Zauważyłem, że pewna gra ma osobliwy licznik życia, który zamiast się zatrzymać 999, zyskał nową cyfrę - następną liczbą była setka koron👑00 . Po 👑99przyszedł korony sto crownty ( 👑👑0) i ostatni numer, po 👑👑9, była korona sto crownty korona lub 👑👑👑, co byłoby 1110 w systemie dziesiętnym.

Twoim zadaniem jest napisanie programu lub funkcji, która wyświetli ten licznik.

Biorąc pod uwagę liczbę całkowitą z zakresu [0,1110](włącznie na obu końcach), wypisz ciąg trzech znaków gdzie

• każda postać jest z listy 0123456789👑
• korona (👑) może pojawiać się tylko jako postać znajdująca się najbardziej na lewo lub gdy po lewej stronie znajduje się korona
• gdy liczba ta jest odczytywana jako liczba dziesiętna, ale z liczeniem korony jako 10, odzyskujesz pierwotną liczbę

Przypadki testowe

   0 → "000"
  15 → "015"
 179 → "179"
 999 → "999"
1000 → "👑00"
1097 → "👑97"
1100 → "👑👑0"
1108 → "👑👑8"
1110 → "👑👑👑"

Zamiast korony możesz użyć dowolnego znaku nie dziesiętnego. Aby zachęcić do ładnego drukowania, znak korony (sekwencja bajtów UTF8 „\ 240 \ 159 \ 145 \ 145”) liczy się jako jeden bajt zamiast czterech. Twój program nie musi działać dla liczb spoza prawidłowego zakresu.

To jest golf golfowy , więc wygrywa najkrótsza odpowiedź, mierzona w bajtach!

05AB1E , 20 18 bajtów

₄‹i₄+¦ëTð.;„1 „  :

Wykorzystuje miejsca na korony.

Wypróbuj online lub sprawdź wszystkie przypadki testowe .

Wyjaśnienie:

₄‹i               # If the (implicit) input is smaller than 1000:
   ₄+             #  Add 1000 to the (implicit) input
     ¦            #  And remove the leading 1 (so the leading zeros are added)
                  #   i.e. 17 → 1017 → "017"
  ë               # Else:
   Tð.;           #  Replace the first "10" with a space " "
                  #   i.e. 1010 → " 10"
                  #   i.e. 1101 → "1 1"
                  #   i.e. 1110 → "11 "
       „1 „  :    #  Replace every "1 " with "  " (until it no longer changes)
                  #   i.e. " 10" → " 10"
                  #   i.e. "1 1" → "  1"
                  #   i.e. "11 " → "   "

JavaScript (Node.js) , 50 bajtów

f=(n,p=1e3)=>n<p?(n+p+'').slice(1):'#'+f(n-p,p/10)

Wypróbuj online!

TIO oparto na odpowiedzi Arnaulda. Pokaż 👑 jako #.

JavaScript (ES6),  62 46  44 bajty

Zaoszczędzono 2 bajty dzięki @nwellnhof

Wyprowadza korony jako xpostacie.

n=>(n+1e4+'').replace(/1+0/,'xxx').slice(-3)

Wypróbuj online!

W jaki sposób?

$10000$/1+0/xxx

Przykłady:

$$\begin{align}&0 &\rightarrow &\text{ "}\color{red}{\text{10}}\text{000"} &\rightarrow &\text{ "xxx000"} &\rightarrow &\text{ "000"}\\ &123 &\rightarrow &\text{ "}\color{red}{\text{10}}\text{123"} &\rightarrow &\text{ "xxx123"} &\rightarrow &\text{ "123"}\\ &1023 &\rightarrow &\text{ "}\color{red}{\text{110}}\text{23"} &\rightarrow &\text{ "xxx23"} &\rightarrow &\text{ "x23"}\\ &1103 &\rightarrow &\text{ "}\color{red}{\text{1110}}\text{3"} &\rightarrow &\text{ "xxx3"} &\rightarrow &\text{ "xx3"}\\ &1110 &\rightarrow &\text{ "}\color{red}{\text{11110}}\text{"} &\rightarrow &\text{ "xxx"} &\rightarrow &\text{ "xxx"} \end{align}$$

Shakespeare Programming Language, 763 692 690 689 683 bytes

,.Ajax,.Ford,.Page,.Act I:.Scene I:.[Enter Ajax and Ford]Ford:Listen tothy!Ajax:You big big cat.Scene V:.Ajax:Remember the remainder of the quotient betweenI twice the sum ofa cat a big big cat.Ford:You be the quotient betweenyou twice the sum ofa cat a big big cat.Ajax:You be the sum ofyou a pig.Be you nicer zero?If solet usScene V.Ford:You big big cat.[Exit Ajax][Enter Page]Page:Recall.Ford:You be I.Scene X:.Page:Recall.Am I nicer zero?If notopen heart.If notlet usScene L.Ford:You big big big big big cat.Speak thy.Am I worse a cat?If soyou zero.Scene L:.[Exit Page][Enter Ajax]Ford:You be the sum ofyou a pig.Is you nicer a cat?[Exit Ajax][Enter Page]Ford:If solet usScene X.

Try it online!

Uses " " instead of crowns. At the cost of 4 more bytes, this could be modified to show a "visible" character instead.

Explanation:

,.Ajax,.Ford,.Page,.Act I:.Scene I:.[Enter Ajax and Ford]

    Boilerplate, introducing the characters.

Ford:Listen tothy!

    Input a value to Ajax.

Ajax:You big big cat.

    Set Ford's value to 4 (we will be pushing 4 digits from Ajax onto Ford's personal stack).

Scene V:.Ajax:Remember the remainder of the quotient betweenI twice the sum ofa cat a big big cat.Ford:You be the quotient betweenyou twice the sum ofa cat a big big cat.

    DIGIT-PUSHING LOOP: Push Ajax's last digit onto Ford's stack; divide Ajax by 10.

Ajax:You be the sum ofyou a pig.Be you nicer zero?If solet usScene V.

    Decrement Ford; loop until Ford is 0.

Ford:You big big cat.

    Set Ajax's value to 4 (we will pop 3 digits from Ford's stack in the next loop).

[Exit Ajax][Enter Page]Page:Recall.Ford:You be I.

    Pop the top value off Ford's stack, and store that into Page.
    Here, Page will contain 0 if there are no crowns to be drawn,
    and 1 if there are crowns to be drawn.

Scene X:.Page:Recall.Am I nicer zero?If notopen heart.If notlet usScene L.

    DIGIT-DRAWING LOOP: Pop the top value off of Ford's stack and set Ford equal to that value.
    If there are no crowns to be drawn, output Ford's literal value here, and skip the crown-drawing section.

Ford:You big big big big big cat.Speak thy.Am I worse a cat?If soyou zero.

    Draw crown.
    If we are drawing crowns, and Ford contains 0 here, then we are now done drawing crowns, and thus we store 0 into Page.
    (Put in one more "big" for the crown to look like an @ symbol.)

Scene L:.[Exit Page][Enter Ajax]Ford:You be the sum ofyou a pig.Is you nicer a cat?[Exit Ajax][Enter Page]Ford:If solet usScene X.

    Decrement Ajax; loop until Ajax is 1 (i.e. 3 times).

Python 2, 53 bytes

Hats off to Arnauld for -22 bytes. Recursion still wins, though.

lambda k:re.sub("1+0","CCC",`k+10000`)[-3:]
import re

Try it online!

Python 2, 51 bytes

This instead implements tsh's recursive method. Saved 2 bytes thanks to ovs.

f=lambda n,p=1000:n/p and'C'+f(n-p,p/10)or`n+p`[1:]

Try it online!

Python 2, 52 bytes

lambda n:['%03d'%n,'%3s'%`n`.lstrip('1')[1:]][n>999]

Try it online!

Retina 0.8.2, 41 bytes

\b((.)|..)\b
$#2$*00$&
T`d`_#`(?=....)1+0

Try it online! Uses #s instead of 👑s. Link includes test cases. Explanation:

\b((.)|..)\b
$#2$*00$&

Pad 1- and 2-digit numbers to three digits.

T`d`_#`(?=....)1+0

Change leading 1s of 4-digit numbers to #s and delete the next 0.

Jelly, 19 bytes - 0 = 19

<ȷ¬ȧDi0ɓ⁶ẋ⁹Ḋ;+ȷDḊṫʋ

A full program printing the result using a space character as the crown.
(As a monadic Link a mixed list of integer digits and space characters is yielded)

Try it online! Or see the test-suite.

...maybe a recursive implementation will be shorter.

How?

<ȷ¬ȧDi0ɓ⁶ẋ⁹Ḋ;+ȷDḊṫʋ - Main Link: integer, N    e.g. 1010       or   10
 ȷ                  - literal 1000                  1000            1000
<                   - less than?                    0               1
  ¬                 - logical not                   1               0
    D               - to decimal list               [1,0,1,0]       [1,0]
   ȧ                - logical and                   [1,0,1,0]       0
      0             - literal zero                  0               0
     i              - first index - call this I     2               1  (0 treated as [0] by i)
       ɓ            - new dyadic chain with swapped arguments - i.e. f(N, I)
        ⁶           - literal space character       ' '             ' '
          ⁹         - chain's right argument        2               1
         ẋ          - repeat                        [' ',' ']       [' ']
           Ḋ        - dequeue                       [' ']           []
                  ʋ - last four links as a dyad - i.e. f(N, I)
             +ȷ     -   add 1000                    2010            1010
               D    -   to decimal list             [2,0,1,0]       [1,0,1,0]
                Ḋ   -   dequeue                     [0,1,0]         [0,1,0]
                 ṫ  -   tail from index (I)         [1,0]           [0,1,0]
            ;       - concatenate                   [' ',1,0]       [0,1,0]
                    - implicit print                " 10"           "010"

Python 2, 40 bytes

lambda n:'%3s'%`10000+n`.lstrip('1')[1:]

Try it online!

Implements an idea similar to Mr. Xcoder's regex-based answer but without a regex. We remove leading 1's in 10000+n as well as the next character, then pad with spaces to length 3. The result is similar to ovs's solution using lstrip but without needing two cases.

Clean, 87 bytes

Doesn't output crowns (uses c).

import StdEnv,Text
$n#i=3-n/1000-n/1100-n/1110
=("ccc"+lpad(""<+n rem(10^i))i'0')%(i,9)

Try it online!

$ n                   // function $ of `n`
 # i =                // define `i` as (the number of digits that aren't crowns)
  3 -                 // three minus
  n / 1000 -          // 1 if first digit is crown
  n / 1100 -          // 1 if second digit is crown
  n / 1110            // 1 if third digit is crown
 = (                  // the string formed by
  "ccc" +             // prefixing three crowns to
  lpad (              // the padding of
   "" <+ n rem (10^i) // non-crown digits of `n`
  ) i '0'             // with zeroes
 ) % (i, 9)           // and removing the extra crowns

Clean, 99 - 3 = 96 bytes

This one has crowns.

import StdEnv,Text
$n#i=3-n/1000-n/1100-n/1110
=("👑👑👑"+lpad(""<+n rem(10^i))i'0')%(i*4,99)

Try it online!

Japt, 20 bytes

A naïve (and slightly drunk!) port of Arnauld's solution. Uses " for crown.

U+A³ s r"^21*0"_çQÃÅ

Try it

Java 10, 84 83 bytes

n->{for(int p=100,t;p>0;n-=t%12*p,p/=10)System.out.printf("%c",t=n/p>9?46:48+n/p);}

Port of @tsh' C comment.
Uses . instead of crowns.

Try it online.

Alternative approach (84 (87-3) bytes):

n->f(n,1000)String f(int n,int p){return n<p?(n+p+"").substring(1):"👑"+f(n-p,p/10);}

Port of @tsh' JavaScript's answer.

Try it online.

APL (Dyalog Unicode), 32 bytes

1e3∘{⍵<⍺:1↓⍕⍵+⍺⋄'C',(⍵-⍺)∇⍨⍺÷10}

Try it online!

Prefix direct function.

Port of @tsh's JS Answer.

How:

1e3∘{⍵<⍺:1↓⍕⍵+⍺⋄'C',(⍵-⍺)∇⍨⍺÷10} ⍝ Main function, arguments ⍵ and ⍺ (⍵ → input, ⍺ → 1000).
     ⍵<⍺:                          ⍝ If ⍵<⍺
         1↓                        ⍝ Drop (↓) the first element (1) of
           ⍕                       ⍝ Format (⍕); 'stringify'
            ⍵+⍺                    ⍝ ⍵+⍺
               ⋄                   ⍝ else
                'C',               ⍝ Concatenate (,) the literal 'C' with
                         ∇⍨        ⍝ Recursive call (∇) with swapped arguments (⍨)
                    (⍵-⍺)  ⍺÷10    ⍝ New arguments; ⍵ → ⍵-⍺; ⍺ → ⍺÷10

PHP, 71 bytes

for($n=$argn,$x=1e4;1<$x/=10;$n%=$n<$x?$x/10:$x)echo$n<$x?$n/$x*10|0:C;

prints C for the crown. Run as pipe with -nR or try it online.

Haskell, 48 bytes

('2'#).show.(+1000)
n#(a:x)|n==a='c':'1'#x|1>0=x

Try it online!

C,  84  58 bytes

Thanks to @tsh for saving 25 bytes and thanks to @ceilingcat for saving a byte!

f(n,p){for(p=1e3;p/=10;)n-=putchar(n/p>9?46:48+n/p)%12*p;}

Try it online!

sed, 39

48 bytes, but score is 39, since each 👑 counts as 1.

/..../s/^1+0/👑👑👑/
s/^/00/
s/.*(...)/\1/

Try it online!

Perl 6, 38 - 9 = 29 bytes

-2 bytes thanks to Jo King

{~m/...$/}o{S/1+0/👑👑👑/}o*+1e4

Try it online!

Inspired by Arnauld's JavaScript solution.

Clean, 96 bytes

I think Super Mario 3D Land, New Super Mao Bros.2, and Super Mario 3D World have this life counter.

import StdEnv,Text
$n#i=3-n/1000-n/1100-n/1110
=("👑👑👑"+lpad(""<+n rem(10^i))i'0')%(i*4,99

I agree with Clean.

Verify that I’m not cheating.