Der wahre Schatz waren die Zahlen, die wir auf dem Weg gemacht haben


23

Ihre Aufgabe ist es, ein Programm, eine Funktion oder ein Snippet (ja, Snippets sind erlaubt) zu schreiben, das einfach eine ganze Zahl ausgibt. Allerdings müssen Sie in der Lage sein , Ihre Vorlage in Präfixe zu trennen , die auch verschiedene ganze Zahlen produzieren. Sie können keine Bytes verwenden, die in früheren Präfixen enthalten waren. Zum Beispiel können wir die Präfixe haben:

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

Regeln

  • Ihr Ziel ist es, zu erstellen, um zu versuchen, die einzigartigsten Ganzzahlen zu erstellen, während sie nahe Null bleiben.
    • Das Punktesystem ist ((number of unique integers)**3)/(sum of absolute values), je höher Ihre Punktzahl, desto besser. Das obige Beispiel .(43)/(1+|-5|+|-1|+9)=64/16=4
  • Es sollten mindestens zwei eindeutige Ganzzahlen vorhanden sein (keine Division durch Null!)
  • Bitte formatieren Sie Ihre Antwort wie folgt:
# Language, \$(4^{3})/16 = 4\$

   1-6/3+0xA (the full program)
  • Erwähnen Sie, wenn Ihr Beitrag aus Ausschnitten besteht, die einen Wert, Funktionen oder vollständige Programme ergeben.
  • Listen Sie jedes der Präfixe und (optional) eine Erklärung für ihre Funktionsweise auf.

2
Die letzte Übermittlungszeile hat anscheinend die 3 gegen eine 2 getauscht (ich habe sie nicht bearbeitet, um dies zu korrigieren, falls ich sie missverstanden habe)
Trichoplax

4
Für mich sieht es so aus, als ob Sie in Python eine nahezu optimale Punktzahl erzielen können, indem Sie schreiben, print 123-len(open(__file__).read())gefolgt von #einem der bisher nicht verwendeten Bytes (mit Ausnahme einiger Steuerzeichen), wobei die Präfixe diese Bytes abschneiden. Stellen Sie 123 so ein, dass der Bereich der Ausgaben um 0 zentriert ist. Um eine bessere Punktzahl zu erzielen, kann der nichtkommentierte Teil des Programms mit nur den 7 verschiedenen Zeichen umgeschrieben werden, die für Python universell sind. Ist das alles gültig?
11.

1
@JoKing Ich beziehe Negative ein, wenn der Unterschied negativ ist.
11.

1
@JoKing Diese Methode kann auch in Sprachen ausgeführt werden, in denen eine Zeichenfolge ohne schließendes Anführungszeichen geschrieben werden kann.
11.

2
Ich kann nicht glauben, dass es so wenigen anderen Leuten aufgefallen ist, die den vorherigen Kommentar positiv bewertet haben. : PI hat gerade bemerkt, dass @trichoplax auch den Tippfehler entdeckt hat.
Peter Cordes

Antworten:


19

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

118-length("+×/ABCDEFGHIJK...

Die 256 Bytes in TI-BASIC gliedern sich folgendermaßen:

  • 240 Bytes sind Ein-Byte-Token, die in Anführungszeichen gesetzt werden können
  • 11 sind der Beginn von Zwei-Byte-Token
  • 2 sind unbenutzt
  • Die restlichen 3 sind die ", und Newline Zeichen , die Pause - Strings.

Die Zeichenfolge enthält alle 237 zulässigen Bytes in der ersten Kategorie, die noch nicht in enthalten sind 118-length(.

Das Programm berechnet 238 verschiedene Ganzzahlen von 118 bis -119. Die Punktzahl ist daher

2383n=119118|n|=952.


1
Müsste die Zeichenfolge nicht mit "das ist ein Byte, das bereits verwendet wurde" enden? Oder warum nicht? Ich bin ein bisschen verwirrt darüber, auch bei anderen Antworten scheint dies so zu passieren, dass ich wahrscheinlich etwas vermisse.
findusl

2
@findusl TI-BASIC neigt dazu, Klammern und Zeichenfolgen automatisch für Sie zu vervollständigen. so könnte man schreiben mean({1,2,3, was sich ergänzen würde mean({1,2,3}). Zeichenfolgen sind auch abgeschlossen: Disp "Hi!Vervollständigt bis Disp "Hi!". Daher ist das endgültige Schlusszitat nicht erforderlich
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

Probieren Sie es online! (alle Präfixe)

Nervenzusammenbruch:

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

Verwenden Sie das <Byte nicht erneut? Oder ist das ein Platzhalter für ein anderes Byte?
Findusl

1
@findusl no - <esc>repräsentiert die Escape-Taste und <c-char>repräsentiert Strg + Zeichen
Türklinke

11

Sprache , 256³ / 16384 = 1024

Der Quellcode der Lösung kann wie folgt beschrieben werden:

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
56025                                                      0x1D bytes
756                                                        0x1E bytes
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2537754615118693246334805                                  0x21 bytes
108                                                        0x22 bytes
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75789                                                      0x25 bytes
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108                                                        0x2A bytes
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75789                                                      0x34 bytes
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699060255345705731106938578982346708                       0x82 bytes
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472416520991134187843376319996477587                       0x84 bytes
108                                                        0x85 bytes
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75789                                                      0x88 bytes
108                                                        0x89 bytes
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4851795238070470150902081892625349                         0x8D bytes
18874626894334054739887341632523361116                     0x8E bytes
114078274655132115171                                      0x8F bytes
12755246066760623071771160639904894849                     0x90 bytes
108                                                        0x91 bytes
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854006569949556303994892977044130020645                    0x9C bytes
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23058177388638020207862110380191510557388                  0xA9 bytes
108                                                        0xAA bytes
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108                                                        0xB7 bytes
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16811989759230232140258944020462709919333792               0xC4 bytes
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330910394430928659216716795154767519342247107424           0xEA bytes
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8934580649635073798851353469178723609411678244474          0xEF bytes
55801305                                                   0xF0 bytes
1506635235                                                 0xF1 bytes
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606761124039196515                                         0xF3 bytes
241233677540146992568986543667825537454115312600825        0xF4 bytes
1506635235                                                 0xF5 bytes
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6513309293583968799362636679031305893812560835614495       0xF7 bytes
40679151345                                                0xF8 bytes
1098337086315                                              0xF9 bytes
175859350926767157582791190333845259132939142561591365     0xFA bytes
1098337086315                                              0xFB bytes
4748202475022713254735362139013821996589386504264297360    0xFC bytes
29655101330505                                             0xFD bytes
128201466825613257877854777753373193907913435615136028720  0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes

Jeder Lauf von identischen Bytes in Kombination mit den vorherigen Läufen bildet ein Präfix, und das vollständige Programm besteht aus allen oben beschriebenen verketteten Strings.

Dies ist eine optimale Punktzahl. (Die Programme geben alle Zahlen von -127 bis einschließlich 128 aus.)

Um diese Antwort zu erstellen, habe ich zuerst die Java-Antwort hier kopiert , um Brainfuck-Programme zum Drucken einer Zeichenfolge zu generieren. Ich habe es mit dieser mainFunktion ausgeführt:

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

Dann habe ich dieses Ruby-Skript ausgeführt, um den hier gezeigten lesbaren Text zu generieren:

#!/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"

Das größte Programm ist 3594571896764310192036774345469579167648804468538578264427 Byte lang (und druckt -90).


1
Requisiten für die Geschwindigkeit und praktische Fähigkeiten zur Problemlösung!
Lirtosiast

8

(1113)/2530=540.5656126482213

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

(Wir danken diesem interessanten Dienstprogramm für die Konvertierung.) Das oben Gesagte ergibt effektiv den folgenden Code:

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

... wodurch ein Proxy erstellt wird, der die Länge eines Eigenschaftsnamens in eine Zahl umwandelt. Das erste Präfix endet ein Zeichen nach dem .und verwendet nur 6 eindeutige Zeichen, mit Ausnahme des .$. Die nächsten 63 Slices hängen 63 weitere Einzelbytezeichen an, die in JS-Objekteigenschaftsnamen gültig sind. Die nächsten 44 Slices hängen zwei Byte gültige Zeichen an, die aus nicht verwendeten Bytes bestehen. Die endgültigen Scheiben hängen &'', |""**""und ,~{}. Die erste Scheibe zurückkehrt -2, die zweite 2, dann -3, dann 3, -4, 4, -5, usw. bis zu den letzten drei Scheiben sind , die 0, 1, und -1.


4

43/4=16

In diesem Fall verwenden wir Brain-Flak-Schnipsel, damit wir uns mit dem Wert befassen, den sie nicht leisten.

Gesamtes Programm:

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

Präfixe und Ausgabe

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

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Erläuterung

Dies ist wahrscheinlich das Beste, was eine Brain-Flak-Antwort bekommen kann, da wir alle Zeichen verbraucht haben, die Brain-Flak interessiert, und wir können die Zahlen nicht näher an Null gruppieren.

Das erste Snippet ist <>: <>Ändert den Stack, wird aber ansonsten zu Null ausgewertet. Wir beginnen also mit Null. Ab hier addieren wir (())dies zu einem. Wir hätten das auch tun ()können, aber wenn wir jetzt eine drücken, können wir uns später mit {}und daran erinnern []. Dies ist so ziemlich der einzige Weg {}und []kann Wert schaffen, also tun wir das. Da es eine Sache auf dem Stapel gibt, die []als 1 gilt, subtrahieren wir 3 von der Gesamtsumme, [[][][]]indem wir uns auf -2 bringen. Zuletzt verwenden wir, {}um das zu platzieren, was wir zuvor gepusht haben und auf -1 gebracht haben.


4

(113)/32=41.59375

Nach zwei Stunden Versuch und Irrtum bekam ich endlich alle ganzen Zahlen von -4bis +5(und 7).

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

Dies ist ein Snippet mit den folgenden Präfix-Snippets:

# -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

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1
Sie können den Wert 2*2auf 22&7(-3 & 6 anstelle von -3 & 4) ändern , um die Wiederholung von zu korrigieren *.
Kevin Cruijssen

@ KevinCruijssen danke für den Vorschlag.
Ovs

Sie können freigeben 'und _verwenden, cmp(cmp,())anstatt cmp('','_')Wenn das hilfreich ist.
Weizen-Zauberer

Sie können den Listenabschnitt auch ein wenig kürzer und damit einfacher verständlich machen --[-1-1-1-1-1], aber das verbessert Ihre Punktzahl nicht wirklich.
Weizen-Zauberer

3

Charge, 176³ / 7744 = 704

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

Das erste Präfix enthält den Doppelpunkt und gibt die Nummer -88 aus. Nachfolgende Präfixe geben aufsteigende Zahlen aus und das endgültige Programm gibt 87 aus.


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

Alle mit Escapezeichen versehenen Zeichen sind Einzelbytes. Ich habe das Programm in dieser Form dargestellt, um das Kopieren und Einfügen zu vereinfachen. Dieser Link führt zum vollständigen Programm ohne maskierte Zeichen. Wie zu sehen ist, gibt es viele Platzhalter für die verwendeten nicht druckbaren Zeichen.

Der verwendete Ansatz ähnelt der TI-BASIC-Antwort von Lirtosiast , obwohl ich meine Antwort geschrieben habe, bevor ich die anderen Einträge durchgesehen habe . Das Kabel funktioniert wie folgt:

-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

Die Zeichenfolge besteht aus jedem Einzelbyte-Zeichen außer 0 (NULL), 127 (Backspace) und den Boilerplate-Zeichen ( -C\|l"). Auf diese Weise kann jedes Zeichen der Zeichenfolge einzeln entfernt werden, um Zeichenfolgen mit einer Länge von 248 bis 0 zu erzeugen. Diese Programme liefern Ergebnisse im Bereich [-124 - 124] mit einer Summe der absoluten Werte von 15500.

Der Code wurde mit diesem Programm generiert :

+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

Und die Summe der Zahlen, die aus jedem der abgeschnittenen Programme generiert wurden, wurde mit diesem Programm gefunden :

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

3

Jelly , 984

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

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Sie können die Präfixe (durch einen doppelten Zeilenumbruch getrennt) hier sehen .

Die gedruckten Zahlen lauten in der Reihenfolge:

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

2

113/53=25.1132075472

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

Präfixe:

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)

Nicht optimal, aber das Beste, was ich persönlich mit dieser speziellen Struktur finden konnte.


2

63/9=24

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

Hier sind die Präfixe und was sie ausgeben

 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

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`

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Legt fest, nwas -20, -19, ..., 18, 19abhängig von der Länge des Typs produziert wird G.

Das kleinste Präfix Gsieht aus wie :: G = G, gefolgt von den sortierten Importen und Definitionen.
Das größte Präfix von Gsieht aus wie , ähnlich vorangestellt.:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`


Ich glaube, ich vermisse etwas in dieser Herausforderung, aber ... diese Antwort (und eine Handvoll anderer) verwendet immer wieder "Bytes, die in früheren Präfixen vorkommen", nicht wahr?
Blackhole

4
@Blackhole So wie ich es verstanden habe (und ich nehme an, dass andere es verstanden haben), können Sie Bytes in demselben Präfix mehrmals verwenden, jedoch nicht in nachfolgenden.
Am

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

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Verwendet wie die anderen Einträge die Länge eines Strings. Zeichen im Start-Snippet sowie \r(Textbereiche spielen damit keine gute Rolle ) und {(Zeichenfolge-Interpolationszeichen) werden von der Zeichenfolge ausgeschlossen.

Das erste Programm ist _Ê-122}$($", dass jedes Programm ein Zeichen länger ist, mit Ausnahme des Hinzufügens \\.


Nicht ganz sicher , dass diese gültig ist, nicht überprüft , dass die Zeichenfolge hat keine Zeichen des Snippets
ASCII-only

Ich wollte Ersatz vorschlagen 122mit , #zaber das würde Ihre Gäste nicht verbessern.
Shaggy

1

(63)/12=18

0x1-2+3|True*~4

Dies ist ein Ausschnitt.

Die einzelnen Präfix-Snippets sind:

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)

Pythons Bitoperatoren behandelt negative ganze Zahlen als Zweierkomplement mit unbegrenzten 1s nach rechts, so -5wird behandelt , als ...11111011, die , wenn sie geODERten mit ...000010gibt ...11111011unverändert.


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 Herausforderung passt nicht zur Formelstruktur von Excel


1

Runic Verzauberungen , 117 3 /3422 = 468,034

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

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Und das Präfix .

Abzüge [-58,58] inklusive.

Verwendet eine ähnliche Taktik wie andere Eingaben, verwendet eine zunehmend längere Zeichenfolge, ermittelt deren Länge, subtrahiert B(Dezimalwert von 66) und gibt das Ergebnis aus. Verwendet ein nicht druckbares Zeichen (wird im Codeblock angezeigt, da dies anscheinend &#128;in HTML ausgewertet wird), um den Bytewert zu verwenden 0x80(und einen ansonsten unbrauchbaren zu verbrauchen 0x00) und einen weiteren Wert zu erzielen. Nur Bytewerte bis zu 127, die in keiner Weise dargestellt werden können , sind 10und 13, da sie dazu führen würden, dass das Programm zwei Zeilen lang ist und vom Befehlszeiger nicht richtig gelesen wird.


1

Charcoal , 95³ / 2256 = 380,04211 96³ / 2304 = 384

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

Das erste Präfix besteht aus den ersten 7 Bytes und gibt die Nummer aus 47. Probieren Sie es online! Nachfolgende Präfixe geben absteigende Zahlen und die endgültigen Programmausgaben aus -48. Probieren Sie es online! Bearbeiten: Erhöht die Punktzahl um 3,95789, dank nur @ ASCII.


Ich denke, Sie sollten in der Lage sein, ein Nicht-ASCII-Zeichen am Anfang oder Ende der Zeichenfolge zu entkommen? (Verwenden Sie das Escape-Zeichen, das aussieht wie ein umgedrehter Grabakzent)
Nur ASCII
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