1에서 255까지의 숫자에 대해 배열의 일부 바이트에서 주어진 숫자를 생성하는 BF 코드와 줄 바꿈을 인쇄하는 프로그램을 작성하십시오.

예를 들어, 출력의 처음 네 줄은 다음과 같습니다.

+
++
+++
++++

승자가 가장 작습니다 source code + output(바이트).

설명 및 수정 :

• BF 프로그램은 랩핑 셀을 사용합니다.

• 출력 BF 프로그램은 숫자를 포함하는 셀인 0이 아닌 유일한 셀로 종료해야합니다.

• 프로그램은 오름차순으로 출력되어야합니다.

• 0으로 프로그램을 출력하는 것은 선택 사항입니다.

• 부정적인 데이터 포인터는 허용되지 않습니다. <첫 번째 포인터에서 아무것도하지 않습니다. (투척하는 것이 더 적절한 경우 의견을 남기십시오)

펄 6 , 224 + 3964 = 5834 4188 바이트

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}

온라인으로 사용해보십시오! (5 월 타임 아웃.을 변경 ^27-13하기 위해^25-12 몇 가지 여분의 출력 비용에 약간의 속도)

가장 짧은 코드를 형식으로 출력합니다 *>[*>*<]>*. 각 코드 *는 +s 또는 -s입니다. 루프가 필요하지 않은 경우 루프를 제거하고 트레일 링을 추가하는 것과 같은 추가 조정이 있습니다 >.

내가 알 수있는 한 출력은이 특정 형식에 가장 적합합니다.

설명:

([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

Malbolge , 28743 바이트 + 7166 출력

너무 창의적이지 않습니까? 나는이 나쁜 소년을 골프로 일하게 될 것입니다.

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BAe(D=<`:9>7[;43270/St2+*N.-m%*#G'&}|#zy?>=<;\rwvon4Ukping-kdLbg`&%$#"!~}|\[ZYXQ9OTMq4JOHlFEJIHGF?cCB;_">7<;:3Wx65.3,+O).'&%I#"'&%|BA@?>|^]sxq7unVrqj0QPle+chg`_%]E[`_XWVzZSXWVUNrqpo2NGLEJIHGF?c=BA:?8\};43WVUTSRQP0p.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/Plkjchg`&G]\aZ~^@?UyY;WPUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I)(h&}$#z@~}v{zs9876tVUqpong-kjihgfe^F\[!_AWVzyxwvutTS54JINGkEDCBfFE'C<;:^!=<54381Uvu32+*N(n&J*#i!&}$#"y?`_{zsxwp6543210/mfed*Kaf_^$Ea`_^WVUyYXQ9OTMqQ3IHlLKDIBA@dcba`@"!=6Z4z2165.R210/(LKJIH"!~}|Bz!x}|{zs9wYutsrk1oQgled*Kg`ed]#[`Y^WVUyYXQ9OTMqQ3IHGLKJCgfedcba`_?>=6Z49816/St210/.'K+*#i!&}$#"y?wv{zs9wvXnm3qSohgfe+*)('Hd]\aZ~^W?[TSXWVONrRKJOHMLEihgfe(>bBA@?>=6Z4z870/.R210/on&Jk)"!~}C#zb~}v{zyxwpo5slqpih.Okdihg`&%$#"!~}|\[ZYXWVUTSRQPO10LEDh+AF?cC<;:^>=6|4X87654-s+*Non,%*#GFEDCB"y~}|{zyxwpo5Vlkjing-kdLbg`&%$#"!~}@?UyY;WVONrLKJn10LEDh+AF?cC%$@?>=6Z:9876v4321*Non,%*#GFEDC#c!xw=uzsr8potslqpong-eMibaf_%]E[`_XWVzZSXWVUNrqponm0/KDIHAF?c=BA:?8\};43WVUTSRQP0p.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/POejcb(Ie^$bD`YX]VzT<XQVUNrqpon10LEDh+AF?cC<;:^>=6|4Xyx05.RsP*/(',+$HGFEDe{z!~w=^]sxq7unmlk10/.-,+cha`_%cb[Z_X|\[=Sw:PUTMq4JOHl/KDhBAFED=a$@9>7<54XWVUTSR210/on&J$)"!~%$#z@awv{zyxwp6Wmlk1ong-Njibgf_%cba`_AWVzyxwvVUTS5Ko2NGkKDCBfedcba`@?>=6|4X8765.Rs10/.'K+k#('~D$#"!x}|{zs9wYotsrqji/mfNjiha'e^Fba`Y^W{zyxwvuUNSRKJIm0FEiIBGF?>=<`@?>=6|4X87654-s+*Non,%*#GFEDC{"!~}|{zyxwpo5Vlkjing-kdLbg`&%$#"!_^@?UyY;WVONrLKJn10LEDh+AFED=<`@?>=6|4Xyx05.RQPONMLK+k#('~D$#"!x}|{zs9wYutsrk1Rhmlejib(fe^$Ea`Y^WV[Tx;QPONMRKonmlk.JIBAF?c=BA:?8\};43WVUTSR21q/.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/Pfkdihg`&^F\a`_X]VzT<XQVUNrqponmlk.JCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I)(h&}$#z@~}v{zs98765VUqpong-kjihgfe^F\[!_AWVzyxwvu8TMLKPINGkEDCBfFE'C<;:^!=<54381U/S-2+0/.-,%I)ih~}${z@?>=<;:9wvoWm3qSing-Ndiha`&^F\a`_X]VzT<XQVUNrR4o2NMFEDCgfedcC%$@?>=6Z4z2165.R210/(LKJIHG'&}e#"y?wv{zs9wYutsrk1oQgled*KJfe^$bD`YX]VzT<XQVUNrqponmlk.JCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I#"'&%|BAya}|{ts9wvXtsrkj0/.Oejchg`&^F\[!_AWVzyxwvutsr54JINGkEDCBfFE'C<;:^!=<54381U/S-2+0/.-,%I)ih~}${z@?>=<;:98voWm3qSing-Ndiha`&^F\a`_X]VzZYR:u8TMRQPImlkjiIBAF?c=BA:?8\};43WVUTSRQ1q/.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/Pfkdihg`&^F\a`_X]VzT<XQVUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I#('~%|B"b~}|{ts9wvXtsrkj0/.-,dLhgfe^$bDZYX]\UyY;WPUNrqponml/KDCHAFED=aA@">=65Yzy70/.R210/on&J$#"!E}e#"y?`_{zsxwp6543210ng-kjcb(I_%cbD`YX]VzZYR:u8TMRQPImlkjiIBAF?c=BA:?8\};43WVUTSRQ1q/.'&%I#('~%|B"!~w|uzyxwp6Wmlkpi/Pfkdihg`&^F\a`_X]VzT<XQVUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I#"'&%|B"b~}|{ts9wvXtsrkj0/.-,dLhgfe^$bDZYX]\UyY;WPUNrqponm0/KDCBf@ED=<`:9>7[;4381Uvu32+*N(n&J*#i!&}$#"y?`_{zsxwp6543210/mfed*Kaf_^$Ea`_^WVUyYXQ9OTMq4JOHlFEDIHG@d'&<A@?>7[ZYXWx65.3,+O/on&Jk)"!~}CBA@?>|{z\xwpun4rkj0QPle+ihgfe^F\[!_AWVzZ<XWPtTS54JINGk.DIBA@dcba`_^]=6Z49816/St210/.'K+*#i!&}$#"y?`|{ts9wvXtsrkj0/.-,dLhgfe^$bDZYX]\UyY;WPUNrqpon10FKDIBf@ED=<`:9>7[;4381U/S-2+0/.-,%I)ih~}${z@?>=<;:987oWm3qSing-Ndiha`&^F\a`_X]Vz=YRQVOsSR4o2NMFEDCgfedcC%$@?>=6Z4z2165.R210/(LKJIH"!~}|Bz!x}|{zs9wYutsrk1oQgled*KJfe^$bD`YX]VzT<XQVUNrqponmlkjiIH*)?D=<`@">=65Yzy70/.R2+0/.'&%I)(h&}$#z@~}v{zs987654rTpong-kjihgfe^F\[!_AWVzyxwvu8TMLKPINGkEDCBfFE'C<;:^!=<54381U/S-2+0/.-,%I)ih~}${z@?>=<;:xwvoWm3qSing-Ndiha`&^F\a`_X]Vz=YRQVOsSR4o2NMFEDCgfedcb%$@?>=6Z4z2165.R210/(LKJIHG'&}e#"y?wv{zs9wYutsrk1oQgled*KJfe^$bD`YX]VzT<XQVUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I)(h&}$#z@~}v{zs987654rTpong-kjihgfe^F\[!_AWVzyxwvu8TMLKPINGkEDCBfFE'C<;:^!=<54381U/S-2+0/.-,%I)ih~}${z@?>=<;:xwvoWm3qSing-Ndiha`&^F\a`_X]VzT<XQVUNrR4o2NMFEDCgfedcb%$@?>=6Z4z2165.R210/(LKJIHG'&}e#"y?wv{zs9wYutsrk1oQgled*KJfe^$bD`YX]VzT<XQVUNrqponml/EDCBAFE>b<A:9>76Z4z870/.R2+0/.'&%I#"'&%|BAya}|{ts9wvXtsrkj0/.-,dLhgfe^$bDZYX]\UyY;WPUNrqponmlkjiIHG@(D=a;_9>7654X81w/.-,+O/(L,+k#"!E}e#"y?`_{zsxwp65432pong-kjcb(I_%cbD`YX]Vzg

온라인으로 사용해보십시오!

브레인 퍽, 77 (75) (73) + = 32,894 32,967 32,969 32,971 바이트

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

온라인으로 사용해보십시오!

출력이 가장 간단합니다

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

설명:

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

Stax , 점수 4751 4783 (812 바이트 + 3971)

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실행 및 디버깅

나는 최적의 출판 프로그램으로 시작했다 .
나는 최대 2 개의 셀을 사용하는 가장 짧은 프로그램으로 제한하기 위해 정규식을 사용했습니다. 그런 다음 후행 <이나 >문자를 자릅니다 . 나는 이것이 프로그램 종료시 외부의 0이 아닌 셀이 없도록 보장하는 보수적 인 방법이라고 생각합니다. 그런 다음 고정 kolmogorov 유형 출력을위한 stax 프로그램을 생성하기 위해 작성한 실험적인 stax 프로그램을 통해 실행했습니다 .
이 프로그램은 문자열 대체를 반복적으로 적용하여 작동합니다. 각 단계에서 가장 자주 발생하는> 1 길이의 하위 문자열을 찾아서 사용하지 않는 문자로 바꿉니다.

숯 , 707 698410 + 3627 = 4334 4325 4037 바이트

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-

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ＵＴ

스페이스 패딩을 끕니다.

≔”...”θ

대한 JonathanAllen의 답변 @ 구성된 큰 압축 된 문자열을 할당합니다 -128.. -15하지만 함께 +하고 -전치 표지판입니다.

Ｇ↘←¹⁴+

+14에서 14의 삼각형을 그리면 1에서 14까지의 정확한 결과가 생성됩니다. 압축 된 문자열의 후행 리턴은 다음 출력을 다음 라인으로 이동하지만 커서는 하단 모서리에 남아 있습니다.

⮌⪪θ⸿

리턴 문자열에서 큰 문자열을 분할하고 각 하위 문자열을 역순으로 인쇄하여 15에서 128 사이의 결과를 생성하십시오.

↑

128의 결과가 무시 된 128의 결과로 겹쳐 쓰여지도록 줄을 위로 이동하십시오.

Ｆθ§⁺+-ι⌕-+ι

문자열 트랜스 포즈를 반복 +하고 -다시 128-241 의 올바른 결과를 생성하도록 반복합니다.

Ｇ→↙¹⁴-

-242에서 255까지의 올바른 결과를 생성하는 14면 의 s의 삼각형을 그립니다.

젤리 , 1224 + 3716 = 4940 바이트

⁾+-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

전체 프로그램.

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어떻게?

후행 테이프 이동이 제거 된 0이 아닌 하나만 남기는 esolang 의 현재 최고 압축은 거의 전적으로 압축 됩니다. BF 프로그램의 하위 집합을 평가하여이 순진한 프로그램을 능가하는 가장 짧은 솔루션을 종료 할 수있는 방법이있을 것입니다. 더 스마트 한 패턴 기반 또는 인수 분해 구축 프로그램으로이를 극복 할 수있는 방법도있을 수 있습니다.

⁾+-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

SuperMarioLang , 231 + 32894 바이트

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

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출력이 brainfuck의 가장 기본이기 때문에이 확실하게 골프를 칠 수는 있지만,이 답변을 작성하는 데 하루 종일 걸렸습니다 (내 세 자녀는 시간을 거의 남겨 두지 않습니다). 이.

파이썬 2, 70 + 8428 = 8498

A__ 덕분에 -2 바이트!
Jonathan Allan에게 -20 바이트 감사합니다! 16에서 9로 전환
하여 두 번째 셀에 숫자를 넣어
-229 바이트 -1000ish 바이트

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

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산출

루비 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}

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각 값을 다른 위치의 초기 자릿수 0 이 포함되지 않은 가장 작은 기준으로 변환 한 다음 해당 기준에서 변환합니다. 127보다 큰 값은 그 역수로 계산됩니다.

비포장, 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}

랩핑되지 않은 셀에서 위와 동일한 방법을 사용합니다.

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자바 스크립트 (Node.js) , 691 + 3627 = 4318

@Neil의 Charcoal answer 와 동일한 접근 방식을 사용 하므로 @JonathanAllan의 Jelly answer 도 기반으로 합니다.

_=>(a=require('zlib').inflateRawSync(Buffer('bZXRsQMhCEX/0wpDB4yNOPbfxoMLKLAvs8lyQBEBDf2I9MHXf+I3X/f9hCJVsckdBgH3gih8YHT80Kpj01zQ7dUZb9osZ0Fzfvra+qiCl6GaqTB9vTdvNn2TjiaVjvBQKG5mHS/mMBGrywc1Vl2eIzqbwFPReVjJo1+I3mLfbAOYY201Sm491vaxDM+Vh9WSaJtgfBDYUHQeVrKkqzMEaYMrWpgbMS7MtIlIry3DgztWahbqNPElykKItD4e1hgN8qyrHA1CJ/dC5Imz3WS5PeDKDdGKDvS8WqBq+reSd4lH2Gtjcr9W95ydm/8kcaaRPsNpMiHS1kjW8qWAOsFmWEdHeiw/kenG1zdqzR0bVXhy0VKVPQVesFNloU4TI5nsp4qzunLytsCheK4B2Gz0UeXobM+M3Q2NcXhx+GlF389GH53eWr31utUf90W02MVu89LZs45IhP/Qc6jx9dYoCs8W1d4vKHHO8wJ9l8y4Vd614g788IzrViKArok/CmguX/KLqSl+c7h3STZcxUYVnvwH','base64'))+'')+`
--[>-<--]>-
`+[...a.split`
`].reverse().map(s=>s.replace(/[-+]/g,c=>c>','?'+':'-')).join`
`

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비공식 술통 16 + 32895 = 32911 바이트

골프 언어를위한 기본 솔루션. 이것은 내가 생각할 수있는 가장 간단한 것입니다.

ÿï((:|\+$;)_\
')

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루비 23 + 32895 = 32918 바이트

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

기준으로. 이것은 내가 생각할 수있는 가장 간단한 솔루션입니다.

레티 나 0.8.2 , 28 + 16640 = 16668 바이트

127$*+128$*-
\+
¶$`+
-
¶$'-

온라인으로 사용해보십시오! 에 대한 출력을 포함합니다 0. +최대 127과 -최대 255 를 사용하여 출력 합니다.

스칼라 , 95 + 16639 = 16734 바이트

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

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분명히 이길 수없는 간단한 대답. -연산자 (바이트 감소)가 255로 줄 바꿈 된다는 사실 만 사용합니다 .

05AB1E , 점수 : 4848 (1219 바이트 소스 코드 + 3629 바이트 출력)

'+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»

출력은 @Neil 의 Charcoal 답변 포트 이므로, 그에게도 투표해야합니다!

온라인으로 사용해보십시오.

설명:

'+                      '# 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)

내 (섹션이 05AB1E 팁을 참조하십시오 어떻게 큰 정수를 압축하는 ) 이유를 이해하는 방법 •тôm...ò´Θ•입니다 183...875.