Hasilkan Brainfuck untuk angka 1–255

34

Tulis sebuah program yang, untuk angka 1 hingga 255, mencetak kode BF yang akan menghasilkan angka yang diberikan dalam beberapa byte array, ditambah baris baru.

Misalnya, empat baris pertama dari output bisa (dan kemungkinan besar akan) adalah:

+
++
+++
++++

Pemenang akan menjadi yang terkecil: source code + output(dalam byte).

Klarifikasi & Revisi :

  • Program BF menggunakan sel pembungkus.

  • Program BF output harus diakhiri dengan satu-satunya sel non-nol menjadi sel yang berisi angka.

  • Program harus berupa output dalam urutan menaik.

  • Mengeluarkan program untuk 0 adalah opsional.

  • Pointer data negatif tidak diperbolehkan. <pada pointer pertama tidak akan melakukan apa-apa. (tinggalkan komentar jika akan lebih tepat untuk melemparkannya)

Tukang batu
sumber
1
@ JoKing Seluruh output dihitung.
Mason
2
Oh saya mengerti, Anda mengatakan bahwa kode tidak harus berakhir pada sel output
Jo King
1
Akan sangat berguna untuk mendapatkan referensi tentang apa yang sebenarnya merujuk pada "BF" dalam konteks Anda, yaitu esolangs.org/wiki/Brainfuck_constants atau yang lain, dll.
HolyAvengerOne
2
@ Alasan Apakah +>++++++++++.program yang valid untuk input 1?
Jonathan Frech
6
+1 untuk tantangan dua-kendala di mana Anda harus menyeimbangkan bermain golf hasil BF versus golf kode untuk output BF. Twist yang menarik ☺
Chronocidal

Jawaban:

15

Perl 6 , 224 + 3964 = 5834 4188 byte

map {say (.[0]~'['~.[3]~'>'~.[1]~'<]')x?.[1],'>'x?.all,.[2]}o*.min({$_>>.abs.sum+6*?.[1]})>>.&{<- +>[.sign>0]x.abs},classify({0+|(grep(*%%1,(((256 X*^4)X+.[0]%256)X/-.[3]))[0]*.[1]+.[2])%256},[X] |(^27-13 xx 3),-7..-1){^256}

Cobalah online! (mungkin batas waktu. Ubah ^27-13ke^25-12 untuk mempercepat sedikit dengan mengorbankan beberapa output tambahan)

Output kode terpendek dalam bentuk *>[*>*<]>*, di mana masing *- masing adalah sejumlah +s atau -s. Ada beberapa tweak tambahan seperti menghapus loop jika tidak diperlukan, serta trailing >s.

Sejauh yang saya tahu, output adalah yang paling golf untuk format khusus ini.

Penjelasan:

([X] |(^27-13 xx 3),-7..-1)        # Define the search space as the cross product of:
                                        # -13 to 13 for:
                                            # Initialisation     +++>
                                            # Change in target   [*>+++<]
                                            # Last change        >+++
                                        # And -7 to -1 for the change in start [-->*<]
  .classify({                  })  # Group them by calculating
                  (256 X*^4)                         # Each of the multiples of 256
                 (          X+.[0]%256)              # Plus the initialisation
                (                      X/-.[3])      # Divided by the change in start
      grep(*%%1,                               )     # Filter out the whole numbers
                                                [0]  # And take the first value
          # This is the amount of times the inner loop will execute
          # Being Nil, converted to 0 if it is an infinite loop
      *.[1]              # Multiply by the change to the target cell
           +.[2]         # And add the final section
     (          )%256    # And modulo the whole lot by 256
                     +|0 # And floor it just to keep the .0 out
classify(                   ){^256}     # Take the corresponding groups in order
   .map(                             )  # And map each to
        *.min({                    })   # Find the minimum by:
               $_>>.abs.sum             # The sum of the absolute values    
                           +6*?.[1]     # Plus 6 if it loops
      >>.*{                   }    # Then map each value to
           <- +>[.sign>0]          # + or - depending on the sign
                         x.abs     # Repeated by the absolute value 
   {                    }o              # And pass this to the next code block
    say                       # Print
        (.[0]~'['~.[3]~'>'~.[1]~'<]')             # The loop section
                                     x?.all       # If it is needed
                                           ,.[2]  # And the final part
Jo King
sumber
12

Malbolge , 28 743 bytes + 7 166 output

Tidak terlalu kreatif, bukan? Saya akan berusaha bermain golf bocah nakal ini.

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Cobalah online!

Krzysztof Szewczyk
sumber
Tapi kenapayyyyyyy?
Yosua
@ Joshua bowling, anakku
Krzysztof Szewczyk
Ini tidak benar-benar golf, tetapi tetap memiliki upvote. Apakah alat kode sumber dibantu / dibuat?
Ethan
/ * Aku benar-benar bingung menjawab * / Ya.
Krzysztof Szewczyk
Gagasan di balik yang ini sama dengan yang ini: codegolf.stackexchange.com/questions/189358/is-it-double-speak/…
Krzysztof Szewczyk
12

Brainfuck, 77 75 73 + 32894 = 32967 32969 32971 byte

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

Cobalah online!

output adalah yang paling sederhana

+
++
+++
++++
...

penjelasan:

++++++[->+++++++<]>+ set cell 2 to 43 (ascii of plus)
>++++++++++ set cell 3 to 10 (ascii of new line)
>+ set cell 4 to 1
[
    >+ increment cell 5
    [
        -<<<.>>> decrement cell 5 and print a plus (content of cell 2)
    ] until cell 5 == 0
    <<.> print a new line (content of cell 3)
    [
        ->+>+<< move value of cell 4 to cell 5 & 6, setting cell 4 to 0
    ]
    >> goto cell 6
    [
        -<<+>> move it's value to cell 4, setting cell 6 to 0
    ]
<<+ increment cell 4
] exit when cell 4 goes beyond 255 because cell contains C uchar meaning 255 + 1 == 0
Jonatjano
sumber
10
Itulah yang ingin saya lihat - kode BF yang menghasilkan kode BF lain!
Dhruv Saxena
10

Stax , skor 4751 4783 (812 byte + 3971)

ç♥←ħòqε↓F"QS₧9(2╤↑▌T~│áZk♣☺nàK╬l•▼2≡→fZ⌂▼├▄<ÖΘá6≈¡K"B∩₧∟µ#°ôQí⌡B2ô§↕*∩)V╕EôD=)O╥T⌠û◘¬dⁿ┤☻∞ô↓♫√○¬z.â\²╕ùHÑ~≡M√☻:EzLƒ→B{O◙ΔΦ_S┼╤g°▓─+dï-┌└α½╥ôRù♠3f½⌐▀Pösúô₧f☻■Aε→τΓ£╒fε▬▬►EÜ%¬╧←y═←{╤╒öú5Ñ╡♀^α☺╨▼$kEÑ■µjh≈↕█Cªü←Z#∟gV↓►S3≥╟╗K‼╞.N|⌠↨╣}5H↕ê;±↓♣≤Tj█'x╒·±ΩßL;ª$Å÷ÑPIδ`◘▌╦┼╡<√▌{òE√PPQ/h@8kq/ÖΓb6╡]≈╤æ░╣{┌‼¢ÜαT├#ΓCN∞*╬⌡↕ÜVX←Ä)◘ù⌂ëøön╗)ôö∙╬⌠☻↨F¢X╓Sż9¡φö^⌂iøFB/┌º▼┤3¶☼Zëôû⌡ôΣfcäéi╣⌠"↨√$,.ë═┴↨Φz⌡τ¢S╜{╨)z:╦@}♦*│P±Æ1x╒ΦP▄◄·╢∙xF╢cá<T╗7;▐≤←÷╛╢;½▲§║│≈⌂ƒ*'F♂☼ùrT╞·╨nG∙=♦`;á$≥┼Ω▬≡aû┼☺╥ò♥R╧╖█▓uìf↕ñ∟φÖ♣°≥←▐G▀╗┘┴÷a*▐♂9╝┬çG─⌠ñ≈☻K·⌐α⌠╡↑!≤≡¡qßτú=Θ≤C░°¡ƒ╛>╨RP○v¡I♦◘╣ô6â₧scÉ♀╣+HO┌☼<♀»?£┴≥ï½.ohaë║ëb┼âù²┌─┬]░ΘQ¥τ┘q▼$v╞Ñ╒æ±tXƒ♪>SC▌LVWª■z↑¶ßΩû↕'L╓BÅï;↑ΦB2.G╞╜&╓π♥1¥0^B0ª≤5e|☼τ5╩╘µåΩ╬◙☻xª└í∞$┐☻∙d▼}╒R⌠AU@Ω♥δÇi0î┴ ↕ù‼☼ƒ┌Aw£╧à7û«W3ùΦ╚A)P○♥Xn⌂øôEΦGB≥╢g[∟a(◘&¥◙─♂→A@┴ö≡↕9PZK║î⌂eóë≥─Åⁿ1‼╢▀ó╙ª▲╒π╗tΣ4○√;■<ä║äqñ8╠T/»q\→↔1ç°ΘZδV♀EçZ▄g┤Å╤ ┴àúJ║wµ$▄«N

Jalankan dan debug itu

Saya mulai dengan program yang diterbitkan optimal .
Saya menggunakan beberapa regex-fu, untuk membatasi ke program terpendek yang menggunakan paling banyak 2 sel. Lalu aku memangkas setiap trailing <atau >karakter. Saya pikir ini adalah cara yang mungkin konservatif untuk memastikan bahwa tidak ada sel non-nol asing pada penghentian program. Kemudian saya menjalankannya melalui program stax eksperimental yang saya tulis untuk menghasilkan program stax untuk output tipe kolmogorov tetap.
Program ini bekerja dengan berulang kali menerapkan penggantian string. Pada setiap langkah, ia mencari> 1 substring panjang yang paling sering terjadi dan menggantikannya dengan karakter yang tidak digunakan.

rekursif
sumber
@ JoKing: Saya pikir saya membahas sel non-nol asing. Harganya lebih dari 200 byte dalam ukuran program brainfuck, tapi saya membuatnya hampir semua kembali dalam hal kompresibilitas.
Rekursif
7

Arang , 707 698 410 + 3627 = 4334 4325 4037 byte

UT≔”}⊞J5±)↷γ²⁼⎇⦃<✂f^⊗L…¬⁻←«θ↥v⊙^≔¶υSψVτ16⁷·9I⌕↘;⦃@Pmt↙ |TL ‹.bE^↷Am⟧←⪫✂«GIχ¤⟲V⁻PÀ$χ¹'$↙‖%S³6◧N=$kHIpQ×ïu|%÷I↖➙⁸≔Wλ¹ê8⌕dNK‽3H∨↥γh➙↘⊙⊕“~Oj↨-⬤…⊟⁺§◨CB℅P⌕KNEAR№K⬤X"¬S⎇⧴V⁻±6⁼✂kι×CÀ⊞‴≡w↓γ=`→P5η1C⊖OSoNυs⊘$M↙êαη↖φ¡¿:θ-γ“rJW%E(7<w¤Uφ´ρHπ←SX↔τ↧%<Tº⎇0gθμ↓⌕;σw⌈pL;Y↘YΠ⊙>ξLzλ↓⁸ι⎚|⌕ΠP″M³⧴⬤¦➙⟧⌕/δ;↥⁻ºJK⌊≡<⊖λ✳Jκ⟲➙ξ⭆|^Σ*βMπ⍘⊟;ÀU÷‹⭆◧�ωκ?σηkYOδO/Bº?lAnaK{*Kaκ◨+↧aSφ0q‖B/φx⊘⌕«³ψü✂‹º≡/yc⁴&J↙S²~⎇z§‖$SP≧”θG↘←¹⁴+⮌⪪θ⸿↑Fθ§⁺+-ι⌕-+ιG→↙¹⁴-

Cobalah online! Tautan adalah untuk mengucapkan versi kode. Penjelasan:

UT

Matikan bantalan ruang.

≔”...”θ

Tetapkan string terkompresi besar yang terdiri dari jawaban @ JonathanAllen untuk -128.. -15tetapi dengan +dan -tanda - tanda diubah.

G↘←¹⁴+

Gambarlah segitiga +s dari sisi 14, yang menghasilkan hasil yang benar untuk 1 hingga 14. Kursor dibiarkan di sudut bawah, meskipun trailing return pada string yang dikompresi akan memindahkan output berikutnya ke baris berikutnya.

⮌⪪θ⸿

Pisahkan string besar pada karakter kembali dan cetak setiap substring dalam urutan terbalik, sehingga menghasilkan hasil 15 hingga 128.

Naikkan garis sehingga hasil untuk 128 ditimpa oleh hasil yang dinegasikan untuk 128.

Fθ§⁺+-ι⌕-+ι

Loop melalui transposing string +dan -kembali lagi sehingga mereka menghasilkan hasil yang benar untuk 128 hingga 241.

G→↙¹⁴-

Gambarlah segitiga -s dari sisi 14, yang menghasilkan hasil yang benar untuk 242 hingga 255.

Neil
sumber
5

Jelly , 1224 + 3716 = 4940 byte

⁾+-ẋ€Ɱ14ZY€U0¦j“6VⱮ×ė7¬(Ị¢ẋṀⱮM⁵Ѭkbvœ⁸½ẋƓ0⁽ṖçḟŻßɓẉḷ0Ƙ¥@ⱮZĊⱮ{ṫṇØ"ỵðẓ⁵!ḳqḄƬiỴƥṇØm@ɗẆḅƥƲ⁴ŀ-5¦€ÑɓZĖ/gPṄḌ!ẹ$ḞıƒĿỵ⁷£Q.%¦ẊiUı-M⁹ƈxṁ,CsḲtÆƇỴṄĿiæEṛⱮẒʠþƘ%ƘƙṾ ('ȥ€½⁵ḥ+,þ@ẇ&ạV|ĊuAYḃfṖƘLƥQtPƬivxHj)Ṇɓ5JṘØẓæĿøɗjḥrñþa®OṅḍṪ¥=ɼġċṫßṬỌƈrUẉçŻ½\=]€ʂ_ⱮṖ¥Ƥȥ6SṡÆcạdn;ṅⱮDɦ⁹ṢAy)~Ḷ`ẒẓMTİṂḋ|ẉ]Wɠ¿⁾Ṣ|ḷ6hẸƒⱮQ1ẏƝC@Ŀ!ʠ⁽ṃ@ƓŒQ3@ƝḊñçcZ\¥3Z¤~çD>ċọuⱮȦAẈⱮ%L3Æ¢ḞtĖė!ƇtñṪɓẓ¥Fṅ⁵shB'wṪẸ¦ṄÞṭ³ʂḶƊ³iȧṂRœŒƤ\r1Çwi6ŀỵɼḃa⁵Ṣ_Q⁸Ẹ'{|\+Æ®|ḤcʂÑ/Ɓz¶ɦÄ!ʂ"Ẋ ẓĠĠ⁷⁵QƝ¶%ṙƇḋ[^j&W×*°ḳçʂSżḊⱮ⁻IȦṄXȥlẋḅ7;⁺ḃİÞÆðLX¢1K£€Ä&X½VȮ(;Q£ḞḢ¹zG+ṅ¹LḥW³ḅd@^ẊḶJ¹T8ṛ($ȧṢzq,Ṫ⁻ȥ{Ṛ"Ḍ®Ä8QḋþɼȮhỵB"Ḍ⁶ȧZ⁵ẒNɓḃȧ¶Ƒð$Ẇ/"Eṭ*I:ØḃL}<KȦ+ṣ¥x&Ṇ£Œṫḋġ0lİḍ¿H£(ỌƝ×^Ḃ°⁽⁼UƭĠḥkQð7ṫƤżȷƘxjƑRḣqƒ$HƬ7ḳ-JµnṇṣðXŻİẉbSu×]bṾ0ƊHßçQh⁸°ƒɦSCñ_⁾ʂC⁼Ġø⁵SAʋƊİ¡⁴ÄḋẸḶwȧZẈĠ7rṀẏẉṖa¤ɱELƝȧẈṣṄk]d⁹øṇÞṡ.ạtƥṢḅ⁺ṂLpÑƘṄṡḍ⁵,Ǥ$Ọ8ṛuṚvAṖÑ1!vƤD3߶ʂа]EÞUĠ€ḋḲ⁸¬r`YḊ0ṙ5ċmṅȯ*ɲU÷pƭẉṭȦB¹ɦSNɱ)]ĠṾʋ³Øḟ23ṭð#ẆuẎṬṫVɠ(ỊỊQɼF}ịƒ$Ẏ_Ṡ'ḳOLc?ṾŀẊẎṆ⁵p"VẏAȮ⁴ⱮȦ®e®Ɱi"ÇJẊ4ñḍḲY]ḌḌẓ⁺ƙ"iṄḅoLṙfOS&}HGɼĖİĠḷuḃ³ṡıḳỊẹzq⁶ƈ£ċHZɱ.#⁶ḟUṗŀȮṘḶḲ]@¶+ḊĖ8ĖṆɗçŻŀ®ṭẇƓḄḷıM@⁷²36ɲṗ¡ḂḊ'Ṿ⁵ėƙṘ-⁺µʠṡṂ[_¤ḥṢ]ṘÐḤ½ḟ4ȷ}E¹Ṙb⁹ḅḢ¹hƬZ§Ẏẹ÷Æ$ḅoĖẉ⁹ịJ.ȥḊẋʋṄȯ1<ẎṄḲṛœ"æ)ḥ8)ḤlñA⁾%⁶LỴ⁶M4Ṙ\`ỵƊȥŀƒ⁷ḌƬƙḳƑ⁴vʂ⁻ðQpñḷḳṄœ>ṪỴƭƙɓ3[&Ḅzḅ<⁾µİṪȧ⁹C>ẹ{ẈÐlC&j?LṆṛ⁽æȤið<⁽$Ḋ7⁻FṡḅɓɱḂJoPŻẆṃṛḂ¹ẓð[1eƘ2T⁶ḟɼ7P~©ṚṙE8RƒṬẹLœẇẸịøḷ*⁾²ÄƓy€VƈɱNSẏẓѶpƲḞḅX⁹ọaœ<aỴTĠ^ðƑṙẊḅOḥŀG4ị¤ÑėÐịʠɗ=YṚċẋĠżẉịṪṁtḳṪ{ṬṃıızD/ĊvȤpḣðСþfÞ⁶ỵỊµṅḷÄ÷Vẇ\Ạ$-§OẠn^ȯfẎlḊd⁶ni¥ẓɱn¶’ṃ“¶><-][+”¤⁷

Program lengkap.

Cobalah online!

Bagaimana?

Hampir sepenuhnya kompresi terbaik saat ini pada esolang yang hanya menyisakan satu non-nol, dengan pita trailing dihapus. Mungkin ada cara untuk mengevaluasi subset dari program BF sedemikian rupa sehingga mereka akan berakhir dan menghasilkan solusi terpendek yang mengalahkan program naif ini. Mungkin juga ada cara untuk mengalahkan ini dengan program berbasis pola atau pembangunan faktorisasi yang lebih cerdas.

⁾+-ẋ€Ɱ14ZY€U0¦j“ ... ’ṃ“¶><-][+”¤⁷ - Link: no arguments
⁾+-                                - list of characters ['+','-']
    €                              - for each:
   ẋ                               -   repeat
     Ɱ14                           -   mapped across [1..14]
        Z                          - transpose
         Y€                        - join each with newline characters
           U0¦                     - reverse the rightmost
                                   -   (now we have ["+\n++\n+++\n ...","... \n---\n--\n-"]
                                ¤  - nilad followed by link(s) as a nilad:
               “ ... ’             -   a really big number compressed as base 250
                       “¶><-][+”   -   list of characters ['\n','>','<','-',']','[','+']
                      ṃ            -   decompress - use as base 7 digits [1,2,3,4,5,6,0]
              j                    - join (the list ["+\n++...","...--\n-"]) with that
                                   - implicit print
                                 ⁷ - a newline character
                                   - implicit print
Jonathan Allan
sumber
Memperbaiki pelanggar (dan menghapus rekaman trailing move)
Jonathan Allan
5

SuperMarioLang , 231 + 32894 byte

)
))++>(>+)*>[!((&(>[!*>-)-[!([!
===+"="==="=#===="=#="====#==#
+++<(       )    !+< !  ( <
+===+ (   - .    #=" #===="
>[!+( (   !(<
"=#++ (   #="
- (++ !.))    )))            <
) +++ #======================"
+ +++
+ ++!
!+<=#
#="

Cobalah online!

Ini pasti bisa lebih golf, karena hasilnya adalah yang paling mendasar untuk brainfuck, tapi saya butuh waktu seharian untuk menulis jawaban ini (ketiga anak saya hanya menyisakan sedikit waktu untuk saya), dan saya bangga bahwa setidaknya saya berhasil mencapai ini.

Charlie
sumber
4

Python 2, 70 + 8428 = 8498

-2 Bytes Berkat A__!
-20 Bytes Terima kasih kepada Jonathan Allan!
-229 Bytes dengan meletakkan nomor di sel kedua
-1000ish bytes dengan beralih dari 16 ke 9

p='+'
i=1
exec"print[p*i,i/9*p+'[>'+p*9+'<-]>'+i%9*p][i>20];i+=1;"*255

Cobalah secara Online!

Keluaran

Zachary Cotton
sumber
4

Ruby 271 + 5363 = 5634

1.upto(255){|n|r=n>(o=n>128?256-n:n)??-:?+;puts o>20?(s=o.to_s(i=(3..9).find{|i|!(s=o.to_s i)[1..-2][s[0]]}).bytes;s[-1]+=s[0]%8;(s[1,9].reverse.map{|c|(c-=s[0])<0??-*-c:c>0??+*c:?-}*?>+'[>'+?+*(s[0]%8)).tr(n>o ?'+-':'','-+')+'[-<'+?+*i+'>]<<]'+(s[-1]>s[0]?'':?>+r)):r*o}

Cobalah online!

Mengkonversi setiap nilai ke basis terkecil yang tidak mengandung nol digit awal di tempat lain, dan kemudian mengonversi dari basis itu. Nilai yang lebih besar dari 127 dihitung sebagai kebalikannya.


Non-Wrapping, 221 + 5888 = 6109

1.upto(255){|n|puts n>20?(s=n.to_s(i=(3..9).find{|i|!(s=n.to_s i)[1..-2][s[0]]}).bytes;s[-1]+=s[0]%8;s[1,9].reverse.map{|c|(c-=s[0])<0??-*-c:c>0??+*c:?+}*?>+'[>'+?+*(s[0]%8)+'[-<'+?+*i+'>]<<]'+(s[-1]>s[0]?'':?>+?-)):?+*n}

Menggunakan pendekatan yang sama seperti di atas, dengan sel yang tidak membungkus.

Cobalah online!

primo
sumber
4

JavaScript (Node.js) , 691 + 3627 = 4318

Menggunakan pendekatan yang sama dengan jawaban @ Neil's Charcoal , dan karenanya juga berdasarkan pada jawaban Jelly JonathanAllan's .

_=>(a=require('zlib').inflateRawSync(Buffer('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','base64'))+'')+`
--[>-<--]>-
`+[...a.split`
`].reverse().map(s=>s.replace(/[-+]/g,c=>c>','?'+':'-')).join`
`

Cobalah online!

Arnauld
sumber
2

Keg Tidak Resmi 16 + 32895 = 32911 byte

Solusi dasar untuk bahasa golf. Ini adalah yang paling sederhana yang bisa saya pikirkan.

ÿï((:|\+$;)_\
')

Cobalah online!

SEBUAH
sumber
Seperti berdiri ini perlu dibalik (meskipun saya bertanya apakah kami dapat output turun)
Jonathan Allan
1
Saya akan menunggu sampai OP menjawab dengan pertanyaan Anda.
A
2

Ruby 23 + 32895 = 32918 byte

256.times{|n|puts ?+*n}

Sebagai garis dasar. Ini adalah solusi paling sederhana yang bisa saya pikirkan.

MegaTom
sumber
0mungkin seharusnya 1(meskipun saya bertanya apakah kami dapat menghasilkan nol juga)
Jonathan Allan
-3: 256.times{|n|puts ?+*n}tidak terlalu penting ...
primo
1

Retina 0.8.2 , 28 + 16640 = 16668 byte


127$*+128$*-
\+
¶$`+
-
¶$'-

Cobalah online! Termasuk output untuk 0. Hanya keluaran menggunakan +s hingga 127 dan -s hingga 255.

Neil
sumber
1

Scala , 95 + 16639 = 16734 byte

object M extends App{(1 to 127).map(x=>println("+"*x));(0 to 127).map(x=>println("-"*(128-x)))}

Cobalah online!

Jawaban sederhana yang jelas tidak akan menang. Hanya menggunakan fakta bahwa -operator (mengurangi satu byte) membungkus kembali ke 255.

troli813
sumber
Scala bukan bahasa golf terbaik, tetapi Anda dapat menyimpan beberapa byte hanya dengan menulis metode alih-alih seluruh aplikasi, yang diizinkan berdasarkan aturan golf yang khas di sini. Selain itu, daripada mencetak hasilnya, biasanya lebih efisien byte untuk hanya mengembalikan daftar. Di Try It Online, Anda dapat meletakkan semua objek memperluas hal-hal jenis aplikasi di header dan footer, untuk menjalankan metode Anda tanpa menghitung byte yang tidak relevan.
Ethan
Dalam hal ini, Anda dapat menyimpan 36 byte dengan: tio.run/…
Ethan
1

05AB1E , skor: 4848 ( kode sumber 1219 bytes + output 3629 bytes)

'+14L×»Â'+'-:•тômG‚ΣP;e3₃ìèÕwƵÜè-½;¨Z±µΛé±V™NkKJžšë₅ušΘ(M₄+ܧ‘мoÕθÚzÇYï#J×¢θýει™₃tQØËв¿U®GƵ´GZ’¯ε¨jjØÛλÄ₅X∍µxθÆvËjS¹∊f˜«VÐZ<ÇĆ’Š2&ØÍäßÍĆlΓV₆ëßê©Œ‡ÛiyĆ=*÷Í´¢‹j,3½íµ'ž4‘û29ôãζм§x…1P|ÛéΣ=~çš5Œ±€Ô“q òǝ?ó¬Æí5¢G‘°êóÿв4LFÍK&zζb2Ó∍æïι8₃4XƵÜÙôt₁‘,Ö…6₅ÞαÇø†c÷Ûλ9…F;ĆA¬iмéλ8ä¶×ƶYΔè¡aû
v=M„ûñ]C₅Õ¶Þ*Ú`Úˆ/₃UιΩW¾eTεvˆ£nYõ¶S¼ÿ{õN9Ω¨£1w‚Ï”Xd;¹OýŒéDнĀvÌ–d=±ΛΣÃÊîD—GR>~ºD‹K¥‘l×yz.éFE1Í©ØM/ƒœOαU‘KΓO‰∍Aм‚œ2нƶþøÌ×¼āHgΩC'Λê¡-߅̾Ā–м–¿<₂δ¡áтgö¬Í~θFíнā‹°ü8[À(xï¸.›*W©¹º₅ÇмδçΛλÉFÕL4†EćÛ´ǝ{тÀ¯†ª™ŽćÉuè¬ƵÀSìFÙη¶1ȸ֛GÜlRv˜jy5mfè∞_åEηŠyo‡xÐ/™¥òÜ#Áx#м6r&₁cÿX۬ƄÌƵ₅∊бγ²Θj∞;6o·¼ýŠΩÚò›c[>ö₅¥=—ªÃ±¿ecSBÐ6Ê!ú¢E¡âìþ߃¿;Ò;„Xoƶ*∍Σǝñ"Tµ†8s®βµ4ìA|«÷γt³+<B¤špTp¸ï7Ëo[>–îiTôó檂?É8zн²ìC1ãl6+ƶå4sЌÚb(°·8ˆ´ˆŸ²ÚÌY3ŸËîÿ‘àUāçh9im„ÝĆm3ŠC×η“åX¨₄|ëPô3O<6Mþ'Ì-s{e`ζQΔ¹œP@l%¥‹èδcsÎcΘÂþ®i₅∞ð¡@`¸¿…BÎN2н>g;ΛSníÐ^Rαθ₆ΣÕ3¹ÐÔCfrQ¦7¨gfŒ|v||þÚÜvz≠pệT˜ǝ=ß·„®¡xи™#?†-Aʒ2åβ₃A¬Ão6ºтõ}Ë.&QηÕ~Δ4€@-5î^a̬.»Èõ4áL¾ò¥n
¶p›éŽžgǝSZγāmεålz₅°dβÂ~λà€Ê%zmŠиˆRη≠éwüǝΛζƵмƶdζ`SÖ₅\≠³äŸj!"(†Üćí“ŠxVöÇe#‡PÏɇ"xð®6ÊεGиe"NÊ›i.k…’Ú8:ǝ/₂ÌÜkãŽo™Áā‚ζΩ«мÁp=}ÂýõλиëÆζиîSÖt¶‚wĀθºd“
₆ŽsLвQ”ÖÜvGõƶiò{÷ÀPy/‹θÑè}¿Á5º˜¯sëØSËƶK_ÍyX∊3Øå4IOθ I+∊ÌñÙçakÞŸŒʒ椱,mεjæ‰O%<ÅtƒVöV=³ÇƶƒC¬‰xðȬM4Ïóä)∍Êfa§õØÂ,“X¾₆₄Ö¦ÈJµÿmȾÎ∍=¡YнŸV!¨J£ü|&¢cUg4e±6w™¼“fÊÙ ,Ž|šP·ùèd}ãŠÅ#GγhYÇN´¼ÁÌMGʒ§Æ1¸‚Δ:j7ΩƵAqá¢<äò´Θ•“-+>][<
“ÅвJs»

Output adalah pelabuhan @ Neil 's Charcoal jawaban , jadi pastikan untuk upvote dirinya juga!

Cobalah online.

Penjelasan:

'+                      '# Push a "+"
  14L                    # Push a list in the range [1,14]
     ×                   # Repeat the "+" that many times as string: ["+","++","+++",...]
      »                  # Join these strings by newlines
       Â                 # Bifurcate it (short for Duplicate & Reverse copy)
        '+'-:            # Replace all "+" for "-"
•тôm...ò´Θ•              # Push compressed integer 18302226724133383998250107335646038608225046109581810887431446835557987256955354954509163336111304735021044106950262344427892947550841899099611054599885158084492762836812161427050275372983896356189873217422270707048679161884382784973706990123491668808316983431947218815813441209357230471947480445527653281307616982417034289994948061000591427114479102114229222423495882782326672492922269629953210111953959859902281658658439835047182218017657439552630082372181376525413759195763958434475193943488791777228373958162363214252781530693967200164833437881609482421594458966138936433283311419810119896020066082377462298326514652481546557215787238749539873039910952003326954299252586309028025200870623285261199142261807190771369911425142504345271105103035478661301795311828767848235694787283635190364512722037791815037475799545052058894119573664059402985074146226606245848663046901585891882552845134633210352731812274795773552227786140415336764040421001184646630833787917147474644077938952053956874031774587527717793206158934471919975714697099518810712871948398923739276321843455690477328633199064849928974478179435369018512187592263559949835435473650276637191671401061097340919482725489354844550472281209666291367830643727358624135371626379451084552903536762775083445643853806852513122856150361979701049267928548063465967555886420646898485890108420374549485423234679327438138302730692296669063696581268627535131608200283731275951916433249161017999011290215932205767570177905442947203826039265793694687731078121685736352831955773450680945121984143563963149079990880719573067270197057276219243821370885160589340870891346257233778661271435191351926058080186177296974642815621539128350975752011448032262905976766027084285390087966682234081285502231618383962136055937741758125210487103109250885525370548106186539295203084216890820183575639032509902729248016346072449636148699098049659529168757116706057794418245039559549604674043961198447420311513558044229534569679723496972989178091506175996419296780639212192856671882116470677803387276814324094247508763467887301684211112080372036284371596072213153957411329532202432808677726223798116216330275138697515009114689489577370759238857602332613821627667530873656034962827810927061440822808985980383150080767015247752949877604372029666921293343149038246728649404223795601960991061986482063744094221616603849190547637439116347239768975065217383194655478092271791087679802480625740835053103772632489195507735140119501008503485917456615266596210333924964188989678201446160111091052524780358620148464886929989973412559470628329156848340802659185674541202787279386158230228148429451357621709967247567009904339076971643378255946241011579618610095231079053137553024558887196808709177094386352264708730475553352082713138948975317023830903305435434148828341201637230241697870602236452176330225025183518037443992277303117971849493548326433875
           "-+>][<\n"    # Push string "-+>][<\n"
                     Åв  # Convert the integer with this string as custom base
                       J # Join all characters together to a single string
s                        # Swap so the triangle of "-" we created it as the top of the stack
 »                       # Join the strings on the stack by newlines
                         # (and output implicitly as result)

Lihat ini ujung 05AB1E saya (bagian Cara kompres bilangan bulat besar ) untuk memahami mengapa •тôm...ò´Θ•adalah 183...875.

Kevin Cruijssen
sumber