Prawdziwym skarbem były liczby, które stworzyliśmy po drodze

23

Twoim zadaniem jest napisanie programu, funkcji lub fragmentu kodu (tak, fragmenty są dozwolone), które po prostu wypisują liczbę całkowitą. Jednakże , trzeba być w stanie oddzielić swojego zgłoszenia do przedrostków, które również produkują różne liczby całkowite. Nie można używać żadnych bajtów, które pojawiały się w poprzednich prefiksach. Na przykład możemy mieć prefiksy:

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

Zasady

  • Twoim celem jest stworzenie, aby spróbować stworzyć najbardziej unikalne liczby całkowite, jednocześnie utrzymując je blisko zera.
    • System punktacji jest taki ((number of unique integers)**3)/(sum of absolute values), że im wyższy wynik, tym lepiej. Powyższe przykładowe wyniki (43)/(1+|5|+|1|+9)=64/16=4 .
  • Powinny istnieć co najmniej dwie unikalne liczby całkowite (bez dzielenia przez zero!)
  • Sformatuj swoją odpowiedź podobnie do:
# Language, \$(4^{3})/16 = 4\$

   1-6/3+0xA (the full program)
  • Wspomnij, jeśli przesłanie składa się z fragmentów, które oceniają na wartość, funkcje lub pełne programy.
  • Wymień każdy z prefiksów i (opcjonalnie) objaśnienie ich działania.
Jo King
źródło
2
Wygląda na to, że w końcowym wierszu zgłoszeń zamieniłem 3 na 2 (nie edytowałem, aby to poprawić, jeśli źle zrozumiałem)
trichoplax
4
Wydaje mi się, że można uzyskać prawie optymalny wynik w Pythonie, pisząc, print 123-len(open(__file__).read())a #następnie jeden z nieużywanych do tej pory bajtów (z wyjątkiem niektórych znaków kontrolnych), z prefiksami odcinającymi te bajty. Dostosuj 123 do dowolnego środka zakresu wyników około 0. Aby uzyskać lepszy wynik, część programu bez komentarza można przepisać, używając tylko 7 różnych znaków, które są uniwersalne dla Pythona. Czy to wszystko jest ważne?
xnor
1
@JoKing Dołączam negatywy, gdy różnica jest ujemna.
xnor
1
@JoKing Metodę tę można również wykonać w językach, w których można napisać ciąg bez cudzysłowu.
xnor
2
Nie mogę uwierzyć, że tak niewielu ludzi to zauważyło i przegłosowało wcześniejszy komentarz. : PI właśnie zauważyła, że ​​@trichoplax też zauważył literówkę.
Peter Cordes

Odpowiedzi:

19

TI-Basic (TI-84 + SD) 2383/14161=952.0

118-length("+×/ABCDEFGHIJK...

256 bajtów w TI-BASIC rozkłada się w następujący sposób:

  • 240 bajtów to jednobajtowe tokeny, które można cytować
  • 11 to początek dwubajtowych tokenów
  • 2 nie są używane
  • Pozostałe 3 są ", znaki, które przerwa ciągi i linią.

Ciąg zawiera wszystkie 237 dozwolonych bajtów w pierwszej kategorii, które nie są jeszcze w 118-length(.

Program oblicza 238 różnych liczb całkowitych, od 118 do -119. Wynik wynosi zatem

2383n=119118|n|=952.

lirtosiast
źródło
1
Czy ciąg nie musiałby kończyć się na „który jest już bajtem, który został już użyty? Albo dlaczego? Nie jestem tym
zaskoczony
2
@findusl TI-BASIC ma tendencję do automatycznego uzupełniania nawiasów i ciągów znaków. abyście mogli napisać mean({1,2,3, co by się wypełniło mean({1,2,3}). ciągi są również uzupełniane: Disp "Hi!uzupełnia do Disp "Hi!". Dlatego ostateczny cytat końcowy nie jest konieczny
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

Wypróbuj online!(wszystkie prefiksy)

Awaria:

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
Klamka
źródło
nie używasz ponownie <bajtu? Czy jest to symbol zastępczy dla innego bajtu?
findusl
1
@findusl no - <esc>reprezentuje klawisz Escape i <c-char>reprezentuje ctrl + char
Doorknob
11

Lenguage , 256³ / 16384 = 1024

Kod źródłowy rozwiązania można opisać w następujący sposób:

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3461439604291557962702078999341076235513663562296408699075 0xFF bytes

Każdy ciąg identycznych bajtów w połączeniu z poprzednimi uruchomieniami stanowi prefiks, a pełny program zawiera wszystkie łańcuchy opisane powyżej połączone.

To optymalny wynik. (Programy drukują wszystkie liczby od -127 do 128 włącznie.)

Aby utworzyć tę odpowiedź, najpierw skopiowałem tutaj odpowiedź Java do generowania programów typu „pieprzenie mózgu” w celu wydrukowania łańcucha. Uruchomiłem go z tą mainfunkcją:

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

Następnie uruchomiłem ten skrypt Ruby, aby wygenerować tekst czytelny dla człowieka pokazany tutaj:

#!/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"

Największy program ma 3594571896764310192036774345469579167648804468538578264427 bajtów (i drukuje -90).

Klamka
źródło
1
Rekwizyty dla szybkości i praktycznej umiejętności rozwiązywania problemów!
lirtosiast
8

(1113)/2530=540,5656126482213

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

(Podziękowania dla tego interesującego narzędzia do konwersji.) Powyższe skutecznie tworzy następujący kod:

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

... który tworzy serwer proxy, który przekształca długość nazwy właściwości na liczbę. Pierwszy prefiks kończy jeden znak po .i używa tylko 6 unikalnych znaków, z wyłączeniem .$. Kolejne 63 wycinki dołączają 63 inne znaki jednobajtowe, które są poprawne w nazwach właściwości obiektu JS. Kolejne 44 wycinki dołączają prawidłowe znaki dwubajtowe złożone z nieużywanych bajtów. Ostateczne plastry dołączyć &'', |""**""i ,~{}. Pierwsze powroty kawałek -2, drugi 2, a potem -3, potem 3, -4, 4, -5, itd. Aż do ostatnich trzech plastrów, które są 0, 1i -1.

Yair Rand
źródło
4

43)/4=16

W tym przypadku korzystamy z urywków mózgu, więc troszczymy się o wartość, a nie o to, co robią.

Kompletny program:

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

Prefiksy i wyjście

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

Wypróbuj online!

Wyjaśnienie

Jest to prawdopodobnie najlepsza odpowiedź, jaką można uzyskać od uderzenia mózgu, ponieważ zużyliśmy już całą postać, o którą troszczy się mózg, i nie możemy grupować liczb bliżej zera.

Pierwszy fragment jest <>, <>zmienia stos ale inaczej ocenia na zero, więc zaczynamy od zera. Stąd dodajemy, (())że to czyni jeden. Moglibyśmy właśnie to zrobić (), ponieważ jest to również jeden, ale jeśli teraz go popchniemy, będziemy mogli później go przywołać za pomocą {}i []. Jest to właściwie jedyny sposób {}i []może tworzyć wartość, więc to robimy. Ponieważ na stosie jest jedna rzecz, []liczy się jako 1, więc odejmujemy 3 od sumy, [[][][]]doprowadzając nas do -2. Na koniec używamy {}tego, który popchnęliśmy wcześniej, co doprowadziło nas do -1.

Kreator pszenicy
źródło
4

Python 2 , 32 34,7 40 (113))/32=41,59375

Po dwóch godzinach prób i błędów w końcu dostałem wszystkie liczby całkowite od -4do +5(i 7).

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

To jest fragment kodu z następującymi fragmentami przedrostka:

# -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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ovs
źródło
1
Można zmienić 2*2na 22&7(-3 i 6 zamiast -3 i 4), aby rozwiązać ten powtórzenia *.
Kevin Cruijssen
@KevinCruijssen dzięki za sugestię.
ovs
Możesz zwolnić 'i _używając cmp(cmp,())zamiast cmp('','_')Jeśli to jest pomocne.
Kreator pszenicy
Możesz również sprawić, że sekcja listy będzie trochę krótsza i łatwiejsza do zrozumienia --[-1-1-1-1-1], ale nie poprawi to twojego wyniku.
Wheat Wizard
3

Partia, 176³ / 7744 = 704

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

Pierwszy prefiks zawiera dwukropek i zwraca liczbę -88. Kolejne prefiksy wyprowadzają rosnące liczby, a wynik końcowy programu 87.

Neil
źródło
3

Pyth (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

Wszystkie znaki ucieczki są jednobajtowe - przedstawiłem program w tej formie, aby ułatwić kopiowanie i wklejanie. Ten link prowadzi do pełnego programu bez znaków specjalnych; jak widać, istnieje wiele symboli zastępczych używanych znaków niedrukowalnych.

Zastosowane podejście jest podobne do odpowiedzi TIR-BASIC'a lirtosiasta , chociaż napisałem odpowiedź przed przeglądaniem innych wpisów. Przewód działa w następujący sposób:

-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

Ciąg składa się z każdego jednobajtowego znaku innego niż 0 (NULL), 127 (backspace) i znaków na płycie głównej (-C\|l" ). W ten sposób każdy znak ciągu można usunąć pojedynczo, aby utworzyć ciągi o długości od 248 do 0. Programy te dają wyniki w zakresie [-124 - 124], z sumą wartości bezwzględnych 15500.

Kod został wygenerowany przy użyciu tego programu :

+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

Za pomocą tego programu znaleziono sumę liczb wygenerowanych z każdego obciętego programu :

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
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3

Galaretka , 984

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

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Widać prefiksy (oddzielone podwójnym znakiem nowej linii) tutaj .

Drukowane liczby są w kolejności:

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 the Outgolfer
źródło
2

Python 2, 113)/53=25.1132075472

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

Prefiksy:

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)

Nie optymalne, ale najlepsze, co mogłem osobiście znaleźć w tej konkretnej strukturze.

Aidan F. Pierce
źródło
2

Python 2 63)/9=24

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

Oto prefiksy i ich wynik

 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
Kreator pszenicy
źródło
2

Czyste , 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`

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Określa, nktóre produkuje w -20, -19, ..., 18, 19zależności od długości typu G.

Najmniejszy prefiks Gwygląda :: G = G, poprzedzony różnymi importami i definicjami.
Największy prefiks Gwygląda podobnie , podobnie jak poprzednio.:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`

Obrzydliwe
źródło
Myślę, że coś mi brakuje w tym wyzwaniu, ale… ta odpowiedź (i garstka innych) używa ponownie „bajtów, które pojawiły się w poprzednich prefiksach”, prawda?
Blackhole
4
@Blackhole Ponieważ ja (i zakładam, że inni) to rozumiem, możesz używać bajtów wiele razy w tym samym prefiksie, ale nie w kolejnych.
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

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Podobnie jak inne wpisy, używa długości ciągu. Znaki w początkowym fragmencie, a także \r(obszary {tekstowe nie bawią się z tym dobrze) i (znak interpolacji łańcucha) są wykluczone z łańcucha.

Pierwszy program polega na tym _Ê-122}$($", że każdy program jest dłuższy o jeden znak, z wyjątkiem momentu dodania \\.

Tylko ASCII
źródło
Nie całkowicie pewien, że to jest ważne, że nie sprawdził, że ciąg nie ma żadnych znaków fragmencie
ASCII tylko
Chciałem zaproponować zastąpienie 122ze #zjednak nie poprawi swój wynik.
Kudłaty
1

Python 3, (63))/12=18

0x1-2+3|True*~4

To jest fragment kodu.

Poszczególne fragmenty prefiksów to:

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)

Bitowe operatory Pythona traktują ujemne liczby całkowite jako uzupełnienie dwóch z nieograniczoną liczbą 1s po prawej stronie, więc -5jest traktowane jako ...11111011, które gdy ORed z ...000010daje ...11111011niezmienione.

pizzapanty184
źródło
1

Przewyższać, (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 wyzwanie nie żeluje dobrze ze strukturą formuły programu Excel

Wernisch
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1

Runic Zaklęcia , 117 3 /3422 = 468.034

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

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I przedrostek .

Wydruki [-58,58] włącznie.

Stosuje podobną taktykę jak inne wpisy, stosując stopniowo dłuższy ciąg, uzyskując jego długość, odejmując B(wartość dziesiętna 66) i drukuje wynik. Wykorzystuje znak, który nie jest drukowany (pokazany w bloku kodu, ponieważ najwyraźniej to właśnie &#128;ocenia w HTML), aby wykorzystać wartość bajtu 0x80(i zużyć inaczej bezużyteczne 0x00) i zdobyć jeszcze jedną wartość. Tylko wartości bajtów do 127, które nie mogą być reprezentowane w jakikolwiek sposób są 10i 13, ponieważ będą one spowodować, że program się dwie linie wysoki i nie być poprawnie odczytane przez wskaźnik instrukcji.

Draco18s
źródło
1

Węgiel drzewny , 95³ / 2256 = 380,04211 96³ / 2304 = 384

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

Pierwszy prefiks to pierwsze 7 bajtów i wyświetla liczbę 47. Wypróbuj online! Kolejne prefiksy wyprowadzają malejące liczby i wyjścia programu końcowego -48. Wypróbuj online! Edycja: Zwiększony wynik o 3,95789 dzięki tylko @ ASCII.

Neil
źródło
Myślę, że powinieneś być w stanie uciec od znaku spoza ASCII na początku lub na końcu łańcucha? (używając znaku ucieczki, tego, który wygląda jak odwrócony grób)
tylko ASCII