음수가 아닌 정수가 주어지면 n솔루션에서 선택한 언어로 프로그램을 출력해야합니다.n 시간이 출력 된 프로그램의 바이트 수만큼 인 프로그램을 출력해야합니다.

규칙

• 솔루션이 출력하는 프로그램의 언어 및 인코딩을 지정해야하며 솔루션의 다른 입력에 대해 다른 언어 또는 인코딩을 선택할 수 없습니다. 출력 프로그램의 언어는 솔루션의 언어와 같거나 같지 않을 수 있습니다.
• 제출 한 언어 범위의 정수만 처리하면되지만이 규칙을 남용하지 마십시오.

이것은 코드 골프 이므로 바이트 단위의 최단 제출이 이깁니다.

예

n4 라고 가정하자 . 내 솔루션은 f_8(가상) 언어 출력으로 프로그램을 출력한다 j3 1s+/2]!mz. 프로그램 출력의 길이는 3이고 출력의 길이는 3 * 4 = 12이므로 솔루션은 입력 4에 적합합니다.

대신 n1이고 내 프로그램 출력 ffffpfpffp(10 바이트) 이라고 가정하십시오 . ffffpfpffp선택한 언어로 된 프로그램 은 10 * 1 = 10 바이트의 출력을 생성해야합니다.

자바 스크립트 (ES6), 38 바이트

n=>`(x="${n}")=>(x+1/7+1e9).repeat(x)`

데모

let f =

n=>`(x="${n}")=>(x+1/7+1e9).repeat(x)`

function update() {
  let n = I.value,
      func = f(n),
      res = eval(func)();
  O.innerText =
    'n = ' + n + '\n\n' +
    'Function (' + func.length + ' bytes):\n' + func + '\n\n' +
    'Function output (' + res.length + ' bytes):\n' + res;
}
update()

<input type=range id=I min=0 max=100 oninput="update()">
<pre id=O>

젤리 , 10 바이트

;“DL+8×x@⁶

온라인으로 사용해보십시오!

입력의 12경우 12DL+8×x@⁶120 공백을 출력합니다. 온라인으로 사용해보십시오!

brainfuck , 348 바이트

--[>+<++++++]>>--[>+<++++++]>++[-<+>]----[>+<----]>---[-<+>]----[>+<----]>-[-<+>]-[>+<---]>++++++[-<+>]-[>+<---]>++++++++[-<+>]--[>+<++++++]>+++[-<+>]<[<],[->.<]>>>>>.<<<.>>.<<<.>>>.<<<.>>>.<<<.>>...>>>.<<....<<.>>>.<<..<<.>>>..<<.......>>>>.<<<..<<.>.....>>>.<<.>>.>>.<<<<<.>>>>.<<.<<.>>.>.<<<.>>.<<<.>>...<<.>>>..>>.<<.>.<<<.>.<<.>>>.>>.<<<..>>>.

온라인으로 사용해보십시오! 또는 Ungolfed 버전 (즉, 내가 작업해야했던 것)을 참조하십시오

기권

나는 인간적으로 가능하다고 생각했던 것보다 더 많은 시간을 보냈습니다. 나는 그녀가이 일을 할 수있게 해 준 여자 친구에게 감사하고 싶다. 뿐만 아니라 나의 구세주 .

어떻게 작동합니까?

실마리 없음.

어떻게 작동합니까?

모든 출력에는 모두 동일한 코드 스 니펫이 있습니다.

[->+>+>+<<<]>>>>-[<<+>>-------]<<+-----[<[.-]>->[->+<<<+>>]>[-<+>]<<]

그것을 세 부분으로 나누겠습니다 a,b,c

a : [->+>+>+<<<]>>>>               THE DUPLICATOR
b : -[<<+>>-------]<<+-----        THE ADJUSTER
c : [<[.-]>->[->+<<<+>>]>[-<+>]<<] THE PRINTER

입력 i은 단항식으로 전면에 고정됩니다.

iabc

(예 : 입력이 10 인 경우 i = '++++++++++')

복제기 - 두 개의 동일한 번호로 입력 분할 m, n입력에 상응하는,

조절기 - n프로그램 길이와 같도록 조절 합니다

프린터m*n -ASCII 문자를 인쇄 합니다

이 예제의 입력은 newline이며 ASCII 값은 10이므로 입력은 10입니다. 다른 작은 숫자를 테스트하려면 원하는 ,많은 숫자로 바꾸십시오 +.

체다 , 10 9 바이트

(*)&" "*9

어색한 양성자 폴리 글롯> _ <

하스켈 , 55 바이트

f n=(++)<*>show$"main=putStr$[1.."++show n++"*2]>>'#':"

온라인으로 사용해보십시오! 사용법 예 : f 1다음 54 바이트 프로그램을 생성합니다.

main=putStr$[1..1*2]>>'#':"main=putStr$[1..1*2]>>'#':"

온라인으로 사용해보십시오! 다음 54 바이트 출력을 생성합니다.

#main=putStr$[1..1*2]>>'#':#main=putStr$[1..1*2]>>'#':

파이썬 3- > HQ9 +, 11 바이트

'Q'.__mul__

해야 했어요

온라인으로 사용해보십시오!

자바 (8), (175) 174 바이트

interface M{static void main(String[]a){System.out.printf("interface M{static void main(String[]a){int i=(88+%s)*%s;for(;i-->0;System.out.print(0));}}",a[0].length(),a[0]);}}

예 :

n=1출력 :

interface M{static void main(String[]a){int i=(88+1)*1;for(;i-->0;System.out.print(0));}}

89 개의 0을 출력하는 (length = 89) :

00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

n=10출력 :

interface M{static void main(String[]a){int i=(88+2)*10;for(;i-->0;System.out.print(0));}}

900의 0을 출력하는 (길이 = 90) :

000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

n=100출력 :

interface M{static void main(String[]a){int i=(88+3)*100;for(;i-->0;System.out.print(0));}}

9100 0을 출력하는 (길이 = 91) :

00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000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설명:

interface M{                                // Class:
  static void main(String[]a){              //  Mandatory main method
    System.out.printf("interface M{static void main(String[]a){
                                            //   Print a new program with:
      int i=(88+%s)*%s;                     //    Integer containing (88*A)+B
      for(;i-->0;System.out.print(0));}}",  //    And print that many zeroes
        a[0].length(),                      //     Where A is the length of the number
                                            //     (0-9 = 1; 10-99 = 2; 100-999 = 3; etc.)
        a[0]);}}                            //     and B is the number itself

RProgN 2 , 7 5 바이트

«•.* 

후행 공간으로

설명

«•.* 
«    # Yield a function from the remaining string.
 •.  # Append a space and stringify, which builds the original program (Because stringifying a function removes noops, which spaces are)
   * # Repeat the implicit input times.

온라인으로 사용해보십시오!

CJam, 8 13 바이트

q_,S\" 8+*S*"

온라인으로 사용해보십시오

생성 된 프로그램은 공백을 출력하므로 구분하기가 어렵습니다.

Ly , 29 바이트

"\"9>("&onu")[<&s&ol>1-]<;"&o

온라인으로 사용해보십시오! (전처리 버그로 인해 현재 작동하지 않습니다) 모두 좋습니다!

Python → TECO, 20 바이트

대답은 파이썬이고 생성 된 코드는 TECO입니다. 파이썬을 반환하는 함수이고 VV12345\VV반복하여 N 개의 시간. TECO에 대한 설명은 여기 를 참조 하십시오 .

'VV12345\VV'.__mul__

PHP, 47 + 1 바이트

<?="<?=str_pad(_,",strlen($argn)+18,"*$argn);";

밑줄 하나 뒤에 공백을 인쇄합니다. ;
로 파이프로 실행 또는로 -F출력 된 프로그램을 실행하십시오 .-f-F

64 자리를 초과하여 입력하는
경우 PHP_INT_MAX(현재) 보다 훨씬 높습니다 .

그러나 PHP_INT_MAX-18 보다 큰 입력에 대해서는 실패합니다 ... 여전히 자격이 있습니까?

PHP → Python 2, 40 + 1 바이트

print "A"*<?=13+strlen($argn),"*",$argn;

반복되는 As 를 인쇄하는 Python 프로그램을 인쇄합니다 . 로 파이프로 실행하십시오 -F.

배치 → 숯, 22 바이트

바이트 형식이므로 어떤 인코딩을 사용해야하는지 잘 모르겠습니다. 다음은 Windows-1252로 해석 된 바이트입니다.

@set/p=Á%1ñªÉñ«Ìñ¹<nul

PC-850과 동일한 바이트 :

@set/p=┴%1±¬╔±½╠±╣<nul

Charcoal의 코드 페이지에서 동일한 바이트 :

@set/p=Ａ%1θ×Ｉθ⁺Ｌθ⁹<nul

결과 숯 프로그램의 Plus(Length(Cast(n)), 9)길이 는 바이트입니다.

Ａ       Assign
 %1      (String representation of n)
   θ      To variable q
        Implicitly print a number of `-`s equal to:
×        Product of:
 Ｉθ       Cast of q to integer
 ⁺        Sum of:
  Ｌθ       Length of q
  ⁹        Integer constant 9

재귀 , 16 15 바이트

++"*"V*15a'"*"'

온라인으로 사용해보십시오!

n=2인쇄물의 입력을위한 것 :

*30"*"  

출력 30 *. 온라인으로 사용해보십시오!

자바 스크립트 (ES8), 43 41 39 바이트

n=>`f=_=>"".padEnd(${n}*(88+f).length)`

그것을 테스트

생성 된 함수의 출력은 *이 스 니펫에서 s 로 대체되는 공백 문자열입니다 .

g=
n=>`f=_=>"".padEnd(${n}*(88+f).length)`

o.innerText=(h=n=>`Function: ${x=g(n)}\nLength:   ${x.length}\nOutput:   "${x=eval(x)().replace(/./g,"*")}"\nLength:   `+x.length)(i.value=10);oninput=_=>o.innerText=h(+i.value)

<input id=i type=number><pre id=o>

R , 46 바이트

function(n)sprintf("cat(rep('a',%d*23),'')",n)

온라인으로 사용해보십시오!

문자열을 반환하는 익명 함수

cat(rep('a',n*23),'')

어떤 인쇄 a(의 그 a뒤에 공백) 23 시간의 n시간. ''그렇지 않으면 cat마지막 공백 문자를 인쇄하지 않기 때문에 필요했습니다 .

C, 94 바이트

main(int c,char**a){if(c>1){c=atoi(a[1]);if(c>0&&c<0xFFFFFF){c*=94;while(c--)printf("r");}}}

이것은 표준 바이트 C가 쓰여야한다고 마지막 \ n을 포함하는 94 바이트입니다. 프로그램 인수가 없거나 <= 0이거나> 0xFFFFF 인 경우 아무 것도 표시하지 않으면 (프로그램의 길이) * (프로그램의 인수)로 'r'문자로 리턴

C:\>nameProg.exe 1
rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

MATLAB (63 바이트)

a=@(n)['repmat(''a'',1,',num2str(n*(15+numel(num2str(n)))),')']

예를 들면 다음과 같습니다.

>> a(5)

ans =

repmat('a',1,80)

과:

>> repmat('a',1,80)

ans =

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa