Le vrai trésor était les chiffres que nous avons faits en cours de route

23

Votre tâche consiste à écrire un programme, une fonction ou un extrait (oui, les extraits sont autorisés) qui génère simplement un entier. Cependant , vous devez pouvoir séparer votre soumission en préfixes qui produisent également des entiers distincts. Vous ne pouvez pas utiliser d'octets apparus dans les préfixes précédents. Par exemple, nous pouvons avoir les préfixes:

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

Règles

  • Votre objectif est de créer pour essayer de créer les entiers les plus uniques, tout en les maintenant proches de zéro.
    • Le système de notation est ((number of unique integers)**3)/(sum of absolute values), où plus votre score est élevé, mieux c'est. L'exemple ci-dessus marque .(43)/(1+|5|+|1|+9)=64/16=4
  • Il doit y avoir au moins deux entiers uniques (pas de division par zéro!)
  • Veuillez formater votre réponse comme suit:
# Language, \$(4^{3})/16 = 4\$

   1-6/3+0xA (the full program)
  • Mentionnez si votre soumission est composée d'extraits qui évaluent une valeur, des fonctions ou des programmes complets.
  • Énumérez chacun des préfixes et (éventuellement) une explication de leur fonctionnement.
Jo King
la source
2
La ligne de soumission finale semble avoir échangé le 3 contre un 2 (je n'ai pas édité pour corriger cela au cas où je me serais mal compris)
trichoplax
4
Il me semble que vous pouvez obtenir un score presque optimal en Python en écrivant print 123-len(open(__file__).read())suivi d' #un de chaque octet non utilisé jusqu'à présent (à l'exception de certains caractères de contrôle), avec des préfixes coupant ces octets. Ajustez 123 à ce qui centre la plage de sorties autour de 0. Pour un meilleur score, la partie non-commentaire du programme peut être réécrite en utilisant seulement les 7 caractères distincts qui sont universels pour Python. Est-ce que tout est valable?
xnor
1
@JoKing J'inclus les négatifs, lorsque la différence est négative.
xnor
1
@JoKing Cette méthode peut également être effectuée dans les langues où une chaîne peut être écrite sans guillemet de fermeture.
xnor
2
Je ne peux pas croire que si peu de gens l'ont remarqué et ont voté pour le commentaire précédent. : PI vient de remarquer que @trichoplax a également repéré la faute de frappe.
Peter Cordes

Réponses:

19

TI-BASIC (TI-84 + CE), 2383/14161=952.0

118-length("+×/ABCDEFGHIJK...

Les 256 octets de TI-BASIC se décomposent ainsi:

  • 240 octets sont des jetons d'un octet qui peuvent être cités
  • 11 sont le début des jetons à deux octets
  • 2 sont inutilisés
  • Les 3 autres sont les caractères ", et les sauts de ligne qui cassent les chaînes.

La chaîne contient les 237 octets autorisés de la première catégorie qui ne figurent pas déjà dans 118-length(.

Le programme calcule 238 entiers distincts, allant de 118 à -119. Le score est donc de

2383n=119118|n|=952.

lirtosiast
la source
1
Ne serait - pas la chaîne doit se terminer par " ce qui est un octet qui a déjà été utilisé ou pourquoi pas je suis un peu confus par ce aussi dans d' autres réponses , cela semble se produire si probablement me manque quelque chose?.
findusl
2
@findusl TI-BASIC a tendance à compléter automatiquement les parenthèses et les chaînes pour vous. afin que vous puissiez écrire mean({1,2,3, ce qui se terminerait mean({1,2,3}). les chaînes sont également complétées: Disp "Hi!complète à Disp "Hi!". Par conséquent, la citation finale n'est pas nécessaire
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

Essayez-le en ligne! (tous les préfixes)

Panne:

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
Poignée de porte
la source
n'utilisez-vous pas <l'octet? Ou s'agit-il d'un espace réservé pour un octet différent?
findusl
1
@findusl no - <esc>représente la touche d'échappement et <c-char>représente ctrl + char
Poignée de porte
11

Lenguage , 256³ / 16384 = 1024

Le code source de la solution peut être décrit comme suit:

183662808731984092                                         0x00 bytes
4775233027031586288                                        0x01 bytes
27                                                         0x02 bytes
81                                                         0x03 bytes
128931291729852829668                                      0x04 bytes
108                                                        0x05 bytes
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27                                                         0x09 bytes
3481144876706026401009                                     0x0A bytes
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27                                                         0x0D bytes
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81                                                         0x0F bytes
2079                                                       0x10 bytes
27                                                         0x11 bytes
56106                                                      0x12 bytes
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93990911671062712827216                                    0x14 bytes
108                                                        0x15 bytes
2808                                                       0x16 bytes
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12755246066760623071771160639904894849                     0x90 bytes
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6513309293583968799362636679031305893812560835614495       0xF7 bytes
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1098337086315                                              0xF9 bytes
175859350926767157582791190333845259132939142561591365     0xFA bytes
1098337086315                                              0xFB bytes
4748202475022713254735362139013821996589386504264297360    0xFC bytes
29655101330505                                             0xFD bytes
128201466825613257877854777753373193907913435615136028720  0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes

Chaque exécution d'octets identiques combinée avec les exécutions précédentes constitue un préfixe, et le programme complet est toutes les chaînes décrites ci-dessus concaténées.

Il s'agit d'un score optimal. (Les programmes impriment tous les nombres de -127 à 128 inclus.)

Pour créer cette réponse, j'ai d'abord copié la réponse Java ici pour générer des programmes brainfuck pour imprimer une chaîne. Je l'ai exécuté avec cette mainfonction:

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

Ensuite, j'ai exécuté ce script Ruby pour générer le texte lisible par l'homme montré ici:

#!/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"

Le plus grand programme a une longueur de 3594571896764310192036774345469579167648804468538578264427 octets (et des impressions -90).

Poignée de porte
la source
1
Des accessoires pour la vitesse et les compétences pratiques en résolution de problèmes!
lirtosiast
8

(1113)/2530=540.5656126482213

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

(Merci à cet utilitaire intéressant pour la conversion.) Ce qui précède s'évalue efficacement dans le code suivant:

(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&''|""**"",~{}

... qui crée un proxy qui convertit la longueur d'un nom de propriété en nombre. Le premier préfixe termine un caractère après le .et n'utilise que 6 caractères uniques, à l'exception du .$. Les 63 tranches suivantes ajoutent 63 autres caractères à un octet qui sont valides dans les noms de propriété d'objet JS. Les 44 tranches suivantes ajoutent des caractères valides de deux octets constitués d'octets inutilisés. Les tranches finales append &'', |""**""et ,~{}. Les premiers retours de tranche -2, la deuxième 2, puis -3, puis 3, -4, 4, -5, etc. jusqu'à trois tranches qui sont 0, 1et -1.

Yair Rand
la source
4

43/4=16

Pour celui-ci, nous utilisons des extraits de brain-flak, donc nous nous soucions de la valeur et non de ce qu'ils font.

Programme complet:

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

Préfixes et sortie

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

Essayez-le en ligne!

Explication

C'est probablement la meilleure réponse d'une brain-flak depuis que nous avons utilisé tout le personnage auquel Brain-flak se soucie, et nous ne pouvons pas regrouper les nombres plus près de zéro.

Le premier extrait est <>, <>modifie la pile mais évalue sinon à zéro, donc nous commençons par zéro. A partir de là, nous ajoutons (())cela en fait un. Nous aurions pu le faire (), car c'est aussi un, mais si nous en poussons un maintenant, nous pourrons le rappeler plus tard avec {}et []. C'est à peu près le seul moyen {}et []peut créer de la valeur, alors nous le faisons. Puisqu'il y a une chose sur la pile qui []compte comme un 1, nous soustrayons donc 3 du total [[][][]]en nous amenant à -2. Enfin, nous utilisons {}pour faire apparaître celui que nous avons poussé plus tôt, ce qui nous amène à -1.

Assistant de blé
la source
4

(113)/32=41.59375

Après deux heures d'essais et d'erreurs, j'ai finalement obtenu tous les entiers de -4à +5(et 7).

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

Il s'agit d'un extrait de code avec les extraits de préfixe suivants:

# -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

Essayez-le en ligne!

ovs
la source
1
Vous pouvez changer le 2*2en 22&7(-3 & 6 au lieu de -3 & 4) pour corriger cette répétition de *.
Kevin Cruijssen
@KevinCruijssen merci pour la suggestion.
OVS
Vous pouvez libérer 'et _en utilisant cmp(cmp,())au lieu de cmp('','_')Si cela est utile.
Wheat Wizard
Vous pouvez également rendre la section de liste un peu plus courte et beaucoup plus facile à comprendre en faisant --[-1-1-1-1-1], mais cela n'améliore pas vraiment votre score.
Wheat Wizard
3

Lot, 176³ / 7744 = 704

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

Le premier préfixe inclut les deux points et affiche le nombre -88. Les préfixes suivants produisent des nombres ascendants et les sorties finales du programme 87.

Neil
la source
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

Tous les caractères échappés sont des octets simples - j'ai présenté le programme sous cette forme pour faciliter la copie et le collage. Ce lien est vers le programme complet sans caractères d'échappement; comme on peut le voir, il existe de nombreux espaces réservés pour les caractères non imprimables utilisés.

L'approche utilisée est similaire à la réponse TI-BASIC de lirtosiast , bien que j'aie écrit ma réponse avant de parcourir les autres entrées. Le cordon fonctionne comme suit:

-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

La chaîne est composée de chaque caractère codé sur un octet autre que 0 (NULL), 127 (retour arrière) et les caractères passe-partout ( -C\|l"). De cette façon, chaque caractère de la chaîne peut être supprimé un par un pour produire des chaînes de longueurs allant de 248 à 0. Ces programmes donnent des résultats dans la plage [-124-124], avec une somme de valeurs absolues de 15500.

Le code a été généré à l'aide de ce programme :

+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

Et la somme des nombres générés à partir de chacun des programmes tronqués a été trouvée avec ce programme :

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
la source
3

Gelée , 984

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

Essayez-le en ligne!

Vous pouvez voir les préfixes (séparés par un double retour) ici .

Les numéros imprimés sont, dans l'ordre:

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 le Outgolfer
la source
2

Python 2, 113/53=25.1132075472

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

Préfixes:

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)

Pas optimal, mais le meilleur que j'ai pu trouver personnellement avec cette structure particulière.

Aidan F. Pierce
la source
2

Python 2 63/9=24

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

Voici les préfixes et ce qu'ils produisent

 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
Assistant de blé
la source
2

Propre , 403/ 400 = 160

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`

Essayez-le en ligne!

Définit nce qui produit en -20, -19, ..., 18, 19fonction de la longueur du typeG .

Le plus petit préfixe d' Gapparence :: G = G, précédé par les importations et définitions assorties.
Le plus grand préfixe d' Gapparence , pareillement précédé.:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`

Οurous
la source
Je pense que je manque quelque chose dans ce défi, mais… cette réponse (et une poignée d'autres) réutilise des "octets apparus dans les préfixes précédents", n'est-ce pas?
Blackhole
4
@Blackhole Comme je l'ai compris (et je suppose que d'autres) l'ont compris, vous pouvez utiliser des octets plusieurs fois dans le même préfixe, mais pas dans les suivants.
Οurous
2

Japt ,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

Essayez-le en ligne!

Comme les autres entrées, utilise la longueur d'une chaîne. Les caractères de l'extrait de départ, ainsi que \r(les zones de texte ne jouent pas bien avec cela) et {(caractère d'interpolation de chaîne) sont exclus de la chaîne.

Le premier programme est _Ê-122}$($", chaque programme est un caractère plus long à l'exception de lorsque vous ajoutez \\.

ASCII uniquement
la source
Je ne suis pas tout à fait sûr que cela soit valide, je n'ai pas vérifié que la chaîne ne contient aucun caractère de l'extrait de
code
J'allais suggérer de le remplacer 122par #zmais cela n'améliorerait pas votre score.
Shaggy
1

Python 3, (63)/12=18

0x1-2+3|True*~4

Ceci est un extrait.

Les extraits de préfixe individuels sont les suivants:

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)

Les opérateurs de Python traitent de bits entiers négatifs comme complément à deux avec un nombre illimité 1de à droite, donc -5est traité comme ...11111011, qui , lorsqu'il est ORed avec ...000010donne ...11111011sans changement.

pizzapants184
la source
1

Exceller, (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 défi ne se gèle pas bien avec la structure de formule d'Excel

Wernisch
la source
1

Enchantments Runiques , 117 trois / 3422 = 468.034

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

Essayez-le en ligne!

Et le préfixe .

Tirages [-58,58] inclus.

Utilise une tactique similaire à celle des autres entrées, utilisant une chaîne progressivement plus longue, obtenant sa longueur, soustrayant B(valeur décimale de 66) et imprime le résultat. Utilise un caractère non imprimable (affiché dans le bloc de code car, apparemment, c'est ce qui est &#128;évalué en HTML) afin d'utiliser la valeur d'octet 0x80(et de consommer une valeur autrement inutilisable 0x00) et de marquer une valeur de plus. Seules les valeurs d'octets jusqu'à 127 qui ne peuvent être représentées d'aucune façon le sont 10et 13, car elles entraîneraient le programme à deux lignes de haut et ne seraient pas lues correctement par le pointeur d'instruction.

Draco18s
la source
1

Charbon de bois , 95³ / 2256 = 380.04211 96³ / 2304 = 384

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

Le premier préfixe est les 7 premiers octets et renvoie le nombre 47. Essayez-le en ligne! Les préfixes suivants produisent des nombres décroissants et les sorties finales du programme -48. Essayez-le en ligne! Edit: augmentation du score de 3,95789 grâce à @ ASCII uniquement.

Neil
la source
Je pense que vous devriez pouvoir échapper à un caractère non ASCII au début ou à la fin de la chaîne? (en utilisant le caractère d'échappement, celui qui ressemble à un accent grave inversé)
ASCII uniquement le