게으른 단어 문제


15

요약

교사는 학생들에게 단어 문제를 준비하라는 지시를 받았습니다. 그녀는 방정식의 목록을 받고 대응하는 단어 문제로 쓰라고 지시합니다. 그러나 그녀는 매우 게으 르기 때문에 창의력을 발휘하지 않습니다. 대신, 그녀는 단순히 문자 그대로 씁니다. 예를 들어, 그녀는 읽을 때 1+1그녀가 쓰고 one plus one, 47 * 2로 변신 것 forty seven times two, 그리고 56.2 / 7.4이 될 것입니다 fifty six point two divided by seven point four.

이 게으른 선생님을 돕기 위해 몇 가지 코드를 작성하십시오.

세부

  • 숫자는 소수점과 음수 부호를 포함 할 수 있습니다.
  • 숫자는 짧은 규모로 작성해야합니다. (즉, 1,000,000,000이다 one billion)
  • 숫자는 999,999,999,999,999,999 ( nine hundred ninety nine quadrillion...nine hundred ninety nine) 까지 올라갈 수 있습니다 .
  • 0 그룹은 제외해야합니다. 예를 들면 1,000,000이다 one million없습니다 one million zero thousand zero hundred.
  • 소수점 이후에는 임의로 많은 숫자가있을 수 있습니다.
  • 소수점 뒤의 숫자는 숫자로 표시되어야합니다. 예를 들면 12.34이다 twelve point three four없습니다 twelve point thirty four.
  • 두 숫자는 항상 연산자로 구분됩니다.
  • 유효한 연산자는 더하기 ( +), 빼기 ( -), 시간 ( *) 및 나누기 ( /)입니다.
  • 괄호가 없습니다.
  • 와 같은 숫자는 에서처럼 출력에 1234선택적으로 포함 할 수 있습니다 .andone thousand two hundred *and* thirty four
  • 입력의 쉼표와 공백은 무시 될 수 있습니다.

입력 : 24 + 65
출력 :twenty four plus sixty five

입력 : 3.33333 - 0
출력 :three point three three three three three minus zero

입력 : 3.6 * 18.18 / 999.0
출력 :three point six times eighteen point one eight divided by nine hundred ninety nine point zero

입력 : 1-1
출력 :one minus one

입력 : 1+-1
출력 :one plus negative one

입력 : 1,000,000,000 + 0.2
출력 :one billion plus zero point two

입력 : 123,000,456,789,012,345.6789
출력 :one hundred twenty three quadrillion four hundred fifty six billion seven hundred eighty nine million twelve thousand three hundred forty five point six seven eight nine

입력 : -4.3 * 7
출력 :negative four point three times seven

입력 : -1-1--1
출력 :negative one minus one minus negative one


1
123,456,789,012,345.6789예제 와 같은 것을 추가 할 수 있습니까? 많은 테스트 사례를 다루어야합니다.
maxb

2
minus대신에 사용할 수 있습니까 negative?
조 왕

3
매스 매 티카의 경우 : 다시 거기에 내장입니다 만, /이다 over음의 수는 minus그 어떤 조작을 필요로하므로.
user202729

2
@ user202729 어메이징 ... 왜 Mathematica에 내장 된 것이 놀랍지 않습니까? :)
Daffy

답변:


15

자바 스크립트 (ES6), (552) 532 바이트

이 불결한 괴물은 코드 골프 지옥의 깊이에서 곧장 온다.

공백이없는 입력 문자열을 예상합니다.

S=>S[R='replace'](/[\d.,]+|./g,s=>1/s[0]?a(+s[S=0]&&14)+s[R](/(\D?)(\d+)/g,(_,s,n)=>s>','?' point'+n[R](/./g,a):j--*n?(u=a(n%10||14),n>99?a(n[0])+' hundred':'')+((n%=100)<13?a(n||14):n<20?(a(n)||u)+'teen':(a(n/10+18)||a(n/10))+'ty'+u)+a(j+27)+(j>1?'illion':''):'',j=s.split`,`.length):a(S='+-*/'.indexOf(s=='-'&&S||s)+34),a=n=>(s='zero0one0two0three0four0five0six0seven0eight0nine0ten0eleven0twelve0thir00fif000eigh00twen0thir0for0fif000eigh00thousand0m0b0tr0quadr0negative0plus0minus0times0divided by'.split`0`[n|0])&&' '+s).trim()

온라인으로 사용해보십시오!


3
거대한 문자열 리터럴을로 바꾸면 18 바이트를 줄일 수 있습니다 btoa`ÍêèÒ‰ÞÒÜ(ÒØkyí¢êô~+ÞÒȱÒǯz}ŠmÒx§{K^ŸG¥z÷§ÒÜ–÷´¶«ÓGâM4z(!ÓKpz}-†*ô~Šô~'ôÓG¢‚4¶.±©ÝÒmÒÚôªæ�¯IÞ�«b½í)–ë4š)î³Kb™ë4v+âuçu×Vò`.replace(111,' ').
kamoroso94

나는 문자열 리터럴을 창의적으로 압축하는 응답을 좋아합니다.
Daffy

6

펄 6 , 434401 387 359 바이트

{~S:g/\d+/{n($//100+64184)x($/>100),$/%100>19&&(n($//10%10+64175),n($!=$/%10)x?$!)||n($/%100+7679),[$,"thousand",|(<m b tr quadr>X~"illion")][+$/.postmatch.words[0].comb(',')]if +$/} /.trans("+,-/*"=><<plus''minus"divided by"times>>).words}o{S:g/\.(\d)+/ point {$0>>.&n}/}o{S:g/[\s|^]0/ zero/}o{S:g/\-(\d)/negative $0/}
my&n=(*+1632+|0).uniname.lc.words[2..*]

온라인으로 사용해보십시오!

확실히 개선의 여지가 있습니다. 나는 그렇게 말하지만, 내가 처리하지 않은 최첨단 사례를 계속 주목합니다 :(. 입력에 공백으로 구분 된 연산자가 있고 숫자의 음수가 구분되지 않는다고 가정합니다.

설명:

my&n=(*+1632+|0).uniname.lc.words[2..*]  # Define a helper function
           # This gets the unicode name, e.g ARABIC-INDIC DIGIT ZERO
           #                              or AEGEAN NUMBER ONE HUNDRED
           # And returns the 3rd word onwards in lowercase e.g. 'zero' or 'one hundred'
{S:g/\-(\d)/negative $0/}  # Turn dashes before numbers to negative
{S:g/[\s|^]0/ zero/}       # Handle zeros
{S:g/\.(\d)+/ point {$0>>.&n}/}  # Replace decimals
{~S:g/\d+/        # Replace series of numbers with
    n($//100+64184)x($/>100)  # The hundreds if the num is bigger than 100
    $/%100>19&&               # If the number is bigger than 19
      (
      n($//10%10+64175),         # The tens number
      n($!=$/%10)x?$!            # And the singles number (if it's not zero)
      )
    ||                        # Else
      n($/%100+7679)             # The name of the number below 20
    ,                      # Then
    [$,"thousand",|(<m b tr quadr>X~"illion")][   # Index into the list of postfixes
      +                         .comb(',')   # The number of commas
       $/.postmatch.words[0]  # in the rest of the number
  if +$/           # All if the number is not 0
.trans("+,-/*"=><<plus''minus"divided by"times>>)  # Translate operators
                                                   # And remove commas
.words}     # And remove all the excess spaces between words

5

루비 + 스위프트 4, 283 279270 바이트

$_=gsub(/(?<=\d)-/,"minus ").gsub(/[*-\/]/,?.=>"point ",?-=>"negative ",?+=>"plus ",?*=>"times ",?/=>"divided by ").gsub(/(?<=^|[^t] )\d+|\d/){`echo "import Foundation
var f=NumberFormatter()
f.numberStyle = .spellOut
print(f.string(from:#$&)!)">.a
swift .a`.tr'-
',' '}

온라인으로 사용해보십시오!

나는 그러한 프랑켄슈타인 솔루션을 제안하는 것에 화를 내야하지만, 한편으로는이 작업에 Swift의 내장 기능을 사용하고 싶어하는 반면, Swift에서 Regexes로 문자열을 처리하는 것은 골프 재앙으로 보입니다.

따라서 Ruby에서 기본 문자열 처리를하기로 결정했지만 숫자를 철자하기 위해 Swift 소스 파일에 저장하고 쉘 명령으로 Swift를 실행하고 출력을 수집합니다.

Swift의 "spellOut"숫자 포맷터는에서와 같이 두 자리 숫자에 불필요한 하이픈을 삽입하는 것을 제외하고 우리가 필요한 것을 거의 정확하게 수행합니다 twenty-two. 사실, 형식의 부동 소수점 출력조차도 integer part point digit digit...좋지만 여기에는주의가 있습니다. 무한 정밀도는 없으며 충분한 숫자 또는 많은 십진수가 있으면 결과가 잘못됩니다. 따라서 정수와 분수 부분을 분리하고 분수를 숫자로 공급해야했습니다.


1
이것은 절대적으로 악마이며 나는 그것을 좋아합니다.
Daffy

4

sfk , 853 바이트

xed -i
"_*_ [part1]_"
+xed
_+_plus_
_\*_times_
"_/_divided by_"
"_- _minus _"
"_-_negative _"
+xed
"_,[keep][19 chars of 0-9,]_quadr@ _"
"_,[keep][15 chars of 0-9,]_tr@ _"
"_,[keep][11 chars of 0-9,]_b@ _"
"_,[keep][digits],[digits],_b@ _"
"_,[keep][digits],_m@ _"
"_,_ thousand _"
+xed
"_ 000[chars]@__"
"_ 000__"
"_ 00[keep][digit]_ _"
"_ 0[keep][2 digits]_ @_"
"_ [digit][keep][2 digits]_[part2]hundred @_"
"_ [ortext] 0[digit]0_ @[part2]_"
"_ [keep][2 digits]_ @_"
"_@_illion _"
+xed
_@11_eleven_
_@12_twelve_
_@1[digit]_@[part2]teen_
_@1_ten_
_@4_forty_
_@[digit]_@[part2]ty_
+xed
_@2_twen_
_@3_thir_
_@4_four_
_@5_fif_
_@6_six_
_@7_seven_
_@8_eigh_
_@9_nine_
+xed
"_0_ zero _"
"_1_ one _"
"_2_ two _"
"_3_ three _"
"_4_ four _"
"_5_ five _"
"_6_ six _"
"_7_ seven _"
"_8_ eight _"
"_9_ nine _"
"_._ point _"
+xed
"_[white]_ _"
+xed
"_[lstart] __"

온라인으로 사용해보십시오!

연산자와 숫자는 하나 이상의 공백 문자로 구분해야합니다.


4

클린 , 766 ... 687 바이트

import StdEnv,Text
m=""
z="zero"
@ =digitToInt
r=reverse
l k=(!!)k o@
^s=l[s:split" ""one two three four five six seven eight nine"]
g s=l[m,m,"twen","thir",s,"fif","six","seven","eigh","nine"]
~['0':t]= ~t
~[a,b,c]= ^""a+" hundred "+ ~[b,c]
~[b,c]|b>'1'=g"for"b+"ty "+ ^""c|c>'2'=g"four"c+"teen"=["ten","eleven","twelve"]!!(@c)
~[c]= ^""c
~_=m
$[]=m
$[x:y]#(h,t)=span(\e=e>'/'||e==',')if(x<'1')y[x:y]
=trim(join" "((case x of'0'=[z];'-'=["negative",$h];'.'=["point":map(^z)h];_=(r[u+v\\u<-r(map~(split[',']h))&v<-[m," thousand":[" "+k+"illion"\\k<-["m","b","tr","quadr"]]]|u>m]))++[?t]))
?['-':t]="minus "+ $t
?['+':t]="plus "+ $t
?['/':t]="divided by "+ $t
?['*':t]="times "+ $t
?t= $t

온라인으로 사용해보십시오!

공백이없는 문자열을 예상합니다.


1

05AB1E , 315 295 282 276 바이트

"+-*/"DˆS¡εDõQi'¢…ë'.¡VYнD_i\'¡×ðë',¡DgUε0›i.•6b©•ð“†ìˆÈŒšï¿Ÿ¯¥Š“©'tKJ#'…§«…€µ‚•„í#®#«…—¿áÓÁÏ#«ìD9£©.•4º»Ÿć'Rþн•ŽH$S£“Œšï¿Ÿ¯¥Š“'tK#«„ty«sõšâðý«õšD®'°¡ðì«sâðý«yèð.•cG3₅¦„¥F•8ô'¾ß«…¡›‡È±°#«õªRXN-<èJëõ}}ðý}Yg<i®'¡×šYθSè'…®šðý}}J}s¯`Ã哉´Øè„ƺߓ#¤… by«¸s¨ì¯`ykè}.ιðý„  ð:„¢…Øè'¢…:

공백없이 입력을받습니다.

온라인으로 시도 하거나 모든 테스트 사례를 확인하십시오 .

설명:

"+-*/"                    # Push string "+-*/"
Dˆ                        # Duplicate it, pop the copy, and push it to the global array
S¡                        # Split the input by any "+", "-", "*", or "/"
ε                         # Map each number to:
 DõQi                     #  If the item is empty (happens for negative numbers)
     '¢…                 '#   Push string "negative"
 ë                        #  Else:
  '.¡                    '#   Split by "."
  VY                      #   Store it in variable `Y`
  н                       #   Take the first number (the integer part)
  D                       #   Duplicate this integer part
  _i                      #   If the integer part is exactly 0:
    \                     #    Discard the duplicated integer part
    '¡×                  '#    Push string "zero"
    ð                     #    Push a space " "
  ë                       #   Else:
   ',¡                   '#    Split by ","
   DgU                    #    Pop and store the amount of items in variable `X`
      ε                   #    Map each part to:
       0i                #     If it's larger than 0:
          .•6b©•          #      Push string "thir"
          ð               #      Push a space " "
          “†ìˆÈŒšï¿Ÿ¯¥Š“  #      Push string "four five six seven eight nine"
          ©               #      Store it in the register (without popping)
           'tK           '#      Remove all "t" (so "eight" becomes "eigh")
          J               #      Join it together with the "thir" and space
          #               #      Split by spaces
          '…§            '#      Push string "teen"
             «            #      And append it to every string in the list
                          #      (We now have ["thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"])
          …€µ‚•„í         #      Push string "one two three"
                 #        #      Split by spaces
          ®               #      Push the string from the register ("four" through "nine")
           #«             #      Split by spaces, and merge both lists together
          …—¿áÓÁÏ         #      Push string "ten eleven twelve"
                 #«       #      Split by spaces, and also merge both lists together
          ì               #      Prepend "one" through "twelve" before "thirteen" through "nineteen"
          D9£             #      Duplicate it, and take the first nine ("one" through "nine")
             ©            #      Store it in the register (without popping)
          .•4º»Ÿć'Rþн•   '#      Push string "twenthirforfif"
          ŽH$             #      Push integer 4433
             S            #      Split to digits: [4,4,3,3]
              £           #      And split the to parts of that size: ["twen","thir","for","fif"]
          “Œšï¿Ÿ¯¥Š“      #      Push string "six seven eight nine"
                    'tK  '#      Remove all "t" (so "eight" becomes "eigh")
                       #« #      Split by spaces, and merge both lists together
          ty             #      Push string "ty"
             «            #      And append it to every string in the list
                          #      (We now have ["twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"])
          s               #      Swap so the list "one" through "nine" is at the top again
           õš             #      Prepend an empty string to that list
             â            #      Create every possible pair of "one" through "nine" with "twenty" through "ninety"
              ðý          #      Join each pair with a space delimiter
          «               #      Merge the "twenty" through "ninety nine" list with "one" through "nineteen"
           õš             #      Prepend an empty string to that list
          D               #      Duplicate the entire list
          ®               #      Push the string from the register ("one" through "nine")
          '°¡            '#      Push string "hundred"
             ðì           #      Prepend it with a space " "
               «          #      Append it to every string in the list
                          #      (We now have ["one hundred","two hundred",...,"nine hundred"])
          s               #      Swap the two lists
           â              #      Create every possible pair of "one hundred" through "nine hundred" with "" through "ninety nine"
            ðý            #      Join each pair with a space delimiter
              «           #      Merge the "one" through "ninety nine" with "one hundred " through "nine hundred ninety nine"
                          #      (We now have ["","one",...,"nine hundred ninety nine"])
          y               #      Get the current number of the map
           è              #      And index it into this list
          ð               #      Push a space " "
          .•cG3₅¦„¥F     #      Push string "quadrilltrill"
                     8ô   #      Split into pieces of size 8: ["quadrill","trill"]
          '¾ß            '#      Push string "ion"
             «            #      Append it to every string in the list
          …¡›‡È±°         #      Push string "billion million thousand"
                 #        #      Split by spaces
                  «       #      And merge both lists together
          õª              #      Append an empty string
            R             #      Reverse the list
                          #      (We now have ["","thousand","million","billion","trillion","quadrillion"])
          X               #      Push variable `X`
           N-             #      Subtract the map-index from it
             <            #      Subtract an additional 1
              è           #      And index it into the list
          J               #      Join the stack together
       ë                  #     Else:
        õ                 #      Push an empty string ""
       }                  #     Close the if-else
      }                   #    Close the map
      ðý                  #    Join the mapped values with space delimiter
  }                       #   Close the if-else
  Y                       #   Push variable `Y`
  g<i                     #   If its length is exactly 2:
     ®                    #    Push the string from the register ("one" through "nine")
     '¡×                 '#    Push "zero"
        š                 #    Prepend it to the list
      Yθ                  #    Push variable `Y` again, and leave the second number (the decimal part)
        S                 #    Split it to digits
         è                #    And index each into the list
      '…®                '#    Push string "point"
         š                #    Prepend it in front of that list
      ðý                  #    Join the list with space delimiter
  }                       #   Close the if
 }                        #  Close the if-else
 J                        #  Join the stack together
}                         # Close the map
s                         # Swap to take the (implicit) input again
¯`                        # Push the global array, and dump it's content (string "+-*/")
  Ã                       # Only keep all "+", "-", "*", and "/", and remove everything else
ε                         # Map each to:
 “‰´Øè„ƺߓ               #  Push string "plus minus times divided"
           #              #  Split by spaces
 ¤                        #  Take the last item (without popping the list)
   by«                   #  Append it with string " by"
       ¸                  #  Wrap it to a list: ["divided by"]
 s                        #  Swap to take the list again
  ¨                       #  Remove the last item
   ì                      #  Prepend it in front of the list: ["plus","minus","times","divided by"]
 ¯`                       # Push the global array, and dump it's content (string "+-*/")
   yk                     #  Push the index in this string for the current map-value `y`
     è                    #  And use that index to index into the string-list
}                         # Close the map
                        # Interweave the list of numbers and list of operators
  ðý                      # Join everything with space delimiter
  ð:                     # Replace every two spaces for a single space
„¢…Øè'¢…:                '# And replace every "negative minus" with "negative"
                          # (and output the result implicitly)

다음과 같은 이유를 이해하려면이 05AB1E 팁 ( 사전을 사용하는 방법 ? , 사전의 일부가 아닌 문자열을 압축하는 방법?큰 정수를 압축하는 방법? ) 을 참조하십시오.

  • ( 사전 사용법은? )- '¢…is "negative"; '¡×이다 "zero"; “†ìˆÈŒšï¿Ÿ¯¥Š“이다 "four five six seven eight nine"; '…§이다 "teen"; …€µ‚•„í이다 "one two three"; …—¿áÓÁÏ이다 "ten eleven twelve"; '°¡이다 "hundred"; '¾ß이다 "ion"; …¡›‡È±°이다 "billion million thousand"; '…®이다 "point"; 하고 “‰´Øè„ƺߓ있다 "plus minus times divided".
  • ( 사전의 일부가 아닌 문자열을 압축하는 방법.•6b©•무엇입니까? )- is "thir"; .•4º»Ÿć'Rþн•이다 "twenthirforfif"; 하고 .•cG3₅¦„¥F•있다 "quadrilltrill".
  • ( 큰 정수를 압축하는 방법? ) – ŽH$입니다 4433.

1

파이썬 2 , 790 774 바이트

lambda T:B("([+/*-])",lambda m:dict(zip("+/*-",S("z"," plus z divided by z times z minus ")))[m.group(0)],B("([+/*-]|^)-",r"\1negative ",B("[^+/*-]+","{}",T))).format(*[J([g[int(S("\.",j)[0])]+S("z",B("y","illion","z thousandz myz byz tryz quadry"))[len(S(",",m))+~i]+(" point "+J(s[int(c)]for c in S("\.",j)[-1]))*("."in j)for i,j in E(S(",",m))if 0<float(j)+(m<"1")])for m in S("[+/*-]+",T)[T[0]=='-':]])
from re import*
E,S,B,P=enumerate,split,sub," ";J=P.join
s,e=S(P,"zero one two three four five six seven eight nine"),[B("urty","rty",j)for i,j in E(c+d for d in S(P,"teen ty")for c in S(P,"twen thir four fif six seven eigh nine"))]
g=s+S(P,"ten eleven twelve")+e[1:8]+[a+(P+b)*(i>0)for a in e[8:]for i,b in E(s)]
g=[(j+" hundred ")*(i>0)+k for i,j in E(s)for k in g]

온라인으로 사용해보십시오!

많은 나쁜 관행. 이 글은 거의 다쳤습니다 ....

공백이 입력되지 않은 유니 코드가 아닌 문자열을 예상합니다.

설명:

# import all functions from re (python regex library)
from re import*

# rename some repeatedly-used functions/variables for reduced bytecount
E,S,B,P=enumerate,split,sub," ";J=P.join

# list the names of 0-9
s=S(P,"zero one two three four five six seven eight nine")
# generate "twenteen" through nineteen and twenty though ninety, changing "fourty" to forty
# using enumerate (E) even though i is not required b/c it's shorter than range(len(x))
# using re.split (S) instead of string.split since it's shorter
e=[B("urty","rty",j)for i,j in E(c+d for d in S(P,"teen ty")for c in S(P,"twen thir four fif six seven eigh nine"))]
# generate 0-999
# 0-9
g=s+
   # 10, 11, 12
   +S(P,"ten eleven twelve")+
                            # remove "twenteen", 13-19
                            +e[1:8]+
                                   # tens' place + ones' place, if ones' place is not zero
                                   +[a+(P+b)*(i>0)                               ]
                                                   # for each tens' place in 20-90
                                                   for a in e[8:]
                                                                  # for each index, value in ones' places 0-9
                                                                  for i,b in E(s)


# hundreds' place if at least 100, plus tens' and ones' place (already calculated and stored in g from before)
g=[(j+" hundred ")*(i>0)+k                          ]
                           # (s) stores names for 0-9, need index to avoid "zero hundred"
                           for i,j in E(s)
                                          # for each hundred, iterate over all values (0-99) already in g
                                          for k in g

# actual function to call. uses previously declared global variable g.
def f(T):
    # gets the numbers in the supplied string (T) by splitting (T) on any operator character
    # remove first item if blank (only happens when staring with a - for negative numbers)
    n=S("[+/*-]+",T)[T[0]=='-':]

    # triply-nested set of re.subs to convert (T) to a sting of where the operators are replaced by their names and numbers are replaced by "{}"
    # EX: "-1-1--1" -> "-{}-{}--{}" -> "negative {}-{}-negative {}" -> "negative {} minus {} minus negative {}"
    # this sub happens last
    # re.sub (B) any operator, with the operators in a group "()" so that they return in match.group
    T=B("([+/*-])",                                                                                                                                        )
                  # an anonymous function to accept match objects (m) from re.sub's search.
                  ,lambda m:
                            # create a dictionary from the combination of operators and their names
                            # like {"+":" plus ",...}
                            # operator names are surrounded by spaces since number names are NOT
                            dict(zip("+/*-",S("z"," plus z divided by z times z minus ")))
                                                                                          # from the constructed dictionary, select the operator matched by re.sub's search and return it for replacement
                                                                                          [m.group(0)],
                                                                                                      # this substitution is second
                                                                                                      # re.sub (B) any operator followed by a minus (-), OR a minus at the beginning of the string
                                                                                                      # operators/start are grouped, trailing minus is not
                                                                                                      ,B("([+/*-]|^)-",                                    )
                                                                                                                      # replace match with the grouped items plus the word "negative"
                                                                                                                      # EX: "-1-1--1" -> "-{}-{}--{}" -> "negative {}-{}-negative {}"
                                                                                                                      ,r"\1negative ",
                                                                                                                                     # this substitution is done first
                                                                                                                                     # replace any sequence of NON-operators with "{}"
                                                                                                                                     # this removes numbers so the names can be inserted later
                                                                                                                                     # EX: "-1-1--1" -> "-{}-{}--{}"
                                                                                                                                     ,B("[^+/*-]+","{}",T))

    # technically the previous construction of (T) and (n) can be placed here to save 5 bytes but my poor eyes can't handle that.
    # insert constructed names back into original string.
    # EX: "-1-1--1" -> "negative {} minus {} minus negative {}" -> "negative one minus one minus negative one"
    print T.format(                                                                                                                                                                                                                     )
                   # string.format needs items in array unpacked, or it will attempt to insert the string representation of the array itself
                   *[                                                                                                                                                                                                                  ]
                     # for each number pulled from (T), generate names and join generated items back together with spaces
                     # EX: "1,456" -> ["1", "456"] -> ["one thousand", "four hundred fifty six"] -> "one thousand four hundred fifty six"
                     J(                                                                                                                                                                                                     )for m in n
                       # split j on periods (.) and take the first item
                       # convert that item into an integer and find the item at that index in g (0-999)
                       [g[int(S("\.",j)[0])]+                                                                                                                                                                              ]
                                            # insert prefix for millions +, split string on "z" (spaces must be preserved for proper separation)
                                            +S("z",B("y","illion","z thousandz myz byz tryz quadry"))
                                                                                                     # left is largest, so take the item at index (total # of groups - current place - 1)
                                                                                                     [len(S(",",m))+~i]+
                                                                                                                       # if group had a period, split string on period and take last item
                                                                                                                       # replace every character in group with number 0-9 name
                                                                                                                       # join them with spaces and add back to rest of group
                                                                                                                       +(" point "+J(s[int(c)]for c in S("\.",j)[-1]))*("."in j)
                                                                                                                                                                                # split number into groups by comma
                                                                                                                                                                                # EX: "123,456" -> ["123","456"]
                                                                                                                                                                                # only return item if j != 0 (avoids returning empty string which will result in too many joined spaces)
                                                                                                                                                                                # OR if m == 0 (avoids not returning anything when should return "zero")
                                                                                                                                                                                for i,j in E(S(",",m))if 0

설명을 쓰는 동안 약 150 바이트가 줄었습니다. 코드 주석 달기 / 검토가 도움이되지 않는다고 절대 말하지 마십시오!

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