Настоящим сокровищем были числа, которые мы сделали по пути


23

Ваша задача - написать программу, функцию или фрагмент (да, фрагменты разрешены), которые просто выводят целое число. Однако вы должны иметь возможность разделять отправку на префиксы, которые также производят разные целые числа. Вы не можете использовать байты, которые появились в предыдущих префиксах. Например, мы можем иметь префиксы:

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

правила

  • Ваша цель - создать, чтобы попытаться создать самые уникальные целые числа, сохраняя их близкими к нулю.
    • Система подсчета очков там ((number of unique integers)**3)/(sum of absolute values), где чем выше ваш балл, тем лучше. Приведенный выше пример оценивает(43)/(1+|5|+|1|+9)=64/16=4 .
  • Должно быть как минимум два уникальных целых числа (без деления на ноль!)
  • Пожалуйста, отформатируйте ваш ответ так:
# Language, \$(4^{3})/16 = 4\$

   1-6/3+0xA (the full program)
  • Укажите, если ваша заявка состоит из фрагментов, которые оценивают значение, функции или полные программы.
  • Перечислите каждый из префиксов и (необязательно) объяснение того, как они работают.

2
Окончательная строка представления, кажется, поменяла 3 на 2 (я не редактировал, чтобы исправить это в случае, если я неправильно понял)
trichoplax

4
Мне кажется, что вы можете получить почти оптимальный результат в Python, написав print 123-len(open(__file__).read())за ним по #одному из каждого байта, который до сих пор не использовался (за исключением некоторых управляющих символов), с префиксами, отбирающими эти байты. Отрегулируйте значение 123 так, чтобы диапазон выходных значений был равен 0. Для лучшей оценки часть программы без комментариев можно переписать, используя только 7 различных символов, универсальных для Python. Это все верно?
xnor

1
@ JoKing Я включаю негативы, когда разница отрицательна.
xnor

1
@JoKing Этот метод также может быть реализован на языках, где строка может быть написана без закрывающей кавычки.
xnor

2
Я не могу поверить, что так мало других заметили это и проголосовали за предыдущий комментарий. : PI только что заметил, что @trichoplax также обнаружил опечатку.
Питер Кордес

Ответы:


19

TI-BASIC (TI-84 + С), 2383/14161знак равно952,0

118-length("+×/ABCDEFGHIJK...

256 байтов в TI-BASIC разбиваются следующим образом:

  • 240 байтов - это однобайтовые токены, которые можно заключать в кавычки
  • 11 - начало двухбайтовых токенов
  • 2 не используются
  • Оставшиеся 3 являются ", символы и символ новой строки , которые нарушают строки.

Строка содержит все 237 допустимых байтов в первой категории, которой еще нет 118-length(.

Программа вычисляет 238 различных целых чисел, в диапазоне от 118 до -119. Счет

2383ΣNзнак равно-119118|N|знак равно+952.


1
Разве строка не должна заканчиваться на «который является байтом, который уже использовался? Или почему бы и нет? Меня это немного смущает и в других ответах, это, кажется, происходит, так что, вероятно, я что-то пропускаю»
findusl

2
@findusl TI-BASIC имеет тенденцию автоматически заполнять скобки и строки для вас. так что вы могли бы написать mean({1,2,3, что бы завершить mean({1,2,3}). Строки также завершены: Disp "Hi!завершает к Disp "Hi!". Поэтому заключительная заключительная цитата не нужна
Конор О'Брайен

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

Попробуйте онлайн! (все префиксы)

Сломать:

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

Вы не используете повторно <байт? Или это заполнитель для другого байта?
findusl

1
@findusl no - <esc>представляет собой клавишу escape и <c-char>представляет ctrl + char
Doorknob

11

Lenguage , 256³ / 16384 = 1024

Исходный код решения можно описать следующим образом:

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Каждый прогон идентичных байтов в сочетании с предыдущими прогонами составляет префикс, а полная программа - все строки, описанные выше, объединенные.

Это оптимальный результат. (Программы печатают все числа от -127 до 128 включительно.)

Чтобы создать этот ответ, сначала я скопировал ответ Java здесь для генерации программ brainfuck для печати строки. Я запустил его с этой mainфункцией:

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

Затем я запустил этот скрипт на Ruby, чтобы сгенерировать читабельный текст, показанный здесь:

#!/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"

Самая большая программа имеет длину 3594571896764310192036774345469579167648804468538578264427 байт (и печатает -90).


1
Опоры для скорости и практического навыка решения проблем!
lirtosiast

8

(1113)/2530знак равно+540,5656126482213

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

(Отдайте должное этой интересной утилите для конвертации.) Выше приведено следующее выражение:

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

... который создает прокси, который преобразует длину имени свойства в число. Первый префикс заканчивается на один символ после .и использует только 6 уникальных символов, исключая .$. Следующие 63 фрагмента добавляют 63 других однобайтовых символа, которые действительны в именах свойств объекта JS. Следующие 44 фрагмента добавляют двухбайтовые допустимые символы, состоящие из неиспользуемых байтов. Конечные ломтики добавить &'', |""**""и ,~{}. Первый срез возвращается -2, второй 2, затем -3, затем 3, -4, 4, -5и т.д. вплоть до последних трех ломтиков , которые 0, 1и -1.


4

43/4знак равно16

Для этого мы используем отрывки мозгов, поэтому нас интересует ценность, а не то, что они делают.

Полная программа:

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

Префиксы и вывод

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

Попробуйте онлайн!

объяснение

Вероятно, это лучшее, что может дать ответ «мозговой злопамятник», поскольку мы израсходовали всех персонажей, которых волнует «мозговой вздор», и не можем сгруппировать числа ближе к нулю.

Первый фрагмент является <>, <>изменяет стек , но в противном случае имеет значение ноль, поэтому мы начинаем с нуля. Отсюда мы добавляем, (())что делает один. Мы могли бы только что сделать (), потому что это тоже один, но если мы нажмем один сейчас, мы сможем вспомнить это позже с {}и []. Это в значительной степени единственный способ {}и []может создать ценность, поэтому мы делаем это. Поскольку в стеке есть одна вещь, которая []считается как 1, мы вычитаем 3 из общей суммы, что [[][][]]приводит к -2. Наконец, мы используем, {}чтобы вытолкнуть тот, который мы нажали ранее, доводя нас до -1.


4

(113)/32знак равно41,59375

После двух часов проб и ошибок я наконец получил все целые числа от -4до +57).

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

Это фрагмент со следующими фрагментами префикса:

# -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
Вы можете изменить 2*2до 22&7(-3 & 6 вместо -3 и 4) , чтобы установить , что повторение *.
Кевин Круйссен

@KevinCruijssen спасибо за предложение.
овс

Вы можете освободить 'и _использовать cmp(cmp,())вместо, cmp('','_')если это полезно.
Пшеничный волшебник

Вы также можете сделать раздел списка немного короче и намного проще для понимания --[-1-1-1-1-1], но это не улучшит ваш счет.
Пшеничный волшебник

3

Пакетный, 176³ / 7744 = 704

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

Первый префикс включает двоеточие и выводит число -88. Последующие префиксы выводят возрастающие числа, а конечные программы выводят 87.


3

Pyth, (2493)/15500знак равно996,0

-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

Все экранированные символы являются однобайтовыми - я представил программу в этой форме для простоты копирования и вставки. Эта ссылка на полную программу без экранированных символов; как можно видеть, есть много заполнителей для непечатаемых символов.

Используемый подход аналогичен ответу TI-BASIC от lirtosiast , хотя я написал свой ответ до просмотра других записей. Шнур работает следующим образом:

-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

Строка состоит из каждого однобайтового символа, кроме 0 (NULL), 127 (backspace) и шаблонных символов ( -C\|l"). Таким образом, каждый символ строки можно удалить по одному за раз, чтобы получить строки длиной от 248 до 0. Эти программы выдают результаты в диапазоне [-124 - 124] с суммой абсолютных значений 15500.

Код был сгенерирован с помощью этой программы :

+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

И сумма чисел, сгенерированных из каждой усеченной программы, была найдена с помощью этой программы :

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

Желе , 984

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

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Вы можете увидеть префиксы (разделенные двойной новой строкой) здесь .

Напечатанные числа в порядке:

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

Python 2, 113/53знак равно+25,1132075472

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

Префиксы:

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)

Не оптимально, но лучшее, что я мог лично найти с этой конкретной структурой.


2

Python 2 63/9знак равно24

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

Вот префиксы и что они выводят

 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

Чистый , 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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Определяет, nчто производит в -20, -19, ..., 18, 19зависимости от длины типа G.

Наименьший префикс Gвыглядит так :: G = G, как и предшествующий сортированный импорт и определения.
Самый большой префикс Gвыглядит так же , как и ранее.:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`


Я думаю, что что-то упустил в этом вызове, но… этот ответ (и несколько других) повторно использует «байты, которые появились в предыдущих префиксах», не так ли?
Blackhole

4
@ Blackhole Как я (и я предполагаю, что другие) это поняли, вы можете использовать байты несколько раз в одном префиксе, но не в последующих.
Οurous

2

Джапт ,2453/15006знак равно+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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Как и другие записи, использует длину строки. Символы в начальном фрагменте, а также \r(текстовые области не очень хорошо с этим {работают ) и (символ интерполяции строки) исключаются из строки.

Первая программа _Ê-122}$($", каждая программа на один символ длиннее, за исключением случаев, когда вы добавляете \\.


Не совсем уверен, что это верно, еще не проверил, что в строке нет символов фрагмента
ASCII-только

Я собирался предложить заменить 122на, #zно это не улучшит ваш счет.
Лохматый

1

Python 3, (63)/12знак равно18

0x1-2+3|True*~4

Это фрагмент.

Отдельные фрагменты префикса:

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)

Битовые операторы Python обрабатывают отрицательные целые числа как дополнение к двум с неограниченным числом 1s справа, поэтому они -5обрабатываются как ...11111011, что при ORed с ...000010дает ...11111011неизменным.


1

Excel, (63)/12знак равно18

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

Prefix вызов плохо сочетается с формулой Excel


1

Рунические Enchantments , 117 3 /3422 = 468.034

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

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И префикс .

Отпечатки [-58,58] включительно.

Использует тактику, аналогичную другим записям, используя постепенно увеличивающуюся строку, получая ее длину, вычитает B(десятичное значение 66) и печатает результат. Использует непечатаемый символ (показанный в блоке кода как, по- видимому, потому что, по-видимому, это то, что &#128;оценивается в HTML), чтобы использовать значение байта 0x80(и использовать непригодное для использования в противном случае 0x00) и получить еще одно значение. Только байтовые значения до 127 , которые не могут быть представлены любым способом , являются 10и 13, потому что они заставляют программу две строки в высоту и не быть правильно прочитан указатель инструкции.


1

Древесный уголь , 95³ / 2256 = 380.04211 96³ / 2304 = 384

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

Первый префикс представляет собой первые 7 байтов и выводит число 47. Попробуйте онлайн! Последующие префиксы выводят нисходящие числа и конечные выходные данные программы -48. Попробуйте онлайн! Редактировать: Увеличено количество очков на 3.95789 благодаря @ ASCII-only.


Я думаю, что вы должны быть в состоянии избежать не-ASCII символа в начале или конце строки? (с использованием escape-символа, похожего на перевернутый серьезный акцент)
только ASCII
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