O verdadeiro tesouro foram os números que fizemos ao longo do caminho

23

Sua tarefa é escrever um programa, função ou snippet (sim, snippets são permitidos) que simplesmente produz um número inteiro. No entanto , você deve poder separar seu envio em prefixos que também produzam números inteiros distintos. Você não pode usar nenhum bytes que tenha aparecido em prefixos anteriores. Por exemplo, podemos ter os prefixos:

1         # 1    (Now we can't use 1)
1-6       # -5   (Now we can't use - or 6)
1-6/3     # -1   (Now we can't use / or 3)
1-6/3+0xA # 9    Final submission

Regras

  • Seu objetivo é criar para tentar criar os números inteiros mais exclusivos, mantendo-os próximos de zero.
    • O sistema de pontuação é que ((number of unique integers)**3)/(sum of absolute values), quanto maior a sua pontuação, melhor. O exemplo acima pontua .(43)/(1+|-5|+|-1|+9)=64/16=4
  • Deve haver pelo menos dois números inteiros únicos (sem dividir por zero!)
  • Formate sua resposta semelhante a:
# Language, \$(4^{3})/16 = 4\$

   1-6/3+0xA (the full program)
  • Mencione se o seu envio é composto por trechos que avaliam um valor, funções ou programas completos.
  • Liste cada um dos prefixos e (opcionalmente) uma explicação de como eles funcionam.
Brincadeira
fonte
2
A linha final de submissão parece ter trocado o 3 por um 2 (não editei para corrigir isso no caso de ter entendido
errado
4
Parece-me que você pode obter uma pontuação quase ideal no Python escrevendo, print 123-len(open(__file__).read())seguido #de um byte não usado até o momento (exceto alguns caracteres de controle), com prefixos cortando esses bytes. Ajuste 123 para centralizar o intervalo de saídas em torno de 0. Para uma pontuação melhor, a parte sem comentários do programa pode ser reescrita usando apenas os 7 caracteres distintos que são universais para o Python. Isso tudo é válido?
xnor 11/02
1
@JoKing Eu incluo negativos, quando a diferença é negativa.
xnor 11/02
1
@JoKing Esse método também pode ser usado em idiomas em que uma string pode ser escrita sem uma citação de fechamento.
xnor 11/02
2
Não acredito que poucas pessoas perceberam isso e votaram positivamente no comentário anterior. : A PI percebeu que o @trichoplax também detectou o erro de digitação.
Peter Cordes

Respostas:

19

TI-BÁSICO (TI-84 + CE), 2383/14161=952,0

118-length("+×/ABCDEFGHIJK...

Os 256 bytes no TI-BASIC são divididos da seguinte maneira:

  • 240 bytes são tokens de um byte que podem ser citados
  • 11 são o início de tokens de dois bytes
  • 2 não são utilizados
  • Os restantes 3 são os ", caracteres, e de nova linha a qual as cadeias ruptura.

A cadeia contém todos os 237 bytes permitidos na primeira categoria que ainda não está dentro 118-length(.

O programa calcula 238 números inteiros distintos, variando de 118 a -119. A pontuação é, portanto,

2383n=-119118|n|=952

lirtosiast
fonte
1
Não seria a string tem que acabar com "o que é um byte que já foi usado Ou por que não estou um pouco confuso com isso também em outras respostas este parece acontecer então provavelmente eu perdi alguma coisa??.
findusl
2
@findusl O TI-BASIC tem a tendência de preencher automaticamente parênteses e cadeias de caracteres para você. para que você pudesse escrever mean({1,2,3, o que seria concluído em mean({1,2,3}). seqüências de caracteres também são concluídas: Disp "Hi!conclui para Disp "Hi!". Portanto, a citação final é desnecessária
Conor O'Brien
13

vim, 14³ / 53 = 51,77

i10<esc>X<c-x>r2hx<c-a>Pa<del>4<c-c>d^s5<c-o>I-<c-o>$<c-o>C6<c-h>7

Experimente online! (todos os prefixos)

Demolir:

i1                  1
0                   10
<esc>X              0
<c-x>               -1  <c-x> decrements
r2                  -2
hx                  2   hx is X, but we've already used that
<c-a>               3   <c-a> increments
P                   -3  paste the - from two steps ago
a<del>4             -4
<c-c>d^             4   <c-c> is an <esc> alternative
s5                  5
<c-o>I-<c-o>$<c-o>  -5  <c-o> allows a single normal mode input
C6                  -6  trailing <c-o> from previous line used here
<c-h>7              -7  <c-h> = backspace
Maçaneta da porta
fonte
você não está reutilizando o <byte? Ou esse é um espaço reservado para um byte diferente?
findusl
1
@findusl no - <esc>representa a tecla Escape e <c-char>representa ctrl + char
Maçaneta da porta
11

Linguagem , 256³ / 16384 = 1024

O código fonte da solução pode ser descrito da seguinte maneira:

183662808731984092                                         0x00 bytes
4775233027031586288                                        0x01 bytes
27                                                         0x02 bytes
81                                                         0x03 bytes
128931291729852829668                                      0x04 bytes
108                                                        0x05 bytes
648                                                        0x06 bytes
81                                                         0x07 bytes
2079                                                       0x08 bytes
27                                                         0x09 bytes
3481144876706026401009                                     0x0A bytes
108                                                        0x0B bytes
2808                                                       0x0C bytes
27                                                         0x0D bytes
17496                                                      0x0E bytes
81                                                         0x0F bytes
2079                                                       0x10 bytes
27                                                         0x11 bytes
56106                                                      0x12 bytes
756                                                        0x13 bytes
93990911671062712827216                                    0x14 bytes
108                                                        0x15 bytes
2808                                                       0x16 bytes
27                                                         0x17 bytes
75789                                                      0x18 bytes
756                                                        0x19 bytes
474552                                                     0x1A bytes
27                                                         0x1B bytes
81                                                         0x1C bytes
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756                                                        0x1E bytes
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2537754615118693246334805                                  0x21 bytes
108                                                        0x22 bytes
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27                                                         0x24 bytes
75789                                                      0x25 bytes
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2046276                                                    0x27 bytes
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108                                                        0x2A bytes
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66829507740256534347823839                                 0x31 bytes
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12755246066760623071771160639904894849                     0x90 bytes
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6513309293583968799362636679031305893812560835614495       0xF7 bytes
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175859350926767157582791190333845259132939142561591365     0xFA bytes
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4748202475022713254735362139013821996589386504264297360    0xFC bytes
29655101330505                                             0xFD bytes
128201466825613257877854777753373193907913435615136028720  0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes

Cada execução de bytes idênticos combinada com as execuções anteriores constitui um prefixo, e o programa completo é todas as seqüências de caracteres descritas acima concatenadas.

Essa é uma pontuação ótima. (Os programas imprimem todos os números de -127 a 128, inclusive.)

Para criar essa resposta, primeiro copiei a resposta Java aqui para gerar programas de foda cerebral para imprimir uma string. Eu o executei com esta mainfunção:

public static void main(String[] args) {
    for (int i = -127; i <= 128; ++i) generate(""+i);
}

Em seguida, executei esse script Ruby para gerar o texto legível por humanos mostrado aqui:

#!/usr/bin/ruby
nums = File.open('f').map do |line|
    line.chomp.sub(/^\+/, '>+').chars.map{|ch|
        '+-><.,[]'.index(ch).to_s(3).rjust(3,?0)
    }.join.to_i 3
end
puts ([0]+nums.sort).each_cons(2).map.with_index{|x,i|"#{x[1]-x[0]} 0x#{i.to_s(16).rjust(2,?0).upcase} bytes"}*"\n"

O maior programa tem 3594571896764310192036774345469579167648804468538578264427 bytes de comprimento (e é impresso -90).

Maçaneta da porta
fonte
1
Suportes para a velocidade e habilidade prática de resolução de problemas!
lirtosiast
8

(1113)/2530=540.5656126482213

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)[!+[]+!+[]+[+[]]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]])().$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}

(Agradeça a este interessante utilitário pela conversão.) O exemplo acima é efetivamente avaliado no seguinte código:

(new Proxy({},{get:(a,p,b,n=p.length+3)=>n%2?(n-1)/2:-n/2})).$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}

... que cria um proxy que converte o comprimento de um nome de propriedade em um número. O primeiro prefixo termina um caractere após o .e usa apenas 6 caracteres exclusivos, excluindo o .$. As próximas 63 fatias anexam 63 outros caracteres de byte válidos nos nomes de propriedades do objeto JS. As próximas 44 fatias acrescentam caracteres válidos de dois bytes compostos por bytes não utilizados. As fatias finais acrescentar &'', |""**"", e ,~{}. A primeira fatia retornos -2, o segundo 2, em seguida -3, em seguida 3, -4, 4, -5, etc. até os últimos três fatias que são 0, 1e -1.

Yair Rand
fonte
4

43/4=16

Para este, usamos trechos de testes cerebrais, por isso estamos preocupados com o valor e não com o que eles fazem.

Programa completo:

<>(())[[][][]]{}

Prefixos e saída

 0 : <>
 1 : <>(())
-2 : <>(())[[][][]]
-1 : <>(())[[][][]]{}

Experimente online!

Explicação

Esta é provavelmente a melhor resposta que um cérebro pode ter, já que usamos todo o personagem que se preocupa com cérebro, e não podemos agrupar os números mais perto de zero.

O primeiro trecho é <>, <>altera a pilha, mas avalia como zero, então começamos com zero. A partir daqui, adicionamos (())isso faz um. Poderíamos ter acabado de fazê-lo (), porque esse também é um, mas se pressionarmos um agora poderemos lembrá-lo mais tarde com {}e []. Essa é praticamente a única maneira {}e []pode criar valor, por isso fazemos isso. Como há uma coisa na pilha que []conta como 1, subtraímos 3 do total, [[][][]]levando-nos a -2. Por fim, usamos o {}pop que pressionamos anteriormente, aumentando para -1.

Assistente de Trigo
fonte
4

(113)/32.=41.59375

Após duas horas de tentativa e erro, finalmente obtive todos os números inteiros de -4até +5(e 7).

0^cmp('','_')&3+1/~9*2-[-[-[[]==[]][[]==[[]]]][[]==[[]]]-[-[[]==[]][[]==[[]]]][[]==[[]]]-[-[[]==[]][[]==[[]]]][[]==[[]]]-[-[[]==[]][[]==[[]]]][[]==[[]]]-[-[[]==[]][[]==[[]]]][[]==[[]]]][[]==[[]]]|6<<7<<7>>8>>8and 5

Este é um trecho com os seguintes trechos de prefixo:

# -1
cmp(cmp,())
# +3 = (-1) & 3
cmp(cmp,())&3
# +7 = 3 + 4
cmp(cmp,())&3+4
# +2 = 3 + (-1) = 3 + floor(4/-10)
cmp(cmp,())&3+4/~9
# +1 = 3 + (-2) = 3 + 2*floor(4/-10)
cmp(cmp,())&3+4/~9*2
# -4 = 1 - 5
cmp(cmp,())&3+4/~9*2-1-1-1-1-1
# -2 = (-4) | 6
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6
# -3 = (-4) | 1 = (-4) | floor(6/4) = (-4) | (6>>2)
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8
#  0 = -3 and 0
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 0
# +5 = -3 and 05
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05
# +4 = 5 ^ 1
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05^True

Experimente online!

ovs
fonte
1
Você pode alterar 2*2para 22&7(-3 e 6 em vez de -3 e 4) para corrigir essa repetição de *.
Kevin Cruijssen 12/02
@KevinCruijssen obrigado pela sugestão.
ovs 12/02
Você pode liberar 'e _usar em cmp(cmp,())vez de cmp('','_')Se isso for útil.
Wheat Wizard
Você também pode tornar a seção da lista um pouco mais curta e muito mais fácil de entender --[-1-1-1-1-1], mas isso não melhora sua pontuação.
Wheat Wizard
3

Lote, 176³ / 7744 = 704

@cmd/cset/a%~z0-110
: !"#$&'()*+,.23456789;<=>?ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`bfghijklnopqruvwxy{|} ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ

O primeiro prefixo inclui dois pontos e gera o número -88. Os prefixos subsequentes geram números ascendentes e o programa final gera 87.

Neil
fonte
3

(2493)/15500=996,0

-C\|l"\x01\x02\x03\x04\x05\x06\x07\x08\t\n\x0b\x0c\r\x0e\x0f\x10\x11\x12\x13\x14\x15\x16\x17\x18\x19\x1a\x1b\x1c\x1d\x1e\x1f !#$%&\'()*+,./0123456789:;<=>?@ABDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijkmnopqrstuvwxyz{}~\x80\x81\x82\x83\x84\x85\x86\x87\x88\x89\x8a\x8b\x8c\x8d\x8e\x8f\x90\x91\x92\x93\x94\x95\x96\x97\x98\x99\x9a\x9b\x9c\x9d\x9e\x9f\xa0¡¢£¤¥¦§¨©ª«¬\xad®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ

Todos os caracteres de escape são bytes únicos - apresentei o programa neste formulário para facilitar a cópia e a colagem. Este link é para o programa completo sem caracteres de escape; como pode ser visto, existem muitos espaços reservados para os caracteres não imprimíveis usados.

A abordagem usada é semelhante à resposta TI-BASIC do lirtosiast , embora eu tenha escrito minha resposta antes de navegar pelas outras entradas. O cabo funciona da seguinte maneira:

-C\|l"...   
  \|        "|"
 C          Character code of the above, i.e. 124
-           Subtract from the above...
    l"...   ... the length of the trailing string - the closing quote is inferred

A sequência é composta por todos os caracteres de byte único que não sejam 0 (NULL), 127 (backspace) e os caracteres padrão ( -C\|l"). Dessa forma, cada caractere da cadeia de caracteres pode ser removido um de cada vez para produzir cadeias de comprimento de 248 a 0. Esses programas produzem resultados no intervalo [-124 - 124], com a soma dos valores absolutos de 15500.

O código foi gerado usando este programa :

+J"-C\\|l\""sCM-U256+CMJ,Z127   
 J"-C\\|l\""                    Assign the boilerplate code to J
                     CMJ        Get the character code of each character in J
                    +   ,Z127   Append 0 and 127
               -U256            Get values in range [0-256) which are not in the above
             CM                 Get each character with code points in the above
            s                   Concatenate into string
+J                              Prepend boilerplate

E a soma dos números gerados de cada um dos programas truncados foi encontrada com este programa :

sm.a.vd>._+J"-C\\|l\""sCM-U256+CMJ,Z127 5   
          +J"-C\\|l\""sCM-U256+CMJ,Z127     The full program string, as above
        ._                                  Get the prefixes of the above
       >                                5   Discard the first 5 (those prefixes which don't include all the boilerplate code)
 m                                          Map each of the above to:
    .vd                                       Evaluate as Pyth code
  .a                                          Take the absolute value of the above
s                                           Take the sum
Sok
fonte
3

Gelatina , 984

LNḂ¡:2µ}“¢¢¢£¤¥¦©¬®½¿€ÆÇÐÑ×ØŒÞßæçðıȷñ÷øœþ !"#$%&'()*+,-./013456789;<=>?@ABCDEFGHIJKMOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|~¶°¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ƁƇƊƑƓƘⱮƝƤƬƲȤɓƈɗƒɠɦƙɱɲƥʠɼʂƭʋȥẠḄḌẸḤỊḲḶṂṆỌṚṢṬỤṾẈỴẒȦĊḊĖḞĠḢİĿṀṄȮṖṘṠṪẆẊẎŻạḅḍẹḥịḳḷṃṇọṛṣṭ§Äẉỵẓȧḃċḋėḟġḣŀṁṅȯṗṙṡṫẇẋẏż”‘

Experimente online!

Você pode ver os prefixos (separados por uma nova linha dupla) aqui .

Os números impressos são, em ordem:

0 1 -1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 -9 9 -10 10 -11 11 -12 12 -13 13 -14 14 -15 15 -16 16 -17 17 -18 18 -19 19 -20 20 -21 21 -22 22 -23 23 -24 24 -25 25 -26 26 -27 27 -28 28 -29 29 -30 30 -31 31 -32 32 -33 33 -34 34 -35 35 -36 36 -37 37 -38 38 -39 39 -40 40 -41 41 -42 42 -43 43 -44 44 -45 45 -46 46 -47 47 -48 48 -49 49 -50 50 -51 51 -52 52 -53 53 -54 54 -55 55 -56 56 -57 57 -58 58 -59 59 -60 60 -61 61 -62 62 -63 63 -64 64 -65 65 -66 66 -67 67 -68 68 -69 69 -70 70 -71 71 -72 72 -73 73 -74 74 -75 75 -76 76 -77 77 -78 78 -79 79 -80 80 -81 81 -82 82 -83 83 -84 84 -85 85 -86 86 -87 87 -88 88 -89 89 -90 90 -91 91 -92 92 -93 93 -94 94 -95 95 -96 96 -97 97 -98 98 -99 99 -100 100 -101 101 -102 102 -103 103 -104 104 -105 105 -106 106 -107 107 -108 108 -109 109 -110 110 -111 111 -112 112 -113 113 -114 114 -115 115 -116 116 -117 117 -118 118 -119 119 -120 120 -121 121 -122 122 123
Erik, o Outgolfer
fonte
2

113/53=25.1132075472

015-8*2+6/3&9%7if[]else 4^ord('\b')

Prefixos:

0                                   # 0

01                                  # 1

015                                 # 13 (octal)

015-8                               # 5

015-8*2                             # -3

015-8*2+6                           # 3

015-8*2+6/3                         # -1

015-8*2+6/3&9                       # 9 (-1 bitwise-AND 9 = 9)

015-8*2+6/3&9%7                     # 2

015-8*2+6/3&9%7if[]else 4           # 4

015-8*2+6/3&9%7if[]else 4^ord('\b') # 12 (4 bitwise-XOR 8 -- OR would also work)

Não é o ideal, mas o melhor que eu pessoalmente poderia encontrar com essa estrutura específica.

Aidan F. Pierce
fonte
2

63/9=24

False-False**0&1+[[]<[[]]][[]<[]]^~3>>2

Aqui estão os prefixos e o que eles produzem

 0 : False-False
-1 : False-False**0
 1 : False-False**0&1
 2 : False-False**0&1+[[]<[[]]][[]<[]]
-2 : False-False**0&1+[[]<[[]]][[]<[]]^~3
-3 : False-False**0&1+[[]<[[]]][[]<[]]^~3>>2
Assistente de Trigo
fonte
2

3

import StdEnv,Data.GenDefault,Gast.GenLibTest
derive genShow G
derive gDefault G
n = sum[(bitnot o bitnot) one \\ n <-: s]-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one
s = (hd o show o d) undef
d :: G -> G
d g = gDefault{|*|} g
:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`

Experimente online!

Define o nque é produzido -20, -19, ..., 18, 19dependendo do comprimento do tipo G.

O menor prefixo de Gaparência :: G = G, precedido pelas importações e definições variadas.
O maior prefixo de Gaparência é similarmente precedido.:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`

Furioso
fonte
Acho que estou perdendo algo neste desafio, mas ... esta resposta (e algumas outras) está reutilizando "bytes que apareceram em prefixos anteriores", não é?
Blackhole
4
@ Blackhole Como eu (e presumo que outros) entendam, você pode usar bytes várias vezes no mesmo prefixo, mas não nos seguintes.
Οurous
2

2453/15006=980.0163268026123

_Ê-122}$($"	
 !#%&')*+,./03456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^`abcdefghijklmnopqrstuvwxyz|~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ

Experimente online!

Como as outras entradas, usa o comprimento de uma sequência. Caracteres no snippet inicial, bem como \r(as áreas de texto não se saem bem com isso) e {(caractere de interpolação de string) são excluídos da string.

O primeiro programa é que _Ê-122}$($"cada programa tem um caractere a mais, com exceção de quando você adiciona \\.

Somente ASCII
fonte
Não tenho certeza absoluta de que isso é válido, não verifiquei se a sequência não possui nenhum caractere do snippet
somente ASCII
Eu ia sugerir a substituição 122com #z, mas que não iria melhorar a sua pontuação.
Shaggy
1

(63)/12=18

0x1-2+3|True*~4

Este é um trecho.

Os fragmentos de prefixo individuais são:

0          # gives 0

0x1        # gives 1 (hex syntax)

0x1-2      # gives -1 (1 minus 2)

0x1-2+3    # gives 2 (-1 plus 3)

0x1-2+3|True # gives 3 (True == 1; 2 bitwise-OR 1 == 3)

0x1-2+3|True*~4 # gives -5 (~4 == -5; True*~4 == -5; 2|-5 == -5)

Os operadores bit a bit do Python tratam números inteiros negativos como complemento de dois com 1s ilimitados à direita, -5sendo tratados como ...11111011, o que quando ORed com ...000010fornece ...11111011inalterado.

pizzapants184
fonte
1

(63)/12=18

=0              0
=01             1
=01*2           2
=01*2-5         -3
=01*2-5+8       5
=01*2-5+8/4     -1

Prefix O desafio não combina bem com a estrutura da fórmula do Excel

Wernisch
fonte
1

Enchantments Rúnicas , 117 3 /3422 = 468.034

>"ul'B-$;	 !#%&()*+,.:0123456789@<=?ACDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijkmnopqrstvwxyz{|}~€

Experimente online!

E o prefixo .

Imprime [-58,58] inclusive.

Usa uma tática semelhante a outras entradas, usando uma sequência progressivamente mais longa, obtendo seu comprimento, subtraindo B(valor decimal de 66) e imprimindo o resultado. Usa um caractere não imprimível (mostrado no bloco de código, porque aparentemente é o que é &#128;avaliado em HTML) para utilizar o valor de byte 0x80(e consumir um que não seria utilizável 0x00) e obter mais um valor. Somente valores de bytes de até 127 que não podem ser representados de maneira alguma são 10e 13, porque causariam o programa com duas linhas de altura e não ser lido corretamente pelo ponteiro de instrução.

Draco18s
fonte
1

Carvão vegetal , 95³ / 2256 = 380,04211 96³ / 2304 = 384

I⁻⁴⁸L´´ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~

O primeiro prefixo são os primeiros 7 bytes e gera o número 47. Experimente online! Os prefixos subsequentes geram números decrescentes e o programa final é gerado -48. Experimente online! Editar: aumento de pontuação em 3,95789 graças a @ ASCII-only.

Neil
fonte
Eu acho que você deve conseguir escapar de um caractere não ASCII no início ou no final da string? (usando o caractere de escape, aquele que se parece com um sotaque grave)
ASCII-only