Harta yang sebenarnya adalah angka yang kami buat di sepanjang jalan

23

Tugas Anda adalah menulis sebuah program, fungsi, atau snippet (ya, snippet diperbolehkan) yang hanya menghasilkan integer. Namun , Anda harus dapat memisahkan kiriman Anda ke dalam awalan yang juga menghasilkan bilangan bulat yang berbeda. Anda tidak dapat menggunakan byte yang muncul di awalan sebelumnya. Sebagai contoh, kita dapat memiliki awalan:

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

Aturan

  • Tujuan Anda adalah membuat untuk mencoba dan membuat bilangan bulat paling unik, sekaligus menjaganya tetap mendekati nol.
    • Sistem penilaian adalah ((number of unique integers)**3)/(sum of absolute values), di mana semakin tinggi skor Anda, semakin baik. Contoh di atas skor (43)/(1+|5|+|1|+9)=64/16=4 .
  • Setidaknya harus ada dua bilangan bulat unik (tidak ada pembagian dengan nol!)
  • Harap format jawaban Anda mirip dengan:
# Language, \$(4^{3})/16 = 4\$

   1-6/3+0xA (the full program)
  • Sebutkan jika kiriman Anda terdiri dari cuplikan yang mengevaluasi nilai, fungsi, atau program lengkap.
  • Sebutkan setiap awalan dan (opsional) penjelasan untuk cara kerjanya.
Jo King
sumber
2
Baris pengajuan Final tampaknya telah menukar 3 untuk 2 (Saya belum diedit untuk memperbaiki ini jika saya salah paham)
trichoplax
4
Sepertinya saya seperti Anda bisa mendapatkan skor hampir optimal dalam Python dengan menulis print 123-len(open(__file__).read())diikuti #kemudian satu dari setiap byte yang tidak digunakan sejauh ini (kecuali beberapa karakter kontrol), dengan awalan memotong byte tersebut. Sesuaikan 123 dengan pusat apa pun kisaran output sekitar 0. Untuk skor yang lebih baik, bagian non-komentar dari program dapat ditulis ulang hanya menggunakan 7 karakter berbeda yang bersifat universal untuk Python. Apakah ini semua valid?
xnor
1
@JoKing saya memasukkan negatif, ketika perbedaannya negatif.
xnor
1
@ JoKing Metode ini juga dilakukan dalam bahasa di mana string dapat ditulis tanpa kutipan penutup.
xnor
2
Saya tidak percaya begitu sedikit orang yang memperhatikannya dan mengungguli komentar sebelumnya. : PI hanya memperhatikan bahwa @trichoplax melihat kesalahan ketik juga.
Peter Cordes

Jawaban:

19

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

118-length("+×/ABCDEFGHIJK...

256 byte dalam TI-BASIC memecah dengan demikian:

  • 240 byte adalah token satu byte yang dapat dikutip
  • 11 adalah awal dari token dua-byte
  • 2 tidak digunakan
  • Sisanya 3 adalah ", karakter, dan baris baru yang istirahat string.

String berisi semua 237 byte yang diizinkan dalam kategori pertama yang belum masuk 118-length(.

Program ini menghitung 238 bilangan bulat yang berbeda, mulai dari 118 ke -119. Skor karenanya

2383n=119118|n|=952.

lirtosiast
sumber
1
Bukankah string harus diakhiri dengan "yang merupakan byte yang sudah digunakan? Atau mengapa tidak? Saya agak bingung dengan ini juga dalam jawaban lain ini sepertinya terjadi jadi mungkin saya kehilangan sesuatu.
findusl
2
@findusl TI-BASIC memiliki kecenderungan untuk secara otomatis menyelesaikan kurung dan string untuk Anda. sehingga Anda bisa menulis mean({1,2,3, yang akan melengkapi mean({1,2,3}). string juga selesai: Disp "Hi!selesai untuk Disp "Hi!". Oleh karena itu kutipan penutupan akhir tidak perlu
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

Cobalah online! (semua awalan)

Kerusakan:

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
Gagang pintu
sumber
apakah Anda tidak menggunakan kembali <byte? Atau apakah itu placeholder untuk byte yang berbeda?
findusl
1
@findusl no - <esc>mewakili kunci keluar dan <c-char>mewakili ctrl + char
Doorknob
11

Bahasa , 256³ / 16384 = 1024

Kode sumber solusi dapat dijelaskan sebagai berikut:

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
1514835                                                    0x1F bytes
20439                                                      0x20 bytes
2537754615118693246334805                                  0x21 bytes
108                                                        0x22 bytes
2808                                                       0x23 bytes
27                                                         0x24 bytes
75789                                                      0x25 bytes
756                                                        0x26 bytes
2046276                                                    0x27 bytes
20439                                                      0x28 bytes
12812877                                                   0x29 bytes
108                                                        0x2A bytes
729                                                        0x2B bytes
2079                                                       0x2C bytes
27                                                         0x2D bytes
1512648                                                    0x2E bytes
40920957                                                   0x2F bytes
1689866867948183303767857                                  0x30 bytes
66829507740256534347823839                                 0x31 bytes
2808                                                       0x32 bytes
27                                                         0x33 bytes
75789                                                      0x34 bytes
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108                                                        0x7D bytes
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699060255345705731106938578982346708                       0x82 bytes
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108                                                        0x85 bytes
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75789                                                      0x88 bytes
108                                                        0x89 bytes
729                                                        0x8A bytes
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27                                                         0x8C bytes
4851795238070470150902081892625349                         0x8D bytes
18874626894334054739887341632523361116                     0x8E bytes
114078274655132115171                                      0x8F bytes
12755246066760623071771160639904894849                     0x90 bytes
108                                                        0x91 bytes
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27                                                         0x93 bytes
75789                                                      0x94 bytes
756                                                        0x95 bytes
2046276                                                    0x96 bytes
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130998471427902694074356211100885179                       0x9B bytes
854006569949556303994892977044130020645                    0x9C bytes
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23058177388638020207862110380191510557388                  0xA9 bytes
108                                                        0xAA bytes
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108                                                        0xB7 bytes
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75789                                                      0xBA bytes
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16811989759230232140258944020462709919333792               0xC4 bytes
108                                                        0xC5 bytes
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453923723499216267786991488552493167822012492              0xD4 bytes
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1506635235                                                 0xF1 bytes
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606761124039196515                                         0xF3 bytes
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1506635235                                                 0xF5 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

Setiap run byte identik yang dikombinasikan dengan run sebelumnya merupakan awalan, dan program lengkapnya adalah semua string yang dijelaskan di atas digabungkan.

Ini adalah skor optimal. (Program-program mencetak semua angka dari -127 hingga 128 inklusif.)

Untuk membuat jawaban ini, pertama, saya menyalin jawaban Java di sini untuk menghasilkan program brainfuck untuk mencetak string. Saya menjalankannya dengan mainfungsi ini :

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

Lalu, saya menjalankan skrip Ruby ini untuk menghasilkan teks yang dapat dibaca manusia yang ditampilkan di sini:

#!/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"

Program terbesar adalah 3594571896764310192036774345469579167648804468538578264427 byte panjang (dan cetakan -90).

Gagang pintu
sumber
1
Alat peraga untuk keterampilan pemecahan masalah yang praktis dan cepat!
lirtosiast
8

(1113)/2530=540.5656126482213

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

(Kredit ke utilitas yang menarik ini untuk konversi.) Yang di atas secara efektif mengevaluasi ke dalam kode berikut:

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

... yang membuat proksi yang mengubah panjang nama properti menjadi angka. Awalan pertama berakhir satu karakter setelah ., dan hanya menggunakan 6 karakter unik, tidak termasuk .$. 63 irisan berikutnya menambahkan 63 karakter single-byte lainnya yang valid dalam nama properti objek JS. 44 irisan berikutnya menambahkan karakter valid dua byte yang terdiri dari byte yang tidak digunakan. Irisan akhir menambahkan &'', |""**"", dan ,~{}. Potongan pengembalian pertama -2, kedua 2, kemudian -3, kemudian 3, -4, 4,-5 , dll hingga tiga irisan terakhir yang 0, 1, dan -1.

Yair Rand
sumber
4

43/4=16

Untuk yang ini kami menggunakan potongan-potongan otak-otak sehingga kami peduli dengan nilai bukan apa yang mereka lakukan.

Program lengkap:

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

Awalan dan keluaran

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

Cobalah online!

Penjelasan

Ini mungkin jawaban terbaik yang bisa didapatkan oleh brain-flak karena kami telah menggunakan semua karakter yang dipikirkan Brain-flak, dan kami tidak dapat mengelompokkan angka yang mendekati nol.

Cuplikan pertama adalah <>, <>mengubah tumpukan tetapi mengevaluasi menjadi nol, jadi kami mulai dengan nol. Dari sini kita tambahkan (())ini bikin satu. Kita bisa saja melakukannya (), karena itu juga satu, tetapi jika kita mendorong satu sekarang kita akan dapat mengingatnya nanti dengan {}dan []. Ini adalah satu-satunya cara {}dan []dapat menciptakan nilai jadi kami melakukannya. Karena ada satu hal di stack yang []dihitung sebagai 1, jadi kita kurangi 3 dari total dengan [[][][]]membawa kita ke -2. Terakhir kami gunakan {}untuk memunculkan yang kami dorong sebelumnya membawa kami ke -1.

Wisaya Gandum
sumber
4

(113)/32=41.59375

Setelah dua jam coba-coba akhirnya saya mendapatkan semua bilangan bulat dari -4ke +5(dan 7).

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

Ini adalah cuplikan dengan cuplikan awalan berikut:

# -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

Cobalah online!

ovs
sumber
1
Anda dapat mengubah 2*2ke 22&7(-3 & 6 bukannya -3 & 4) untuk memperbaiki pengulangan itu *.
Kevin Cruijssen
@KevinCruijssen terima kasih atas sarannya.
Ovs
Anda dapat membebaskan 'dan _menggunakan cmp(cmp,())bukan cmp('','_')jika itu membantu.
Wheat Wizard
Anda juga bisa membuat bagian daftar sedikit lebih pendek dan lebih mudah dipahami dengan melakukan --[-1-1-1-1-1], tetapi itu tidak benar-benar meningkatkan skor Anda.
Wheat Wizard
3

Batch, 176³ / 7744 = 704

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

Awalan pertama meliputi titik dua dan menampilkan angka -88. Awalan berikutnya menghasilkan angka naik dan keluaran akhir program 87.

Neil
sumber
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

Semua karakter yang lolos adalah byte tunggal - Saya telah mempresentasikan program dalam bentuk ini untuk kemudahan penyalinan dan menempel. Link ini ini ke program lengkap tanpa karakter yang lolos; seperti yang dapat dilihat ada banyak placeholder untuk karakter yang tidak patut digunakan.

Pendekatan yang digunakan mirip dengan jawaban TI-BASIC lirtosiast , meskipun saya menulis jawaban saya sebelum menjelajahi entri lain. Kabelnya berfungsi sebagai berikut:

-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

String dibuat dari setiap karakter byte tunggal selain 0 (NULL), 127 (backspace), dan karakter boilerplate (-C\|l" ). Dengan begitu, setiap karakter string dapat dihapus satu per satu untuk menghasilkan string dengan panjang 248 hingga 0. Program-program ini menghasilkan kisaran [-124 - 124], dengan jumlah nilai absolut 15500.

Kode dihasilkan menggunakan program ini :

+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

Dan jumlah angka yang dihasilkan dari masing-masing program terpotong ditemukan dengan program ini :

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
sumber
3

Jelly , 984

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

Cobalah online!

Anda dapat melihat awalan (dipisahkan oleh baris ganda ganda) di sini .

Angka-angka yang dicetak adalah, dalam urutan:

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
sumber
2

Python 2, 113/53=25.1132075472

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

Awalan:

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)

Tidak optimal, tetapi yang terbaik yang bisa saya temukan secara pribadi dengan struktur khusus ini.

Aidan F. Pierce
sumber
2

Python 2 63/9=24

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

Berikut adalah awalan dan apa yang mereka hasilkan

 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
Wisaya Gandum
sumber
2

Bersih , 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`

Cobalah online!

Menentukan nyang menghasilkan -20, -19, ..., 18, 19tergantung pada panjang jenis G.

Awalan terkecil Gseperti :: G = G, didahului oleh berbagai macam impor dan definisi.
Awalan terbesar Gseperti , juga didahului.:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`

Suram
sumber
Saya pikir saya kehilangan sesuatu dalam tantangan ini, tapi ... jawaban ini (dan beberapa orang lain) menggunakan kembali "byte yang telah muncul di awalan sebelumnya", bukan?
Blackhole
4
@ Lubang Bor Seperti yang saya (dan saya asumsikan orang lain) memahaminya, Anda dapat menggunakan byte beberapa kali dalam awalan yang sama, tetapi tidak pada yang berikutnya.
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

Cobalah online!

Seperti entri lainnya, gunakan panjang string. Karakter dalam cuplikan awal, juga \r(textareas tidak bermain bagus dengan ini) dan{ (karakter interpolasi string) dikecualikan dari string.

Program pertama adalah _Ê-122}$($", setiap program lebih panjang satu karakter dengan pengecualian saat Anda menambahkan \\.

Khusus ASCII
sumber
Tidak sepenuhnya yakin ini valid, belum memeriksa bahwa string tidak memiliki karakter cuplikan
ASCII
Saya akan menyarankan untuk mengganti 122dengan #ztetapi itu tidak akan meningkatkan skor Anda.
Shaggy
1

Python 3, (63)/12=18

0x1-2+3|True*~4

Ini cuplikan.

Cuplikan awalan individu adalah:

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)

Operator bitwise Python memperlakukan bilangan bulat negatif sebagai pelengkap dua dengan 1s yang tidak terbatas di sebelah kanan, sehingga -5diperlakukan sebagai ...11111011, yang ketika ORed dengan ...000010memberi ...11111011tidak berubah.

pizzapants184
sumber
1

Unggul, (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 tantangan tidak cocok dengan struktur rumus Excel

Wernisch
sumber
1

Pesona rahasia , 117 3 /3422 = 468,034

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

Cobalah online!

Dan awalannya .

Mencetak [-58,58] inklusif.

Gunakan taktik yang sama seperti entri lainnya, menggunakan string yang semakin lama semakin panjang, mendapatkan panjangnya, mengurangi B(nilai desimal dari 66), dan mencetak hasilnya. Menggunakan karakter non-cetak (ditunjukkan dalam blok kode karena tampaknya itulah yang &#128;dievaluasi dalam HTML) untuk memanfaatkan nilai byte 0x80(dan menggunakan yang tidak dapat digunakan 0x00) dan mencetak satu nilai lagi. Hanya nilai byte hingga 127 yang tidak dapat direpresentasikan dalam mode apa pun 10dan 13, karena mereka akan menyebabkan program menjadi dua baris dan tidak dibaca dengan benar oleh penunjuk instruksi.

Draco18s
sumber
1

Arang , 95³ / 2256 = 380.04211 96³ / 2304 = 384

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

Awalan pertama adalah 7 byte pertama dan menampilkan nomor 47. Cobalah online! Awalan berikutnya menghasilkan angka-angka menurun dan keluaran program akhir -48.Cobalah online! Sunting: Peningkatan skor sebesar 3,95789 berkat hanya @ ASCII.

Neil
sumber
Saya pikir Anda harus dapat melarikan diri dari karakter non-ASCII pada awal atau akhir dari string? (menggunakan karakter melarikan diri, yang terlihat seperti aksen kuburan terbalik)
ASCII-hanya