Nol dalam interval

14

Tugas Anda adalah menulis sebuah fungsi atau program yang mengambil dua bilangan bulat non-negatif idan k( ik), dan mencari tahu berapa banyak nol yang akan Anda tulis jika Anda menulis semua bilangan bulat dari ihingga k(inklusif) dalam basis pilihan Anda pada sebuah karya dari kertas. Keluarkan bilangan bulat ini, jumlah nol, ke stdout atau serupa.

-30%jika Anda juga menerima argumen ketiga b, basis integer untuk menuliskan angka-angka. Setidaknya dua basis harus ditangani untuk mencapai bonus ini.

  • Anda dapat menerima input di pangkalan apa pun yang Anda suka, dan Anda dapat mengubah pangkalan di antara kotak uji.
  • Anda dapat menerima argumen i, kdan secara opsional bdalam urutan yang Anda suka.
  • Jawaban harus menangani setidaknya satu pangkalan yang bukan unary.

Kasus uji (dalam basis 10):

i k -> output
10 10 -> 1
0 27 -> 3
100 200 -> 22
0 500 -> 92

Ini adalah kode-golf; byte paling sedikit menang.

Filip Haglund
sumber
2
Jika Anda dapat menggunakan basis apa pun yang Anda inginkan dari kasus ke kasus, tidak bisakah Anda melakukannya di basis k dan mencetak 0 atau 1, tergantung pada apakah i = 0?
StephenTG
4
Anda mungkin ingin mengecualikan unary sebagai basis, atau masalah ini sepele: dapatkan input, cetak 0.
Mego
Bisakah Anda menambahkan beberapa test case untuk pangkalan lain?
Morgan Thrapp
3
Saya pikir ini akan lebih menarik jika argumen dasar diperlukan. "Basis pilihanmu" aneh bagiku.
Alex A.
1
Ya, @AlexA. tetapi sudah terlambat untuk mengubahnya sekarang, 10 jawaban.
Filip Haglund

Jawaban:

17

Jelly, 1 byte

¬

Ini menggunakan basis k+2, dalam hal ini ada 0 iff tunggal iadalah 0. Dibutuhkan dua argumen, tetapi menerapkan logika TIDAK untuk hanya yang pertama.

Jika kami tidak ingin menipu:

7 byte - 30% = 4,9

-1.1 poin oleh @Dennis

rb⁵$¬SS

Ini mendapat bonus.

             dyadic link:
r            inclusive range
 b⁵$           Convert all to base input.
    ¬          Vectorized logical NOT
     S         Sum up 0th digits, 1st digits, etc.
      S        Sum all values
lirtosiast
sumber
7
Ini adalah program Jelly kedua yang saya tulis di ponsel saya.
lirtosiast
13
Sial, 1 byte? Beri kami kesempatan.
Rɪᴋᴇʀ
2
Ini dapat dengan mudah dilakukan dalam beberapa byte dalam bahasa lain. Saya katakan tetap pada versi yang tidak curang.
ETHproductions
13
@ ETHproductions Aturan pertanyaan secara eksplisit memungkinkan melakukan hal ini. Cheaty atau tidak, itu adalah jawaban dari aturan yang diminta.
Dennis
8

Python 2, 36 byte

lambda a,b:`range(a,b+1)`.count('0')

Kredit untuk ikan lumpur untuk trik ``.

Dantal
sumber
1
Selamat Datang di Programming Puzzles & Code Golf! Ini jawaban pertama yang bagus. :)
Alex A.
Wow! I didn't know it's working!
Dantal
6

05AB1E, 3 1 byte

Uses base k+2 like the Jelly answer, Code:

_

Explanation:

_  # Logical NOT operator

3 byte non-cheating version:

Code:

Ÿ0¢

Explanation:

Ÿ    # Inclusive range
 0¢  # Count zeroes

The bonus gives me 3.5 bytes due to a bug:

ŸB)0¢

Explanation:

Ÿ      # Inclusive range
 B     # Convert to base input
  )    # Wrap into an array (which should not be needed)
   0¢  # Count zeroes

Uses CP-1252 encoding.

Adnan
sumber
How does this work?
lirtosiast
@ThomasKwa Explanation added
Adnan
5

Japt, 3 bytes

+!U

Uses base k+2, as the Jelly answer. There is a zero iff i==0. Test it online!

Better version, 10 8 bytes

UòV ¬è'0

This one uses base 10. Test it online!

Bonus version, 14 12 bytes - 30% = 8.4

UòV msW ¬è'0

Sadly, with the golfing I did, the bonus is no longer worth it... Test it online!

How it works

UòV msW ¬è'0   // Implicit: U = start int, V = end int, W = base
UòV            // Create the inclusive range [U..V].
    msW        // Map each item by turning it into a base-W string.
        ¬      // Join into a string.
         è'0   // Count the number of occurances of the string "0".
ETHproductions
sumber
5

ES6, 91 86 - 30% = 60.2 bytes

(i,k,b=10)=>([...Array(k+1-i)].map((_,n)=>(i+n).toString(b))+'0').match(/0/g).length-1

Or save 3 (2.1) bytes if b doesn't need to default to 10.

Best non-bonus version I could do was 65 bytes:

(i,k)=>([...Array(k+1).keys()].slice(i)+'0').match(/0/g).length-1

Edit: Saved 5 bytes by using @edc65's zero-counting trick.

Neil
sumber
As I don't manage to get votes for my answer, I'll upvote yours (at least there my name inside)
edc65
4

Seriously, 10 bytes

'0,,u@xεjc

Explanation:

'0,,u@xεjc
'0,,u       push "0", i, k+1
     @x     swap i and k+1, range(i, k+1)
       εjc  join on empty string and count 0s

Try it online!

With bonus: 11.9 bytes

'0,,u@x,╗`╜@¡`Mεjc

Try it online!

Explanation:

'0,,u@x,╗`╜@¡`MΣc
'0,,u@x             push "0", range(i, k+1)
       ,╗           push b to register 0
         `   `M     map:
          ╜@¡         push b, push string of a written in base b
               Σc   sum (concat for strings), count 0s
Mego
sumber
3

CJam, 12 10 3 bytes

li!

This uses the shortcut @ThomasKwa does.

If this is not allowed, then here is a 10 byte answer.

q~),>s'0e=

Nice and short! Works like @Mego's Seriously answer.

Thanks @Dennis!

Had fun writing my first CJam answer!

Try it here!

TanMath
sumber
3

T-SQL, 394 Bytes (No bonus)

I figure 'why not', right?

DECLARE @i INT, @k INT SET @i = 100 SET @k = 200  WITH g AS (SELECT @i AS n UNION ALL SELECT n+1 FROM g WHERE n+1<=@k ) SELECT LEN(n) AS c FROM (SELECT STUFF((SELECT REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(n, 1, ''), 2, ''), 3, ''), 4, ''), 5, ''), 6, ''), 7, ''), 8, ''), 9, ''), ' ', '') FROM g FOR XML PATH ('')) ,1,0,'') n ) a OPTION (maxrecursion 0)

And the friendly one:

-- CG!

DECLARE @i INT, @k INT 
SET @i = 100
SET @k = 200

WITH g AS 
(
    SELECT @i AS n
    UNION ALL
    SELECT n+1 FROM g WHERE n+1<=@k
)

SELECT LEN(n) AS c FROM 
(
    SELECT 
        STUFF((SELECT REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(n, 1, ''), 2, ''), 3, ''), 4, ''), 5, ''), 6, ''), 7, ''), 8, ''), 9, ''), ' ', '')
FROM g FOR XML PATH ('')) ,1,0,'') n
) a

OPTION (maxrecursion 0)
Nelz
sumber
that's dedication.
colsw
3

Ruby, 46 - 30% = 32.2 bytes

You could probably golf this more, but at least I get the 30% bonus!

->i,k,b{((i..k).map{|a|a.to_s b}*"").count ?0}

...or without the bonus (27 bytes.)

->i,k{([*i..k]*"").count ?0}

Tips are welcome, still learning this whole "Ruby" thing.

snail_
sumber
Nice answer, you don't need the splat operator when using map, this could save 1 byte. (i..k) is as good as [*i..k] in the first case.
G B
2

Brachylog, 26 bytes

,{,.e?}?:1frcS:0xlI,Sl-I=.

Takes input as a list [i,k].

Explanation

,{    }?:1f                § Unify the output with a list of all inputs which verify the
                           § predicate between brackets {...} with output set as the input
                           § of the main predicate

  ,.e?                     § Unify the input with a number between i and k with the ouput
                           § being the list [i,k]

           rcS             § Reverse the list and concatenate everything into a single
                           § number (we reverse it to not lose the leading 0 if i = 0 when
                           § we concatenate into a single number). Call this number S.

              :0xlI        § Remove all occurences of 0 from S, call I the length of this new
                           § number with no zeros

                   ,Sl-I=. § Output the length of S minus I.
Fatalize
sumber
2

Julia, 48 bytes - 30% = 33.6

f(i,k,b)=sum(j->sum(c->c<49,[base(b,j)...]),i:k)

This is a function that accepts three integers and returns an integer. One of the arguments specifies the base, so this qualifies for the bonus.

Ungolfed:

function f(i, k, b)
    # For each j in the inclusive range i to k, convert j to base
    # b as a string, splat the string into a character array, and
    # compare each character to the ASCII code 49 (i.e. '1'). The
    # condition will only be true if the character is '0'. We sum
    # these booleans to get the number of zeros in that number,
    # then we sum over the set of sums to get the result.
    sum(j -> sum(c -> c < 49, [base(b, j)...]), i:k)
end

Implementing the bonus yields a score just barely better than the not implementing it (34 bytes):

f(i,k)=sum(c->c<49,[join(i:k)...])
Alex A.
sumber
2

Seriously, 2 bytes

This might be taking the Jelly answer trick to the limit, but here is a simple 2 byte Seriously answer.

,Y

Try it online!

TanMath
sumber
2

Pyth, 6.3 bytes, with bonus (9 bytes - 30%)

/sjRQ}EE0

Explanation:

  jRQ     - [conv_base(Q, d) for d in V]
     }EE  - inclusive_range(eval(input), eval(input))
 s        - sum(^, [])
/       0 - ^.count(0)

Try it here

Or 7 bytes without the bonus:

/`}EE\0

Explanation:

  }EE   - inclusive_range(eval(input), eval(input))
 `      - repr(^)
/    \0 - ^.count("0")

Try it here

Or use a test suite

Blue
sumber
I think getting the bonus is worth it: /sjRQ}EE0
FryAmTheEggman
Ehh, it's the same code with a base conversion, I'm pretty sure you know what you're doing, just the problem of a bonus forcing you to try different stuff and count... :P
FryAmTheEggman
2

PHP, 50 Bytes

supports decimal only

<?=substr_count(join(range($argv[1],$argv[2])),0);

supports decimal and binary with Bonus 63

<?=substr_count(join(array_map([2=>decbin,10=>""][$argv[3]],range($argv[1],$argv[2]))),0);

supports decimal,hexadecimal, octal and binary with Bonus 77.7

<?=substr_count(join(array_map([2=>decbin,8=>decoct,10=>"",16=>dechex][$argv[3]],range($argv[1],$argv[2]))),0);

supports base 2 - 36 with Bonus 78.4

<?=substr_count(join(array_map(function($i){return base_convert($i,10,$_GET[2]);},range($_GET[0],$_GET[1]))),0);
Jörg Hülsermann
sumber
Nice collection! Care to do a version 3a including base 64? :D
Titus
@Titus How is the order of base 64? Why not en.wikipedia.org/wiki/Ascii85 or make a little more with all printable ascii chars
Jörg Hülsermann
2

JavaScript (ES6), 50 (71 - 30%)

(n,k,b)=>eval("for(o=0;n<=k;++n)o+=n.toString(b)").match(/0/g).length-1

No bonus, base k+2 is 10 bytes (i,k)=>+!i

No bonus, unary is 8 bytes (i,k)=>0

TEST

f=(n,k,b)=>eval("for(o=0;n<=k;++n)o+=n.toString(b)").match(/0/g).length-1

function go() {
  var i=I.value.match(/\d+/g)
  R.textContent = f(i[0],i[1],i[2])
}

go()
i,k,b:<input id=I value='0,500,10' oninput="go()">
<span id=R></span>

edc65
sumber
If you move the o='0' before the loop your code continues to work even when k<i.
Neil
@Neil nice, but the spec says (i ≤ k). Update I tried this but in fact it does not work for k<i
edc65
Well, it worked for me (and I know the spec guarantees that i <= k, but your code crashes when k < i; by comparison my code only crashes when k < i - 1 !)
Neil
@Neil uh ok now I get it. It does not give a sensible answer but at least does not crash
edc65
1
@ForcentVintier anyway after your input I revised the code saving some bytes
edc65
1

Jolf, 7 bytes

Replace with \x11. Try it here!

Zl♂sjJ0
   sjJ  inclusive range between two numeric inputs
  ♂      chopped into single-length elements
Zl    0  and count the number of zeroes
        implicitly printed
Conor O'Brien
sumber
1

Lua 74 bytes

z,c=io.read,""for a=z(),z()do c=c..a end o,b=string.gsub(c,"0","")print(b)

There's gotta be a more effective way to do this...

I thought I was really onto something here:

c,m,z=0,math,io.read for a=z(),1+z()do c=c+((m.floor(a/10))%10==0 and 1 or a%100==0 and 1 or a%10==0 and 1 or 0) end print(c)

But alas... It keeps getting longer and longer as I realize there's more and more zeroes I forgot about...

Skyl3r
sumber
1

APL, 22 bytes

{+/'0'⍷∊0⍕¨(⍺-1)↓⍳⍵}

This is a monadic function that accepts the range boundaries on the left and right and returns an integer.

Ungolfed:

           (⍺-1)↓⍳⍵}  ⍝ Construct the range ⍺..⍵ by dropping the first
                      ⍝ ⍺-1 values in the range 1..⍵
       ∊0⍕¨           ⍝ Convert each number to a string
{+/'0'⍷               ⍝ Count the occurrences of '0' in the string

Try it here

Alex A.
sumber
1

Haskell, 29 bytes

i#k=sum[1|'0'<-show=<<[i..k]]

I'm using base 10.

Usage example: 100 # 200 -> 22

How it works: turn each element in the list from i to k into it's string representation, concatenate into a single string, take a 1 for every char '0' and sum those 1s.

nimi
sumber
1

MATL, 7 (10 bytes − 30% bonus)

2$:i:qYA~z

Try it online!

This works in release 11.0.2, which is earlier than this challenge.

Explanation

2$:      % implicitly input two numbers and generate inclusive range
i:q      % input base b and generate vector [0,1,...,b-1]
YA       % convert range to base b using symbols 0,1,...,b-1. Gives 2D array
~        % logical negation. Zeros become 1, rest of symbols become 0
z        % number of nonzero elements in array
Luis Mendo
sumber
1

Matlab: 27 bytes

@(q,w)nnz(num2str(q:w)==48)

creates a vector from lower number to larger one, then converts all numbers to string and counts all the '0' symbols.

brainkz
sumber
1

Python 3, 52.

Tried to implement the bonus, but it doesn't seem to be worth it.

lambda a,b:''.join(map(str,range(a,b+1))).count('0')

With test cases:

assert f(10, 10) == 1
assert f(0, 27) == 3
assert f(100, 200) == 22
assert f(0, 500) == 92
Morgan Thrapp
sumber
1
I literally never heard about the assert statement before this comment. Thanks mate!
sagiksp
1

Perl 6, 23 bytes

{+($^i..$^k).comb(/0/)}
  1. creates a Range ( $^i..$^k )
  2. joins the values with spaces implicitly ( .comb is a Str method )
  3. creates a list of just the zeros ( .comb(/0/) )
  4. returns the number of elems in that list ( + )

Usage:

my &zero-count = {…}

for (10,10), (0,27), (100,200), (0,500), (0,100000) {
  say zero-count |@_
}
1
3
22
92
38895
Brad Gilbert b2gills
sumber
You know, that comment at the end of your code makes it seem longer...
ETHproductions
@ETHproductions I usually do that so that if I come up with more than one way to do things that I can see if it is shorter than others. I just keep adding more ways to do it until I come up with what I think is the shortest way.
Brad Gilbert b2gills
1

Mathematica, 39 bytes, 27.3 with bonus

Count[#~Range~#2~IntegerDigits~#3,0,2]&
A Simmons
sumber
1

C# 112 Bytes

int z(int i,int k)=>String.Join("",Enumerable.Range(i,k-i+1)).Count(c=>c=='0')
  1. Create a string with numbers from the first number up to the last number
  2. Count the zero characters in the string
lee
sumber
Welcome to PPCG! I'm not super familiar with C# but I think you could probably save a few bytes if you removed some of the spaces.
0 '
thank you 0, you are right but only a couple bytes. I believe my edited answer removes all the spaces I can. :)
lee
1

PHP, 84 bytes *.7=58.8 (bases 2 to 36)

for(;($v=$argv)[2]>$a=$v[1]++;)$n+=substr_count(base_convert($a,10,$v[3]),0);echo$n;

or

for(;($v=$argv)[2]>$v[1];)$n+=substr_count(base_convert($v[1]++,10,$v[3]),0);echo$n;

takes decimal input from command line arguments; run with -r.

Titus
sumber
For fun: <?=0 supports unary and alphabetic. ;)
Titus
1

PowerShell, 56 54 51 48 42 bytes

param($i,$k)(-join($i..$k)-split0).count-1

Takes input, creates a range with $i..$k then -joins that together into a string, followed by a regex -split command that separates the string into an array by slicing at the 0s. We encapsulate that with ().count-1 to measure how many zeros. That's left on the pipeline, and output is implicit.

Saved 6 bytes thanks to @ConnorLSW

Try it online!


Base-handling in PowerShell is limited and doesn't support arbitrary bases, so I'm not going for the bonus.

AdmBorkBork
sumber
param($i,$k)(-join($i..$k)-split'0').Length-1 works for me, -3, or use .Count-1 to save even more, haven't tested that yet though.
colsw
@ConnorLSW Thanks! Don't need the quotes around '0', so that trimmed off a few more.
AdmBorkBork
nice one, I always forget powershell handles numbers like that.
colsw
0

Java 8, 102 bytes - 30% = 71.4

Why not.

(i,k,b)->{int j=0;for(;i<=k;i++)for(char c:Integer.toString(i,b).toCharArray())if(c==48)j++;return j;}

Without the bonus, 96 bytes (so the bonus actually improves my score!):

(i,k)->{int j=0;for(;i<=k;i++)for(char c:String.valueOf(i).toCharArray())if(c==48)j++;return j;}

This implements the following:

interface Function {
    public int apply(int i, int k, int b);
}
HyperNeutrino
sumber
@mbomb007 The problem is that formatting it this way renders the answer as 102 bytes on the Leaderboard in the question.
HyperNeutrino
That's a flaw with the leaderboard, not the post. Look at how most of the other answers are doing it the same way.
mbomb007
@mbomb007 I'm looking at the answers and I see a ton of different formats, some of which work with the leaderboard, some of which don't.
HyperNeutrino
0

Clojure, 50 49 bytes

#(count(re-seq #"0"(apply str(range %(inc %2)))))

Oh regex is shorter than filtering. Original:

#(count(filter #{\0}(apply str(range %(inc %2)))))

Very basic, uses the set of character \0 to remove others and counts how many were found.

NikoNyrh
sumber