *

Output: 911

Input:
#
#  #
#  #  ##
#  #  ##  ###
#  #  ##  ###  #####
#  #  ##  ###  ##*##  ########
#  #  ##  ###  #####  ########

Output: 523

Input:
           #
           *
           #
           #
           #
           #
           #
           #
#########  #  #

Output: 281

Input:
                        ########*
                        #########
                        #########
                        #########
                        #########
                        #########
                        #########
                        #########
#  #  #  #  #  #  #  #  #########

Output: 999

CJam，34 31字节

qN/W%zSf-La%{_{s'*&}#_)@@=}3*;\

这要求将输入填充为带有空格的矩形。

在线尝试！或者，运行所有测试用例。

说明

qN/      e# Read input and split into lines.
W%z      e# Rotate 90° counter-clockwise.
Sf-      e# Remove all spaces from the rows.
La%      e# Split into buildings. We've now got a 3D array of rooms, where the first
         e# dimension is the building, the second the room number and the third is the
         e# the floor number.
{        e# Run this block three times. At each stage it will find the index of the "*"
         e# along the current dimension and leave the element at that index on the stack
         e# for the next round...
  _      e#   Duplicate the current array.
  {      e#   Find the index of the first element where this block yields something
         e#   truthy...
    s    e#     Flatten into a single string.
    '*&  e#     Set intersection with "*".
  }#
  _)     e#   Duplicate the index and increment it, because the results should be 1-based.
  @@=    e#   Pull up the array and the other copy of the index and select the
         e#   corresponding element.
}3*
;\       e# We've now got the building, room and floor index on the stack, as well as the
         e# "*" character itself. We discard the character and swap the room and the floor
         e# floor number. When the three indices are printed back-to-back at the end of
         e# the program, that will yield the desired result.

Pyth，34个字节

LxKh/#\*b\*jkhM[//<hJ_.zyJd2xJKycK

示范

这使用了我以前从未使用过的打高尔夫球技巧：分配给K函数（y）中的变量（），以保存该函数的部分结果。

说明：

LxKh/#\*b\*jkhM[//<hJ_.zyJd2xJKycK
L                                     Define y(b):      (b is a list of strigs)
    /#\*b                             Filter b for strings containing '*'
   h                                  Take the first such string
  K                                   Store it in K
 x       \*                           And return the index of '*' in that string.
                      .z              Take the input as a list of strings
                     _                Reverse it (bottom to top)
                    J                 Store in J
                   h                  Take the bottommost row
                        yJ            Find y(J). This is the index in whichever
                                      row of J has the * of the *. Also store
                                      that row in K.
                  <                   Slice J up to that index.
                 /        d           Count the number of spaces
                /          2          Divide by 2. This is the building number.
                            xJK       Take the index in J of K. This is the floor.
                                cK    Chop K on whitespace.
                               y      Find the index in whatever element of K has
                                      the * of the *. This is the room number.
                                      This also overwrites K, but we don't care.
               [                      Gather the above into a list.
             hM                       Convert 0-indexing to 1-indexing.
           jk                         Concatenate. Print implicitly.

的JavaScript（ES6），142个 136字节

h=>h.split`
`.reverse(r=0).map((t,i,l)=>r?0:(f=i+1,b=1,l[o=0].slice(0,r=t.indexOf`*`+1).replace(/  /g,(_,s)=>o=++b&&s+2),r-=o))&&[b]+f+r

@nderscore节省了6个字节！

说明

h=>
  h.split`
`                            // get each line of the input string
  .reverse(                  // reverse the lines to make getting the ground floor easy
    r=0)                     // initialise r to 0
  .map((t,i,l)=>             // for each line of the reversed input string
    r?0:(                    // if the marked room has not been found yet:
      f=i+1,                 // f = floor number
      b=1,                   // b = building number, default to 1
      l[o=0].slice(0,        // get the substring of 0 to the marked room, default o to 0
        r=t.indexOf`*`+1)    // r = absolute index of room + 1 (or 0 if not found)
      .replace(/  /g,(_,s)=> // count the spaces between buildings
        o=++b&&s+2),         // increment b, o = index of marked room's building
      r-=o                   // make r relative to the room's building
    )
  )
  &&[b]+f+r                  // output the result ([b] casts b to a string)

测试

var solution = h=>h.split`
`.reverse(r=0).map((t,i,l)=>r?0:(f=i+1,b=1,l[o=0].slice(0,r=t.indexOf`*`+1).replace(/  /g,(_,s)=>o=++b&&s+2),r-=o))&&[b]+f+r

<textarea id="input" rows="8" cols="60">#
#  #
#  #  ##
#  #  ##  ###
#  #  ##  ###  #####
#  #  ##  ###  ##*##  ########
#  #  ##  ###  #####  ########</textarea><br />
<button onclick="result.textContent=solution(input.value)">Go</button>
<pre id="result"></pre>

awk，70岁

!i{i=index($0,"*")}i{$0=substr($0,0,i);f++}END{print NF f length($NF)}

例：

Input:
#
#  #
#  #  ##
#  #  ##  ###
#  #  ##  ###  #####
#  #  ##  ###  ##*##  ########
#  #  ##  ###  #####  ########

While no * was found, do nothing.
#
#  #
#  #  ##
#  #  ##  ###
#  #  ##  ###  #####
A * is found in column 14. From now on, truncate and increment the floor counter.
#  #  ##  ###  ##*    f=1
#  #  ##  ###  ###    f=2
Awk automatically splits $0 into space separated fields, counted by builtin NR.
In the end, NR and f hold hotel and floor number.
The room number is the length of the last hotel.

C，131个 130 119 113字节

b,f,i,j=111;main(c){for(;c=~getchar();)c&32?f+=10,b=i=0:++i<j?c%3?f=j,j=i:c&2?b+=50-b%50:++b:0;printf("%d",b+f);}

在stdin上输入；输入内容不得包含换行符。假设2的补码。

取消高尔夫：

// Declare variables (default type is int) and initialize, by default to 0:
b,    // Building number (multiplied by 100, 0-based) + room number (0-based)
f,    // Floor (111-based, multiplied by 10)
i,    // Current column of input character within line (1-based)
j = 111; // Column of asterisk character once found (1-based), 111 before then
main (c)    // Declare main function and variable c to hold input character
{
  for (;    // Loop on input
    c = ~getchar();  // Read a character into c, bitwise inverted to break
                     // EOF (numeric value -1). This means that following 
                     // operations (on the ASCII value of the input) are 
                     // also inverted.
    )
    c & 32 ?              // Newline?
      f += 10, b = i = 0 :  // Increment floor; reset building, room, column
      ++i < j ?             // Increment column; before asterisk, or asterisk not yet found?
        c % 3 ?               // Asterisk character?
          f = j, j = i :        // Reset floor and record column
          c & 2 ?               // Space character?
            b += 50 - b % 50 :    // Increment building and reset room
            ++b :                 // Otherwise, # character; increment room
        0;                    // After asterisk; do nothing
  printf("%d", b + f);  // Write out results
}

Stackgoat，73个字节[非竞争]

Stackgoat是一种基于堆栈的语言，与山羊无关。

y'#ZGDYZG'*iVXsV@"\\*"ZGN2/1+y'#ZG' ZG'q:Nq'*i-yXsq'*i@"[#*]+"M0M1-@'*i1+

这是一种相当新的语言，所以请让我知道他们是否有任何问题。弄清楚这个问题我很头痛，所以这和我打高尔夫球一样多。

说明

该程序由房间号的每3位数字组成3个部分

y'#ZG    // Remove all # from input
D        // Duplicate
YZG      // Remove all spaces
'*i      // Index of *
V        // Reverse stack
Xs       // Split on spaces
V@       // Unreverse, item at *'s index
"\\*"ZG  // Remove all *s
N        // Get length
2/1+     // Divide by 2, add 1

y'#ZG    // Remove all #
' ZG     // Remove all spaces
'q:      // Store in q
N        // Get length
q'*i     // *'s index in q
-        // Subtracted from length

yXs      // Split on newlines
q'*i     // Get index of * in q
@        // Get indexed-th line
"[#*]+"M // Match all buildings
0M       // Get *'s building no.
1-       // Subtract one
@        // nth building at right line
'*i      // *'s index
1+       // Added to one

红宝石，103

->n{r=x=b=0
n.lines{|s|(t=s=~/\*/)&&(x=t;r=($`.reverse+' ')=~/ /)
r+=10;b=s[0..x].count" "}
b*50+r+101}

取消测试程序

g=->n{
  r=x=b=0
  n.lines{|s|                  #for each line in n
    (t=s=~/\*/)&&              #if the line contains an asterisk
      (x=t                     #record its position in x. $` is a special variable containing the part of the string to the left of the last match made.
      r=($`.reverse+' ')=~/ /) #reverse $` and search for the index of the first space to find room number (before the search a space is appended in case it is 1st building.)
    r+=10                      #increment r by 10 for the floor number (obviously this will have been reset to the row ith the asterisk by the previous line)
    b=s[0..x].count" "}        #count the number of spaces left of x in the current row to find building number (loop will exit with calc from bottom row, which is the correct one.) 
b*50+r+101}                    #multiply number of spaces by 50 to get 1st digit, add r for 2nd and 3rd digit. Then add 101 to correct 1st and 3rd digits from 0-indexed to 1-indexed.

puts g["
*"]



puts g["
#  #  *  #  #"]


puts g["
#####
#####
####*
#####
#####"]

puts g["
         #####
         #####           ######
         #####           ######
#  #  #  #####  #  #  #  ######  *"]



puts g["
#
#  #
#  #  ##
#  #  ##  ###
#  #  ##  ###  #####
#  #  ##  ###  ##*##  ########
#  #  ##  ###  #####  ########"]




puts g["
           #
           *
           #
           #
           #
           #
           #
           #
#########  #  #"]



puts g["
                        ########*
                        #########
                        #########
                        #########
                        #########
                        #########
                        #########
                        #########
#  #  #  #  #  #  #  #  #########"]

JavaScript（ES6），121

x=>x.split`
`.reverse().map((r,f,h,a=r.indexOf`*`)=>x=~a?(h=h[0].slice(0,a).split`  `).length+[f+1]+-~h.pop().length:x)|x

少打高尔夫球和解释

H=x=>x.split`\n` // split in lines
  .reverse() // reverse, so we can scan bottom up
  .map( (r,f,h) => // exectute for each line
         // r is the current row
         // f in the row index, so that f+1 is the floor number
         // h is the reversed array, h[0] is the bottom floor
       ~(a=r.indexOf`*`) // a is the position of '*' in the line, if found - else 0
       && ( // if a >= 0
         h = h[0]        // bottom floor line 
             .slice(0,a) // ... truncated at position of '*'
             .split`  `, // ... and splitted at '  ', as an array
         x = h.length  // the array len is the building number
             + [f+1]   // floor number, using [] to force string concatenation
             + -~ h.pop().length // the length of the last array element is the number 
                                 // of chars in the block before '*'
                                 // increment by 1 to get the room number
      )
  )
  && x // return the found value

测试

H=x=>x.split`
`.reverse().map((r,f,h,a=r.indexOf`*`)=>x=~a?(h=h[0].slice(0,a).split`  `).length+[f+1]+-~h.pop().length:x)|x

// test
console.log=x=>O.textContent+=x+'\n';

;[['*',111],['#  #  *  #  #',311],
[`#####
#####
####*
#####
#####`,135],
[`         #####
         #####           ######
         #####           ######
#  #  #  #####  #  #  #  ######  *`,911],
[`#
#  #
#  #  ##
#  #  ##  ###
#  #  ##  ###  #####
#  #  ##  ###  ##*##  ########
#  #  ##  ###  #####  ########`,523],
[`           #
           *
           #
           #
           #
           #
           #
           #
#########  #  #`,281],
[`                        ########*
                        #########
                        #########
                        #########
                        #########
                        #########
                        #########
                        #########
#  #  #  #  #  #  #  #  #########`,999]]  
.forEach(t=>{
  var i=t[0],k=t[1],r=H(i)
  console.log(i+'\n' + (k!=r?'Error '+r+' expected '+k:'Ok '+r)+'\n')
})

<pre id=O></pre>

Python 2.7、153168 个字符

我喜欢这个挑战！如果可以使用Python列表作为输入（请参见testsuite以获取示例），则此解决方案有效。

编辑2016年1月5日：添加了一行（10个字符）以将字符串拆分为多行。

简短说明：

• t 是房间所在的行（从顶行开始算作数组索引= 0）；
• i 是房间在其行中的索引；
• 建筑物的计算是根据直到底行底部的两倍空格i;
• 底数是行数减去t;
• room是反向字符串中第一个双空格的索引，该字符串从i开始到有该房间的楼层开始，并附加一个空格以覆盖该房间位于第一座建筑物中的情况。

码：

def r(l):
 l=h.split(",")
 a,w,s="*","  ",str
 t=l.index(filter(lambda c:a in c,l)[0])
 i=l[t].find(a)
 return s(l[-1][:i].count(w)+1)+s(len(l)-t)+s((l[t][i::-1]+w).find(w))

测试套件：

cases = [
         (["*"], 111),
         (["#  #  *  #  #"], 311),
         (["#####","#####","####*","#####","#####"], 135),
         (["         #####","         #####           ######","         #####           ######","#  #  #  #####  #  #  #  ######  *"], 911),
         (["#","#  #","#  #  ##","#  #  ##  ###","#  #  ##  ###  #####","#  #  ##  ###  ##*##  ########","#  #  ##  ###  #####  ########"], 523),
         (["           #","           *","           #","           #","           #","           #","           #","           #","#########  #  #"], 281),
         (["                        ########*","                        #########","                        #########","                        #########","                        #########","                        #########","                        #########","                        #########","#  #  #  #  #  #  #  #  #########"], 999)
        ]

for idx,(hotel,roomnr) in enumerate(cases):
    output=r(hotel)
    if str(output)==str(roomnr):
        result="SUCCESS"
    else:
        result="FAILURE!!!"
    print "Case {} gives output: {}. Correct output is: {}. Result: {}".format(idx,output,roomnr,result)

蜗牛，32字节

A
\*{l\#,|r,9d.,{|=~r,9{|l.,\ \#

C，142个 138 137字节

#include <stdio.h>
f,b,x,p=110;main(c){while(~(c=getchar()))c<11?f+=c,b=x=0:x++<p?++b,c&2?c&8?f=p,p=x:0:(b+=50-b%50):0;printf("%d",b+f);}

（123 119 118字节+ 19 #include线）

我从ecatmur窃取了价值合并的想法，但是我以完全不同的方式合并了它们（最后节省了8个字节）。这也做出了相同的假设EOF == -1。

输入来自stdin，在最后一行的最后一个建筑物之后不能有空格或换行符，因此输入示例如下：

printf "##\n##       #\n##  ##*  #\n##  ###  #" | ./hotel
# or for better visualisation:
printf "##\n##       #\n##  ##*  #\n##  ###  #" | tee /dev/fd/2 | ./hotel;echo ""

分解：

// Globals initialise to 0
f,     // floor number * 10 + shift
b,     // building number * 100 + room number
x,     // current column
p=110; // will store column of * (must start >= 11*9-2, and 110 will be used later)
main(c){
    while(~(c=getchar()))              // For each character until EOF
        c<11                           //  Is \n? (10)
            ?f+=c,                     //   Add 10 to floor number
             b=x=0                     //   Reset building, room, column
            :x++<p                     //  Else, is column <= *?
                  ?++b,                //   Add to room number
                   c&2                 //   Is # or *?
                      ?c&8             //   If *:
                          ?f=p,p=x:0   //    Set floor to 110, set p to column
                      :(b+=50-b%50)    //   If ' ': go to next building
                  :0;
    printf("%d",b+f);                  // Result is building+room+floor+shift
}

Haskell中，128个是125字节

l=length
f h|b<-snd$break(elem '*')$lines h,q<-fst(span(<'*')$b!!0)++"*"=l(last$words$q)+10*l b+l(words$take(l q)$last b)*100

用法示例：f "# # * # #"-> 311。

这个怎么运作：

b<-snd$break(elem '*')$lines h      -- split the input into a list of lines
                                    -- and assign b to the lines starting with
                                    -- the one that includes * up to the end,
                                    -- i.e. drop leading lines without the *
q<-fst(span(<'*')$b!!0)++"*"        -- assign q to the line with the *, but strip
                                    -- off all chars after the *

l(last$words$q)                     -- the room on floor number is the length of
                                    -- the last word of q
10*l b                              -- the floor number is 10 times the length of b
l(words$take(l q)$last b)*100       -- the hotel number is 100 times the number of
                                    -- words in the last line cut down to the
                                    -- length of q

                                    -- add for final room number

Lua，165字节

l={}i=1while(l[i-1]~="")do l[i]=io.read()o=l[i]:find"%*"x=o or x y=o and i or y i=i+1 end print(#l[i-2]:sub(1,x):gsub("%S+%s*","#")*100+(i-y-1)*10+#l[y]:match"#-%*")

不打高尔夫球

l={}
i=1
while(l[i-1]~="")do
    l[i]=io.read()
    o=l[i]:find"%*" --find "*", and record:
    x=o or x        --position and
    y=o and i or y  --current floor
    i=i+1
end
print(#l[i-2]:sub(1,x):gsub("%S+%s*","#")*100 --[[Take last string of list, and then
                                                  take the substring up until the 
                                                  asterisk. Substitute any substrings
                                                  that include nonspace characters 
                                                  (%S+) followed by a minimum of 0 space 
                                                  characters (%s*) with one character
                                                  (in this code snippet I chose # for no 
                                                  particular reason.) Then take the length 
                                                  of this string, with the # operator. 
                                                  The %S+%s* regex and gsub do the bulk 
                                                  of the magic.
                                                ]]
      +(i-y-1)*10                             --[[Total number of lines minus '*' floor 
                                                  minus one.
                                                ]]
      +#l[y]:match"#-%*")                     --[[Find the substring on the asterisk floor
                                                  with '#' symbols preceding an asterisk.
                                                ]]

CoffeeScript，110字节和JavaScript，121字节

(s)->s.split('\n').reverse().map((f,g)->f.split('  ').map((h,i)->r=h.indexOf('*');s=''+i+g+r if r>-1));111+1*s

可读的

(s)->
    s.split '\n' 
        .reverse()
        .map (f,fi)->
            f.split('  ')
                .map (h,hi)->
                    ri = h.indexOf('*')
                    s = ''+hi+fi+ri if ri>-1
    111+1*s

Javascript中的基本相同：

(s)=>{s.split('\n').reverse().map((f,g)=>f.split('  ').map((h,i)=>{r=h.indexOf('*');r>-1?s=''+i+g+r:''}));return 111+1*s}

Java，231字节

String a(String a){String[]b=a.split("\n");int i=0,c=b.length,m;String x,k=" ";for(;i<c;i++){x=b[c-1].substring(0,b[i].indexOf("*")+1);m=x.length();a=m>0?""+(m-x.replace(k+k,k).length()+1)+(c-i)+(m-x.lastIndexOf(k)-1):a;}return a;}

去高尔夫

 String a(String a) {
     String[] b = a.split("\n");                                // Split the input into floor lines
     int i = 0, c = b.length, m;                                // i=floor line counter c= number of floor lines
     String x, k = " ";
     for (; i < c; i++) {                                       // Loop through floor lines
        x = b[c - 1].substring(0, b[i].indexOf("*") + 1);       // x = part of bottom floor line up to '*' position in current line (Empty string when no '*')
        m = x.length();                                         // m = length of floor line part
        a = m > 0 ? "" + (m - x.replace(k + k, k).length() + 1) // if m>0 ('*' is on this line) set a=building no+floor no+room no.   building no calculated by replacing double space in x with single space and compare length to x (+1) 
              + (c - i)                                         // floor number is total floor lines (c) - floor line loop counter (i)
              + (m - x.lastIndexOf(k) - 1) : a;                 // room number is m ('*' position in x) - position of last space in x (-1)
     }
     return a;                                                  // return the result at the end.
  }

Powershell，154个字节

param($s)filter s{$s|sls $_ -a|% M*|% Le*}(($s-split'
')[-1]|% s*g 0('(?m)^.*\*'|s)|sls '^|  '-a|% M*).Count,('(?ms)(?<=\*.*)$'|s).Count,('#*\*'|s)-join''

少打高尔夫的测试脚本：

$f = {

param($s)

filter s{
    $s|sls $_ -AllMatches|% Matches|% Length
}                                   # select an array of lengths of all matches of the string $s by pattern $_

$hpos='(?m)^.*\*'|s                 # horizontal position of the room in the source string
$basement=($s-split"`n")[-1]        # basement floor string

$building=($basement|% substring 0 $hpos|sls '^|  ' -AllMatches|% Matches).Count
                                    # truncate the basement to the position of the room
                                    # and count all double spaces or a 'start of string'
$floor=('(?ms)(?<=\*.*)$'|s).Count  # count all 'end of line' after the room
$room='#*\*'|s                      # count all #, preceding the room, and room itself

$building,$floor,$room-join''


}

@(

,(@"
*
"@, 111)

,(@"
#  #  *  #  #
"@,311)

,(@"
#####
#####
####*
#####
#####
"@, 135)

,(@"
        #####
        #####           ######
        #####           ######
#  #  #  #####  #  #  #  ######  *
"@, 911)

,(@"
#
#  #
#  #  ##
#  #  ##  ###
#  #  ##  ###  #####
#  #  ##  ###  ##*##  ########
#  #  ##  ###  #####  ########
"@, 523)

,(@"
        #
        *
        #
        #
        #
        #
        #
        #
#########  #  #
"@, 281)

,(@"
                        ########*
                        #########
                        #########
                        #########
                        #########
                        #########
                        #########
                        #########
#  #  #  #  #  #  #  #  #########
"@, 999)

) | % {
    $n,$expected = $_
    $result = &$f $n
    "$($result-eq$expected): $result"
}

输出：

True: 111
True: 311
True: 135
True: 911
True: 523
True: 281
True: 999

05AB1E，34 个字节

|€SζJðмõ¡εεR'*k>]DOZ©k>;ò®«sĀ€ƶ˜à«

带有字符串列表的邮编目前存在错误。€SζJ可能只是ζ05AB1E的旧Python旧版本，但由于某种原因，在Elixir重写版本中不再起作用。

在线尝试或验证所有测试用例。

说明：

|                    # Take the input split by newlines
                     #  i.e. "   ###\n#  ###\n#  ###  ##\n#  ##*  ##"
                     #   → ["   ###","#  ###","#  ###  ##","#  ##*  ##"]
 €S                  # Convert each to a list of characters
   ζ                 # Zip, swapping rows and column
    J                # Join them together to a string again
                     #  → [" ###","    ","    ","####","####","###*","    ","    ","  ##","  ##"]
     ðм              # Remove all spaces
                     #  → ["###","","","####","####","###*","","","##","##"]
       õ¡            # Split on empty strings
                     #  → [["###"],[],["####","####","###*"],[],["##","##"]]
ε                    # Map each building to:
 ε                   #  Map each column of the building to:
  R                  #   Reverse the column
   '*k              '#   Get the 0-indexed index of "*"
      >              #   Increase it by 1 to make it 1-indexed
]                    # Close both maps
                     #  → [[0],[],[0,0,1],[],[0,0]]
 D                   # Duplicate the resulting list
  O                  # Sum each building
                     #  → [0,0,1,0,0]
   Z                 # Get the max (without popping)
                     #  → 1
    ©                # Store this max in the register (without popping)
     k>              # Get the index (+ 1) of this max in the sum-list
                     #  → 3
       ;             # Halve it
                     #  → 1.5
        ò            # Round it up to the nearest integer (bankers rounding)
                     #  → 2
         ®           # Retrieve the value from the register again
          «          # Merge the two digits together
                     #  → 21
           s         # Swap so the duplicate list is at the top again
            Ā        # Trutify (0 remains 0, every other integer becomes 1)
                     #  → [[0],[],[0,0,1],[],[0,0]]
             €       # For each building:
              ƶ      #  Multiply the integer with the 1-indexed index
                     #  → [[0],[],[0,0,3],[],[0,0]]
               ˜     # Flatten the list
                     #  → [0,0,0,3,0,0]
                à    # Pop the list, and get the max
                     #  → 3
                 «   # Merge it with the other two digits (and output implicitly)
                     #  → 213

飞镖，165字节

F(List s)=>s.indexWhere((a)=>a.contains('*'));f(String s){var b=s.split('\n'),v=F(b),c=b[v].split('  '),w=F(c),x=c[w].indexOf('*')+1;return '${w+1}${b.length-v}$x';}

在线尝试！