Tugas Anda adalah menulis sebuah fungsi atau program yang mengambil dua bilangan bulat non-negatif idan k( i≤ k), 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.

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

Python 2, 36 byte

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

Kredit untuk ikan lumpur untuk trik ``.

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.

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".

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.

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

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!

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)

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.

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.

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)...])

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!

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

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);

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>

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

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...

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

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.

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

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.

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

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

Mathematica, 39 bytes, 27.3 with bonus

Count[#~Range~#2~IntegerDigits~#3,0,2]&

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

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.

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.

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);
}

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.