编写一个程序，对于从1到255的数字，将打印出BF代码，该代码将在数组的某个字节中生成给定的数字，并加上换行符。

例如，输出的前四行可能（并且很可能会）是：

+
++
+++
++++

获胜者将最小：（source code + output以字节为单位）。

澄清与修订：

• BF程序使用包装单元。

• 输出BF程序必须以唯一的非零单元格为包含数字的单元格终止。

• 程序必须以升序输出。

• 输出为0的程序是可选的。

• 不允许使用负数据指针。<在第一个指针将什么都不做。（如果更合适，请发表评论）

Perl 6，224 + 3964 = 5834 4188字节

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}

在线尝试！（可能会超时。将更^27-13改为^25-12可以稍微加快速度，但会增加一些输出）

以形式输出最短的代码*>[*>*<]>*，其中每个*都是一定数量的+s或-s。还有一些额外的调整，例如在不需要时删除循环以及尾随>s。

据我所知，此特定格式的输出最多。

说明：

([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字节+ 7166输出

不太有创意，不是吗？我要去打高尔夫球这个坏男孩。

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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

在线尝试！

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

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

在线尝试！

输出是最简单的

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

说明：

++++++[->+++++++<]>+ 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，得分4751 4783（812 字节 + 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

运行并调试

我从最佳的已发布程序开始。
我使用了一些regex-fu，将其限制为最多使用2个单元的最短程序。然后我修剪了所有结尾<或>字符。我认为这是确保程序终止时没有多余的非零单元的一种保守方法。然后，我通过编写的实验性stax程序运行该程序，以为固定的kolmogorov型输出生成stax程序。
该程序通过重复应用字符串替换来工作。在每个步骤中，它都会查找> 1个长度最频繁出现的子字符串，并将其替换为未使用的字符。

木炭，707 698 410 + 3627 = 4334 4325 4037个字节

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-

在线尝试！链接是详细版本的代码。说明：

ＵＴ

关闭空格填充。

≔”...”θ

分配一个大的压缩字符串，该字符串由@JonathanAllen对-128.. 的答案组成，-15但带有+和-换位。

Ｇ↘←¹⁴+

在+第14 边的s上绘制一个三角形，可生成1到14的正确结果。光标位于左下角，尽管压缩字符串上的尾随返回将使下一个输出移至下一行。

⮌⪪θ⸿

在返回字符上分割大字符串，并以相反的顺序打印每个子字符串，从而生成结果，范围从15到128。

↑

向上移动一行，以使128的结果被128的取反结果覆盖。

Ｆθ§⁺+-ι⌕-+ι

遍历字符串转置+并-再次返回，以便它们生成128到241的正确结果。

Ｇ→↙¹⁴-

画出-边14 的s的三角形，可生成242到255的正确结果。

果冻，1224 + 3716 = 4940字节

⁾+-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

完整程序。

在线尝试！

怎么样？

当前最好成绩的上几乎完全压缩esolangs其中仅单个非零离开，尾随除去胶带移动。可能有一种方法可以评估BF程序的子集，使它们可以终止并产生最短的解决方案，而该解决方案仅能胜过该天真的程序。也许还有一种方法可以通过基于模式或因数分解的更智能程序来克服。

⁾+-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字节

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

在线尝试！

这种输出肯定是最基本的，因为输出是进行脑力劳动的最基本条件，但是我花了整整一天的时间写出这个答案（我的三个孩子给我留下的时间很少），我为至少能够成功实现而感到自豪这个。

python 2，70 + 8428 = 8498

-2字节感谢A__！
-20字节感谢乔纳森·艾伦！
将数字放在第二个单元格中
-229字节-从16切换到9 -1000ish字节

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

在线尝试！

输出量

红宝石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}

在线尝试！

将每个值转换为最小的底数，该底数在任何其他位置不包含零的初始数字，然后从该底数转换。大于127的值将作为其倒数进行计算。

不包装，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}

对于非包装单元，使用与上述相同的方法。

在线尝试！

JavaScript（Node.js），691 + 3627 = 4318

使用与@Neil的木炭答案相同的方法，因此也基于@JonathanAllan的Jelly答案。

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

在线尝试！

非官方小桶 16 + 32895 = 32911个字节

高尔夫语言的基准解决方案。这是我能想到的最简单的方法。

ÿï((:|\+$;)_\
')

在线尝试！

Ruby 23 + 32895 = 32918字节

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

作为基线。这是我能想到的最简单的解决方案。

视网膜0.8.2，28 + 16640 = 16668字节

127$*+128$*-
\+
¶$`+
-
¶$'-

在线尝试！包括的输出0。仅使用+s最多127和-s最多255进行输出。

斯卡拉，95 + 16639 = 16734字节

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

在线尝试！

一个简单的答案显然不会成功。仅使用-运算符（减少字节）回绕到255 的事实。

05AB1E，得分：4848（1219 字节源代码+ 3629字节输出）

'+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»

输出是@Neil的木炭答案的端口，因此请确保也对他进行！

在线尝试。

说明：

'+                      '# 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)

请参阅此05AB1E技巧（如何压缩大整数）理解为什么•тôm...ò´Θ•是183...875。