답변:
pΙ2○3V?hG]‘Γ
설명:
pΙ2○3V?hG]‘ Push compressed string of the first half of the string (this is the kind of thing this compressor was made for)
Γ mirror, swap chars & palindromize with one character overlap
"(<>.".∞ηJÀ24£.∞
설명
"(<>." # push this string
# STACK: "(<>."
.∞ # intersected mirror
# STACK: "(<>.<>)"
η # compute prefixes
J # join to string
# STACK: "((<(<>(<>.(<>.<(<>.<>(<>.<>)"
À # rotate left
24£ # take the first 24 chars
# STACK: "(<(<>(<>.(<>.<(<>.<>(<>."
.∞ # intersected mirror
# STACK: "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
웹 사이트에 접속할 수 없었습니다.이 모든 것을 수동으로 알아 내야했습니다. -.-
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+(+(+!+[]+[+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+[!+[]+!+[]]+[+[]])+[])[+!+[]]
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+(+[]+[])[([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[+[]]+(![]+[])[+!+[]]+(![]+[])[+!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(![]+[])[!+[]+!+[]+!+[]]]()[+!+[]+!+[]]
+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[+!+[]+!+[]]+(![]+[])[+!+[]+!+[]]])[+!+[]+(+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+[])]
대부분의 문자는 상대적으로 쉽습니다.
([]["fill"]+[])[13] // '('
([]["fill"]+[])[14] // ')'
JS가 1.1e + 21과 같은 과학적 표기법 번호를 만들고 문자열로 변환 한 후 점 문자를 가져 와서 전체 정지를 얻습니다.
대괄호는 더 어렵습니다. 이탤릭 html 객체 문자열을 생성하고 꺾쇠 괄호를 훔치려면 함수를 실행해야합니다. 이 문제의 주요 요령은 이탤릭체 단어에 대해 'c'를 얻는 것입니다.
여기에는 대괄호를 얻는 데 주로 사용되는 함수에 대한 개선 범위가 많이 있습니다. 기울임 꼴이 더 좋으면 문자 'c'를 얻는 더 쉬운 방법을 기억하지만 이전 파일을 찾아서 찾아야합니다.
console.log( (!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[+!+[]+!+[]]+(![]+[])[+!+[]+!+[]]])[+!+[]+(+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+[])]
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s=>"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
Kevin Cruijssen 덕분에 -2 바이트
RIP C #
;
은 람다 답변으로 계산되지 않습니다. 그리고 ()
하나의 문자 (사용하지 않는로 대체 할 수 null
개체). 두 번째 관련 메타 게시물.
s=>
완벽하게 좋습니다.
삼? 4? @KritixiLithos 덕분에 바이트
i(<>.<>)òÙxlHÄ$xGòxÍî
i(<>.<>) 'insert (<>.<>)
ò ò 'recursively
Ù 'duplicate this (bottom line) down
x 'delete the first character
l 'and break when there's only one character left
H 'go to the top of the buffer
Ä 'duplicate that line up
$x 'delete the last character
G 'and go back to the bottom of the buffer
x 'delete the extra ) left by the loop
Íî 'and remove all newlines
hx
그냥로 대체 할 수 있습니다 X
.
dkHxògJ
1 바이트보다 짧습니다 dkVHgJ0x
.
Íî
것은 너무 일
i(<>.<>)òÙxlHÄ$xGòddÍî
23
lX
을 위해 xl
D : 너무
cat(substring('(<>.<>)',-4:6,2:12),sep='')
R에서 부분 문자열이 작동하는 방식을 이용하는 매우 간단한 프로그램입니다. 따라서 substring('(<>.<>)',-4:6,2:12)
다음 벡터를 생성합니다.
> substring('(<>.<>)',-4:6,2:12)
[1] "(<" "(<>" "(<>." "(<>.<" "(<>.<>" "(<>.<>)" "<>.<>)" ">.<>)"
[9] ".<>)" "<>)" ">)"
cat
빈 구분 기호를 사용하면 필요한 형식으로 STDOUT에 출력됩니다.
간단한 puts()
솔루션 보다 2 4 바이트가 적습니다 . 우아한 재귀 솔루션이 있어야하지만 지금까지는 오버 헤드가 너무 커집니다.
f(i){for(i=1;i++<12;)printf("%.*s",i,"(<>.<>)"+i/7*i%7);}
나는 결코 codegolf를 수행하지 않았으므로 이것이 나의 첫 번째 것입니다. 의견을 보내 주셔서 감사합니다.
++++[>+++++<-]>[>++>+++>+++>++>++[<]>-]>.>.<.>.>++.<<.>.>.>>++++++.<<<<.>.>.>>.<<<.<.>.>.>>.<<<.>.<<.>.>.>>.<<<.>.>+.<<.>.>>.<<<.>.>.<.>>.<<<.>.>.>.<<<.>.>.<<.>.>.<.>.
내 의견으로는 가장 확실한 해결책을 찾았습니다. 먼저 셀 1-5를 "(). <>"문자 중 하나로 설정했습니다. 그런 다음 올바른 셀로 이동하여 캐릭터를 출력합니다.
업데이트 : 문자가 "테이프"에 표시되는 순서를 변경하여 프로그램을 더 짧고 효율적으로 만듭니다.
업데이트 2 : 잠시 후 내 게시물을 다시 방문하여 셀을 설정하기 위해 짧은 시퀀스를 사용하면 바이트를 절약 할 수 있음을 깨달았습니다.
g=(i=2)=>i>12?'':'(<>.<>)'.slice(i-14,i)+g(i+1)
f=(i=11)=>i?f(i-1)+'(<>.<>)'.slice(i-13,i+1):''
// Both of the above work and are the same size
document.write('<pre>Actual g(): ' + g() + '\nActual f(): ' + f() + '\nExpected: (<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)</pre>');
Rod의 Python 답변 과 동일한 방법을 사용합니다 .
A(<>ι(<ιι.ι.<ι.ιι.‖B
패턴을 감지 한 Destructible Lemon 에게 감사합니다 (-4).
AST :
Program
├A: Assign
│├'(<>': String '(<>'
│└ι: Identifier ι
├Print
│└'(<': String '(<'
├Print
│└ι: Identifier ι
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
├Print
│└ι: Identifier ι
├Print
│└'.<': String '.<'
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
├Print
│└ι: Identifier ι
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
└‖B: Reflect butterfly
└Multidirectional
어떤 ‖B
일은 기본적으로 시각적으로 캔버스를 palindromize입니다.
-a
.
Print
s가 있습니다. 무슨 의미인지 확실하지 않습니다.
나는 이것이 절름발이이지만 다른 솔루션보다 훨씬 짧다는 것을 알고 있습니다 ...
f(){puts("(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)");}
여기 다소 까다로운 버전이 있습니다 ...
char*e="(<>.<>)<>.<>)>.<>).<>)<>)>)";f(i,n){for(n=2;n<8;n++)for(i=0;i<n;i++)putchar(e[i]);puts(&e[i]);}
완전히 까다로운 재귀 버전 ...
char*e="(<>.<>)";main(i,j){if(i<6){for(++i,j=0;j<i;putchar(e[j++]));main(i);}else if(i<12)printf(e+i++-6),main(i);}
배트맨을보고 빈 응시의 군중 아담 웨스트의 명예에 ...
s=>{string b="NanananaNanaBATMAN!NanaBATMAN!NANANanaBATMAN!NANAnanaNanaBATMAN!NANABATMAN!NanaBATMAN!NANABATMAN!naNABATMAN!NANABATMAN!naNANAnaNANABATMAN!naNANANABATMAN!naNABATMAN!naNANAnanaNA",r="",t;for(int i=0;i<176;)if(b[i]=='B'){r+="<>";i+=7;}else{t=b.Substring(i,4);if(t=="Nana")r+="(";else if(t=="nana")r+="<";else if(t=="NAna")r+=">";else if(t=="NANA")r+=".";else r+=")";i+=4;}return r;}
raze"(<>.<>)"{(y-5)_x}/:(!)11
streetster 덕분에 -2 바이트
설명을 편집하십시오.
언어는 오른쪽에서 왼쪽으로 해석됩니다.
작동 원리
raze"(<>.<>)"{(y-5)_x}/:(!)11
(!)11 /yields indices [0-10], right parameter of function (y)
/: /each right: loop right parameter of a dyadic function
{ } /function
"(<>.<>)" /left parameter of function (x)
(y-5) /subtract 5 from y (indices)
_ /remove first y chars from x, or last y chars if negative
raze /flatten string output to produce the final string
_x
1 바이트를 위해, 변화 til 11
에 대한 (!)11
또 다른 1 바이트 절약 = 2 바이트는 :) 저장
map take[2..7]++map drop[1..5]>>=($"(<>.<>)")
하드 코딩 된 솔루션을 능가하는 최초의 Haskell 답변!
이 답변의 작동 방식은 문자열에 적용 할 함수 목록을 작성하는 것 (<>.<>)
입니다. 먼저 왼쪽과 가운데를 만들어
map take[2..7]
크기 2에서 7까지 모든 접두사를 제공합니다. 그런 다음 우리는
map drop[1..5]
크기 6에서 2까지의 모든 접미사를 제공합니다.
함수 목록을 얻은 후에는 더 짧은 monadic bind ( >>=
)를 사용합니다 concatMap
. 우리가 연결하는 함수는 ($"(<>.<>)")
입력을 문자열에 적용하는 것 (<>.<>)
입니다.
이것은 문자열을 만듭니다.
µ
x->"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
지루하지만 Java에서 이것을 더 짧게 만드는 방법은 없습니다. 임시 문자열을 초기화하는 것은 이미 11 바이트입니다. ( String t="something";
)이며 substring
몇 번 사용 하면 확실히 너무 많은 바이트가 필요합니다.
리터럴 리턴에 대한 가장 짧은 대안은 아마도 다음과 같습니다 (58 바이트).
x->"(<(x(x.(x.<(x.x(x.x)x.x)>.x).x)x)>)".replace("x","<>")
=>
추가하려는 경우 "뚱뚱한 화살표"함수 ( )를 사용하여 JavaScript에서도 작동 합니다.
골프가 진행 중입니다.
>++++++++[<+++++>-]>++++++[<++++++++++>-]>++++++[<++++++++++>-]<++>>++++++++[<+++++>-]<++++++<<<.>.<.>.>.<<.>.>.>.<<<.>.>.>.<<.<.>.>.>.<<.>.<<.>.>.>.<<.>.<<+.>.>.>.<<.>.<<.>>.>.<<.>.<<.>>>.<<.>.<<.>.>.<<.>>.<<.
import Data.List
import Control.Arrow
concat$tail$init$uncurry(++)$inits&&&tail.tails$"(<>.<>)"
이것은 출력이있는 문자열 리터럴에 필요한 49 바이트보다 훨씬 길지만 구조 를 활용하기 위해 최선을 다할 수 있습니다. 평소와 같이, 나는 화살표를 좋아 inits &&& tails
하고 입력 하위 문자열 목록과 입력 하위 문자열 목록의 튜플을 생성합니다. 그런 다음 튜플을에 전달 하여이 두 튜플 요소를 하나의 목록에 모으고 ++
해당 목록은 concat
하나의 큰 문자열로 묶습니다. tail
및 init
호출은 중복되지 않도록 (<>.<>)
중간 (의해 생성 된 하나 inits
에 의해 다른 tails
길이를 2 이상의 계좌로의 하위 복용) 및 시작과 끝에서 불필요한 괄호 놓는다.
“(<>.<>)”ḣJœ|ṫJ$ḊṖ
“(<>.<>)”ḣJœ|ṫJ$ḊṖ Main link. No arguments.
“(<>.<>)” Set the argument and return value to s := "(<>.<>)".
J Yield all indices of s, i.e., [1, 2, 3, 4, 5, 6, 7].
ḣ Dyadic head; yield s's prefixes of lengths 1 to 7.
$ Combine the two links to the left into a chain.
J Indices; yield [1, 2, 3, 4, 5, 6, 7].
ṫ Dyadic tail; yield s's postfixes of lengths 1 to 7.
œ| Multiset union; concatenate the results to both sides,
discarding the copy of s. (s is both a prefix and a postfix.)
Ḋ Dequeue; remove the first prefix, i.e., "(".
Ṗ Pop; remove the last postfix, i.e., ")".
main(n){while(write(++n<13,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}
이 종료은 표준 쓰기 때문에 TIO에 타임 아웃 후에 입력이 실패하지만 것이다 단말기에서 종료. 이 프로그램은 특정 평가 순서 (정의되지 않은 동작)에 의존하며 gcc와 같이 작동하지 않습니다.
바이트를 절약 한 아이디어에 대한 @Steadybox에 감사합니다!
$ cat crowd.c
main(n){while(write(++n<12,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}
$ clang -o crowd crowd.c 2> /dev/null
$ ./crowd 0> /dev/null
(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)
main(n){while(++n<13)write(1,n/7*(n-7)+"(<>.<>)",n/7?14-n:n);}
하나의 바이트를 추가로 사용하면 솔루션의 이식성이 훨씬 향상됩니다.
(<
거기에 없습니다.
현재 파이썬 2 답변을 치는 것을 저항 할 수 없었습니다.
print "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
(<>.<>)
."(<" + "(<>" + "(<>." + ... + "(<>.<>)" + "<>.<>)" + ">.<>)" + ... + ">)"