असली खजाना हम रास्ते में बनाई गई संख्याएँ थे


23

आपका कार्य एक प्रोग्राम, फ़ंक्शन या स्निपेट लिखना है (हाँ, स्निपेट्स की अनुमति है) जो केवल एक पूर्णांक को आउटपुट करता है। हालाँकि , आपको अपने सबमिशन को अलग- अलग पूर्णांक बनाने में सक्षम होना चाहिए जो अलग पूर्णांक भी बनाते हैं। आप पिछले उपसर्गों में दिखाई देने वाले किसी भी बाइट का उपयोग नहीं कर सकते हैं। उदाहरण के लिए, हम उपसर्ग कर सकते हैं:

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

नियम

  • आपका लक्ष्य सबसे अनोखा पूर्णांक बनाने और बनाने का प्रयास करना है, जबकि उन्हें शून्य के करीब रखना है।
    • स्कोरिंग प्रणाली है ((number of unique integers)**3)/(sum of absolute values), जहां आपका स्कोर जितना अधिक होगा, उतना बेहतर होगा। ऊपर के उदाहरण स्कोर (43)/(1+|-5|+|-1|+9)=64/16=4
  • कम से कम दो अद्वितीय पूर्णांक (शून्य से कोई विभाजन नहीं होना चाहिए!)
  • कृपया अपने उत्तर को इसी तरह प्रारूपित करें:
# Language, \$(4^{3})/16 = 4\$

   1-6/3+0xA (the full program)
  • उल्लेख करें यदि आपका सबमिशन स्निपेट्स से बना है जो किसी मूल्य, फ़ंक्शन या पूर्ण कार्यक्रमों का मूल्यांकन करता है।
  • उपसर्गों में से प्रत्येक को सूचीबद्ध करें और (वैकल्पिक रूप से) एक स्पष्टीकरण कि वे कैसे काम करते हैं।

2
अंतिम प्रस्तुत लाइन 2 (मैं मामले में यह सही करने के लिए मैंने गलत समझा संपादित नहीं की है) के लिए 3 बदली है प्रकट होता है
trichoplax

4
यह मुझे ऐसा लगता है कि आप पायथन में एक पास-इष्टतम स्कोर प्राप्त कर सकते हैं, उसके print 123-len(open(__file__).read())बाद लिख सकते हैं, #तब तक प्रत्येक बाइट का उपयोग नहीं किया जाता है (कुछ नियंत्रण वर्णों को छोड़कर), उपसर्गों के साथ उन बाइट्स को काटते हुए। 123 को समायोजित करें जो कुछ भी हो, जिसमें आउटपुट की सीमा लगभग 0. है। बेहतर स्कोर के लिए, प्रोग्राम के गैर-टिप्पणी वाले हिस्से को केवल 7 अलग-अलग पात्रों का उपयोग करके फिर से लिखा जा सकता है, जो कि पायथन के लिए सार्वभौमिक हैं। क्या यह सब मान्य है?
xnor

1
@ जॉकिंग मैं नकारात्मक शामिल करता हूं, जब अंतर नकारात्मक होता है।
xnor

1
@JoKing यह विधि उन भाषाओं में भी की जाती है जहाँ एक स्ट्रिंग को बिना किसी बंद उद्धरण के लिखा जा सकता है।
xnor

2
मुझे विश्वास नहीं हो रहा है कि कुछ अन्य लोगों ने इसे देखा और पहले की टिप्पणी को गलत ठहराया। : PI ने अभी देखा कि @trichoplax ने टाइपो को भी देखा।
पीटर कॉर्ड्स

जवाबों:


19

TI-बेसिक (TI-84 + सीई), 2383/14161=952.0

118-length("+×/ABCDEFGHIJK...

टीआई-बेसिक में 256 बाइट्स टूट जाते हैं:

  • 240 बाइट्स एक-बाइट टोकन हैं जिन्हें उद्धृत किया जा सकता है
  • 11 दो-बाइट टोकन की शुरुआत हैं
  • 2 अप्रयुक्त हैं
  • शेष 3 रहे हैं ", और न्यू लाइन जो तोड़ तार अक्षर।

स्ट्रिंग में पहले से ही नहीं पहले श्रेणी में सभी 237 स्वीकार्य बाइट्स शामिल हैं 118-length(

कार्यक्रम 118 से नीचे -119 तक 238 अलग-अलग पूर्णांकों की गणना करता है। इसलिए स्कोर

2383n=119118|n|=952.


1
क्या स्ट्रिंग को समाप्त नहीं करना होगा "जो कि एक बाइट है जो पहले से ही इस्तेमाल किया गया था? या क्यों नहीं? मैं इससे थोड़ा भ्रमित हूं अन्य उत्तरों में भी ऐसा लगता है कि शायद मुझे कुछ याद आ रहा है।
findusl

2
@findusl TI-BASIC में स्वचालित रूप से आपके लिए कोष्ठक और तार को पूरा करने की प्रवृत्ति है। इसलिए आप लिख सकते हैं mean({1,2,3, जो पूरा होगा mean({1,2,3})। तार भी पूरा हो गया है: को Disp "Hi!पूरा करता है Disp "Hi!"। इसलिए अंतिम समापन उद्धरण अनावश्यक है
कॉनर ओ'ब्रायन

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

इसे ऑनलाइन आज़माएं! (सभी उपसर्ग)

टूट - फूट:

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

क्या आप <बाइट का पुन: उपयोग नहीं कर रहे हैं? या कि एक अलग बाइट के लिए एक प्लेसहोल्डर है?
findusl

1
@findusl नहीं - <esc>भागने कुंजी का प्रतिनिधित्व करता है और <c-char>ctrl + चार का प्रतिनिधित्व करता है
दरवाज़े

11

पेंगुइन , 256³ / 16384 = 1024

समाधान का स्रोत कोड निम्नानुसार वर्णित किया जा सकता है:

183662808731984092                                         0x00 bytes
4775233027031586288                                        0x01 bytes
27                                                         0x02 bytes
81                                                         0x03 bytes
128931291729852829668                                      0x04 bytes
108                                                        0x05 bytes
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4851795238070470150902081892625349                         0x8D bytes
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175859350926767157582791190333845259132939142561591365     0xFA bytes
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4748202475022713254735362139013821996589386504264297360    0xFC bytes
29655101330505                                             0xFD bytes
128201466825613257877854777753373193907913435615136028720  0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes

पिछले रनों के साथ संयुक्त बाइट्स के प्रत्येक रन में एक उपसर्ग होता है, और पूरा कार्यक्रम ऊपर वर्णित सभी स्ट्रिंग है।

यह एक इष्टतम स्कोर है। (कार्यक्रम -127 से 128 समावेशी तक सभी संख्याओं को प्रिंट करते हैं।)

इस उत्तर को बनाने के लिए, पहले, मैंने स्ट्रिंग को प्रिंट करने के लिए ब्रेनफक प्रोग्राम बनाने के लिए जावा उत्तर की प्रतिलिपि बनाई । मैंने इसे इस mainफ़ंक्शन के साथ चलाया :

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

फिर, मैंने इस रूबी स्क्रिप्ट को यहां दिखाए गए मानव-पठनीय पाठ को बनाने के लिए चलाया:

#!/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"

सबसे बड़ा कार्यक्रम 35945718967643101920367743454695797167648804468538578264427 बाइट्स (और प्रिंट -90) हैं।


1
गति और व्यावहारिक समस्या को सुलझाने के कौशल के लिए सहारा!
3

8

(1113)/2530=540.5656126482213

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

( रूपांतरण के लिए इस दिलचस्प उपयोगिता का श्रेय ।) उपरोक्त प्रभावी रूप से निम्नलिखित कोड में मूल्यांकन करता है:

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

... जो एक प्रॉक्सी बनाता है जो एक संपत्ति नाम की लंबाई को एक संख्या में परिवर्तित करता है। पहला उपसर्ग एक वर्ण को समाप्त .करता है, और केवल 6 अद्वितीय वर्णों का उपयोग करता है, को छोड़कर .$। अगले 63 स्लाइस में 63 अन्य एकल-बाइट वर्ण हैं जो जेएस ऑब्जेक्ट प्रॉपर्टी के नाम से मान्य हैं। अगले 44 स्लाइस अप्रयुक्त बाइट्स से बने दो-बाइट मान्य वर्णों को जोड़ते हैं। अंतिम स्लाइस संलग्न &'', |""**"", और ,~{}। पहले टुकड़ा रिटर्न -2, दूसरी 2, तो -3, फिर 3, -4, 4, -5, आदि पिछले तीन स्लाइस जो कर रहे हैं करने के लिए 0, 1और -1


4

43/4=16

इसके लिए हम ब्रेन-फ्लैक स्निपेट्स का उपयोग करते हैं, इसलिए हम इस मूल्य से चिंतित हैं कि वे क्या करते हैं।

पूरा कार्यक्रम:

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

उपसर्ग और आउटपुट

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

इसे ऑनलाइन आज़माएं!

व्याख्या

यह शायद सबसे अच्छा मस्तिष्क-फ्लैक का जवाब है क्योंकि हमने सभी चरित्रों का उपयोग किया है मस्तिष्क-फ्लैक के बारे में परवाह है, और हम संख्याओं को शून्य के करीब नहीं कर सकते हैं।

पहला स्निपेट है <>, <>स्टैक को बदलता है लेकिन अन्यथा शून्य का मूल्यांकन करता है, इसलिए हम शून्य से शुरू करते हैं। यहां से हम इसे जोड़ते हैं (())। हम सिर्फ किया सकता है (), क्योंकि वह भी एक है, लेकिन अगर हम एक एक धक्का अब हम साथ बाद में इसे याद करने में सक्षम हो जाएगा {}और []। यह बहुत ही एकमात्र तरीका है {}और []मूल्य बना सकता है इसलिए हम ऐसा करते हैं। चूंकि स्टैक पर []1 के रूप में एक बात है, इसलिए हम [[][][]]हमें -2 में लाने के साथ कुल से 3 घटाते हैं । अंत में हम {}उस पॉप का उपयोग करते हैं जिसे हमने पहले पुश किया था, जो हमें -1 तक ला रहा है।


4

(1 13)/32=४१.५९,३७५

परीक्षण और त्रुटि के दो घंटे के बाद मैं अंत में से सभी पूर्णांकों मिला -4करने के लिए +5(और 7)।

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

यह निम्नलिखित उपसर्ग स्निपेट के साथ एक स्निपेट है:

# -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

इसे ऑनलाइन आज़माएं!


1
आप उस पुनरावृत्ति को ठीक करने के 2*2लिए 22&7(-3 और 4 के बजाय -3 और 6) को बदल सकते हैं *
केविन क्रूज़सेन

सुझाव के लिए @KevinCruijssen धन्यवाद।
अंडा

आप खाली कर सकते हैं 'और _का उपयोग करके cmp(cmp,())बजाय cmp('','_')अगर वह उपयोगी है।
गेहूं जादूगर

आप सूची अनुभाग को थोड़ा छोटा और बहुत आसान कर सकते हैं --[-1-1-1-1-1], लेकिन ऐसा करने से आपके स्कोर में सुधार नहीं होता है।
गेहूं जादूगर

3

बैच, 176³ / 7744 = 704

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

पहले उपसर्ग में बृहदान्त्र शामिल है और संख्या -88 को आउटपुट करता है। इसके बाद के उपसर्गों में आरोही संख्या और अंतिम कार्यक्रम में आउटपुट 87 आते हैं।


3

(2493)/15,500=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

बच गए पात्रों में से सभी सिंगल बाइट्स हैं - मैंने नकल और चिपकाने में आसानी के लिए इस रूप में कार्यक्रम प्रस्तुत किया है। यह लिंक पूर्ण कार्यक्रम से बच गए पात्रों के बिना है; जैसा कि देखा जा सकता है कि उपयोग किए जाने वाले अप्रमाणित पात्रों के लिए बहुत सारे प्लेसहोल्डर हैं।

उपयोग किया गया दृष्टिकोण इसके समान है lirtosiast के TI-BASIC उत्तर के , हालांकि मैंने अन्य प्रविष्टियों को ब्राउज़ करने से पहले अपना उत्तर लिखा था। कॉर्ड निम्नानुसार काम करता है:

-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

स्ट्रिंग 0 (NULL), 127 (बैकस्पेस), और बॉयलरप्लेट वर्णों ( -C\|l") के अलावा हर एक-बाइट वर्ण से बना है । इस तरह, स्ट्रिंग के प्रत्येक चरित्र को एक बार में 248 से 0.8 की लंबाई के तार उत्पन्न करने के लिए हटाया जा सकता है। इन कार्यक्रमों के परिणामस्वरूप 15500 के पूर्ण मान के साथ सीमा [-124 - 124] होती है।

इस कार्यक्रम का उपयोग करके कोड तैयार किया गया था :

+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

और इस कार्यक्रम के साथ प्रत्येक काटे गए कार्यक्रमों से उत्पन्न संख्याओं का योग :

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

जेली , 984

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

इसे ऑनलाइन आज़माएं!

आप उपसर्गों को देख सकते हैं (एक डबल न्यूलाइन द्वारा अलग) यहां

मुद्रित संख्याएँ क्रम में हैं:

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

1 13/53=२५.११३२०७५४७२

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

उपसर्ग:

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)

इष्टतम नहीं है, लेकिन सबसे अच्छा मैं व्यक्तिगत रूप से इस विशेष संरचना के साथ मिल सकता है।


2

63/9=24

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

यहाँ उपसर्ग हैं और वे क्या उत्पादन करते हैं

 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`

इसे ऑनलाइन आज़माएं!

परिभाषित करता है nजो पैदा करता है-20, -19, ..., 18, 19 प्रकार की लंबाई के आधार पर करता है G

Gजैसा दिखता है :: G = G, वैसा ही सबसे छोटा उपसर्ग , मिश्रित आयात और परिभाषा से पहले होता है। जैसा दिखता है , उसी तरह
का सबसे बड़ा उपसर्ग पूर्व में हुआ।G:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`


मुझे लगता है कि मैं इस चुनौती में कुछ याद कर रहा हूं, लेकिन ... यह उत्तर (और कुछ मुट्ठी भर) "बाइट्स का पुन: उपयोग कर रहा है जो पिछले उपसर्गों में दिखाई दिए हैं", है ना?
ब्लैकहोल

4
@Blackhole जैसा कि मैंने (और मैं दूसरों को मानता हूं) इसे समझा, आप एक ही उपसर्ग में कई बार बाइट्स का उपयोग कर सकते हैं, लेकिन बाद के वाले में नहीं।
1

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

इसे ऑनलाइन आज़माएं!

अन्य प्रविष्टियों की तरह, एक स्ट्रिंग की लंबाई का उपयोग करता है। प्रारंभिक स्निपेट में वर्ण, साथ ही साथ\r (textareas इस के साथ अच्छा नहीं खेलते हैं) और {(स्ट्रिंग प्रक्षेप चरित्र) स्ट्रिंग से बाहर रखा गया है।

पहला कार्यक्रम है _Ê-122}$($", जब आप जोड़ते हैं तो प्रत्येक प्रोग्राम एक वर्ण से अधिक लंबा होता है \\


यह पूरी तरह से सुनिश्चित नहीं है कि यह वैध है, यह नहीं जांचा गया है कि स्ट्रिंग में स्निपेट का कोई वर्ण नहीं है
ASCII-only

मैं सुझाव दे रहा था कि मैं उसके 122साथ जगह बदलूंगा #zलेकिन इससे आपका स्कोर नहीं सुधरेगा।
झबरा

1

(63)/12=18

0x1-2+3|True*~4

यह एक स्निपेट है।

अलग-अलग उपसर्ग स्निपेट हैं:

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)

पायथन के बिटवाइज़ ऑपरेटर नकारात्मक पूर्णांक को दो के पूरक के रूप 1में दाईं ओर असीमित एस के साथ मानते हैं , इसलिए -5उस के रूप में माना जाता है ...11111011, जब ओर्डेड अपरिवर्तित ...000010देता है ...11111011


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 एक्सेल के सूत्र संरचना के साथ चुनौती अच्छी तरह से जेल नहीं करती है


1

रुनिक टोना , 117 3 /3422 = 468.034

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

इसे ऑनलाइन आज़माएं!

और उपसर्ग

प्रिंट [-58,58] समावेशी।

एक समान रणनीति का उपयोग अन्य प्रविष्टियों के रूप में करता है, उत्तरोत्तर लंबी स्ट्रिंग का उपयोग करके, इसकी लंबाई प्राप्त करना, घटाना B(दशमलव मान 66), और परिणाम प्रिंट करता है। एक गैर-मुद्रण वर्ण का उपयोग करता है (कोड ब्लॉक में दिखाया गया है क्योंकि जाहिर है कि &#128;HTML में मूल्यांकन होता है) बाइट मूल्य का उपयोग करने के लिए 0x80(और अन्यथा अनुपयोगी 0x00) और अधिक मूल्य का स्कोर करता है। केवल 127 तक के बाइट मूल्य जो किसी भी फैशन में प्रतिनिधित्व नहीं कर सकते हैं 10और 13, क्योंकि वे प्रोग्राम को दो लाइनों के लम्बे होने का कारण बनेंगे और निर्देश सूचक द्वारा सही ढंग से नहीं पढ़ा जाएगा।


1

चारकोल , 95co / 2256 = 380.04211 96 230/2304 = 384

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

पहला उपसर्ग पहले 7 बाइट्स है और संख्या को आउटपुट करता है 47इसे ऑनलाइन आज़माएं! इसके बाद के उपसर्गों में अवरोही संख्या और अंतिम कार्यक्रम आउटपुट होता है -48इसे ऑनलाइन आज़माएं! संपादित करें: @ ASCII- केवल के लिए 3.95789 द्वारा बढ़ा हुआ स्कोर।


मुझे लगता है कि आपको स्ट्रिंग के शुरू या अंत में एक गैर-एएससीआईआई चरित्र से बचने में सक्षम होना चाहिए? (बच चरित्र का उपयोग करके, जो एक फ़्लिप किए गए गंभीर लहजे की तरह दिखता है)
एएससीआईआई-केवल
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