Question:

Vous recevrez les entiers de début et de fin d'une séquence et devriez lui renvoyer le nombre d'entiers ne contenant pas le chiffre 5. Les numéros de début et de fin doivent être inclus!

Exemples:

1,9 → 1,2,3,4,6,7,8,9 → résultat 8

4,17 → 4,6,7,8,9,10,11,12,13,14,16,17 → résultat 12

50,60 → 60 → résultat 1

-59, -50 → → résultat 0

Le résultat peut en contenir cinq.

Le numéro de départ sera toujours inférieur au numéro de fin. Les deux nombres peuvent être aussi négatifs!

Je suis très curieux de vos solutions et de la façon dont vous le résolvez. Peut-être que quelqu'un d'entre vous trouvera une solution simple de mathématiques pures.

Edit Ceci est un défi code-golf, donc le code le plus court gagne.

JavaScript (ES6), 36 33 octets

Prend les entrées avec la syntaxe de currying (a)(b).

a=>F=b=>b<a?0:!/5/.test(b)+F(b-1)

Formaté et commenté

a =>                 // outer function: takes 'a' as argument, returns F
  F = b =>           // inner function F: takes 'b' as argument, returns the final result
    b < a ?          // if b is less than a
      0              //   return 0
    :                // else
      !/5/.test(b) + //   add 1 if the decimal representation of b does not contain any '5'
      F(b - 1)       //   and do a recursive call to F with b - 1

Cas de test

let f =

a=>F=b=>b<a?0:!/5/.test(b)+F(b-1)

console.log(f(1)(9))
console.log(f(4)(17))
console.log(f(50)(60))
console.log(f(-50)(-59))

Gelée , 8 7 octets

-1 octet grâce à Dennis (sachant que l'indexation dans un nombre traite ce nombre comme une liste décimale)

rAw€5¬S

TryItOnline!

Comment?

rAw€5¬S - Main link: from, to    e.g. -51, -44
r       - range(from, to)        e.g. [-51,-50,-49,-48,-47,-46,-45,-44]
 A      - absolute value         e.g. [51,50,49,48,47,46,45,44]
  w€    - first index of... for €ach (0 if not present)
    5   - five                   e.g. [1,1,0,0,0,0,2,0]
     ¬  - logical not            e.g. [0,0,1,1,1,1,0,1]
      S - sum                    e.g. 5

* La valeur absolue atome Aest nécessaire puisqu'un nombre négatif converti en une liste décimale a des entrées négatives, dont aucune ne sera jamais un 5(l'exemple donné compterait les huit au lieu de deux).

Bash + grep, 17 octets

seq $@|grep -cv 5

Essayez-le en ligne!

2sable , 6 5 octets

Enregistrement d'un octet grâce à Adnan

Ÿ5¢_O

Essayez-le en ligne!

Explication

 Ÿ      # inclusive range
  5¢    # count 5's in each element of the range
    _   # negate
     O  # sum

Remarque: cela fonctionne en raison d'un bogue ¢qui oblige la fonction à s'appliquer à chaque élément au lieu de compter les éléments correspondants dans la liste.

Python2, 59 55 52 51 47 43 42 octets

f=lambda a,b:a<=b and-(`5`in`a`)-~f(a+1,b)

Une solution récursive. Merci à @xnor de me motiver à trouver une solution à l'aide d'opérateurs logiques! Merci également à @JonathanAllan et à @xnor de m'avoir guidé coupé l'octet de 43 à 42!

Autres tentatives à 43 octets

f=lambda a,b:a<=b and-~-(`5`in`a`)+f(a+1,b)
f=lambda a,b:a<=b and 1-(`5`in`a`)+f(a+1,b)

Utilitaires Bash / Unix, 21 octets

seq $*|sed /5/d|wc -l

Essayez-le en ligne!

05AB1E , 8 7 6 octets

Enregistrement d'un octet grâce à Adnan

Ÿ5.å_O

Essayez-le en ligne!

Explication

Ÿ         # inclusive range
 5.å      # map 5 in y for each y in the list
    _     # logical negation 
     O    # sum

Pyth, 9 8 bytes

Saved a byte thanks to FryAmTheEggman!

lf-\5T}E

Explanation:

        Q # Input
      }E  # Form an inclusive range starting from another input
          #   order is reversed, but doesn't matter
 f-\5T    # Filter by absence of '5'
l         # Count the number of elements left

Try it online!

Perl 6, 23 bytes

{+grep {!/5/},$^a..$^b}

Try it online!

How it works

{                     }  # A lambda.
              $^a..$^b   # Range between the two lambda arguments.
  grep {!/5/},           # Get those whose string representation doesn't match the regex /5/.
 +                       # Return the size of this list.

Haskell, 39 bytes

s!e=sum[1|x<-[s..e],notElem '5'$show x]

Try it online! Usage:

Prelude> 4 ! 17
12

Explanation:

             [s..e]                     -- yields the range from s to e inclusive
          x<-[s..e]                     -- for each x in this range
          x<-[s..e],notElem '5'$show x  -- if the char '5' is not in the string representation of x
       [1|x<-[s..e],notElem '5'$show x] -- then add a 1 to the resulting list      
s!e=sum[1|x<-[s..e],notElem '5'$show x] -- take the sum of the list

R, 33 bytes

f=function(x,y)sum(!grepl(5,x:y))

Usage:

> f=function(x,y)sum(!grepl(5,x:y))
> f(40,60)
[1] 10
> f(1,9)
[1] 8
> f(4,17)
[1] 12

Octave, 36 bytes

@(m,n)sum(all(dec2base(m:n,10)'-52))

Try it online!

Groovy, 47 45 43 40 bytes

{a,b->(a..b).findAll{!(it=~/5/)}.size()}

This is an unnamed closure. findAll is similar to adding an if condition in a list comprehension in python.

Try it online!

PHP 7.1, 57 55 bytes

for([,$a,$b]=$argv;$a<=$b;)$n+=!strstr($a++,53);echo$n;

Run with php -r '<code>' <a> <b>

Mathematica, 46 44 42 bytes

Thanks to alephalpha and DavidC for saving 2 bytes each!

Tr@Boole[FreeQ@5/@IntegerDigits@Range@##]&

Unnamed function taking two integer arguments and returning an integer. IntegerDigits@Range@## converts all the numbers between the inputs into lists of digits; FreeQ@5 tests those lists to decide which ones do not contain any 5. Then Boole converts booleans to zeros and ones, and Tr sums the results.

Other solutions (44 and 47 bytes):

Count[Range@##,x_/;IntegerDigits@x~FreeQ~5]&

IntegerDigits@x~FreeQ~5 determines whether the list of digits of a number is free of 5s, and Count[Range@##,x_/;...]& counts how many numbers between the inputs pass that test.

Tr[Sign[1##&@@IntegerDigits@#-5]^2&/@Range@##]&

1##&@@IntegerDigits@#-5 takes the list of digits of a number, subtracts 5 from all of them, and multplies the answers together; Sign[...]^2 then converts all nonzero numbers to 1.

Ruby, 36 35 bytes

->a,b{(a..b).count{|x|!x.to_s[?5]}}

Thx IMP1 for -1 byte

Java 7, 80 78 bytes

int c(int a,int b){int r=0;for(;a<=b;)r+=(""+a++).contains("5")?0:1;return r;}

Ungolfed:

int c(int a, int b){
  int r = 0;
  for (; a <= b; ) {
    r += ("" + a++).contains("5")
          ? 0
          : 1;
  }
  return r;
}

Test code:

Try it here.

class M{
  static int c(int a,int b){int r=0;for(;a<=b;)r+=(""+a++).contains("5")?0:1;return r;}

  public static void main(String[] a){
    System.out.println(c(1, 9));
    System.out.println(c(4, 17));
  }
}

Output:

8
12

PowerShell, 42 41 bytes

param($a,$b)$a..$b|%{$z+=!($_-match5)};$z

Called from the command line as .\no5s.ps1 1 20

Python 2, 61 56 bytes

lambda a,b:len([n for n in range(a,b+1) if not"5"in`n`])

-5 bytes thanks to tukkaaX

Swift 52 bytes

($0...$1).filter { !String($0).contains("5") }.count

Batch, 95 bytes

@set/an=0,i=%1
:g
@if "%i%"=="%i:5=%" set/an+=1
@set/ai+=1
@if %i% leq %2 goto g
@echo %n%

Manually looping saves some bytes because I need the loop counter in a variable anyway.

PHP, 56 bytes

for($i=$argv[1];$i<=$argv[2];)trim(5,$i++)&&$x++;echo$x;

Run like this:

php -r 'for($i=$argv[1];$i<=$argv[2];)trim(5,$i++)&&$x++;echo$x;' 1 9 2>/dev/null;echo
> 8

A version for PHP 7.1 would be 53 bytes (credits to Titus):

for([,$i,$e]=$argv;$i<=$e;)trim(5,$i++)&&$x++;echo$x;

Explanation

for(
  $i=$argv[1];          # Set iterator to first input.
  $i<=$argv[2];         # Loop until second input is reached.
)
  trim(5,$i++) && $x++; # Trim string "5" with the characters in the
                        # current number; results in empty string when
                        # `5` is present in the number. If that is not
                        # the case, increment `$x`

echo$x;                 # Output `$x`

CJam "easy pure mathematics solution", 60

{{Ab5+_,\_5#)<\9e]);_4f>.m9b}%}:F;q~_:z$\:*0>{((+F:-}{F:+)}?

Try it online

It takes the numbers in any order, in an array.

Explanation:

One core problem is to calculate f(n) = the number of non-5 numbers from 1 to n (inclusive) for any positive n. And the answer is: take n's decimal digits, replace all digits after the first 5 (if any) with 9, then replace all digits 5..9 with 4..8 (decrement), and convert from base 9. E.g. 1752 → 1759 → 1648 → 1*9^3+6*9^2+4*9+8=1259. Basically, each digit position has 9 acceptable values, and a 5xxxx is equivalent to a 49999 because there are no more valid numbers between them.

Once we solved this, we have a few cases: if the input numbers (say a and b, a<b) are (strictly) positive, then the result is f(b)-f(a-1). If they are negative, then we can take the absolute values, reorder them and use the same calculation. And if a<=0<=b then the result is f(-a)+f(b)+1.

The program first implements the function F as described above (but applied to each number in an array), then reads the input, converts the numbers to the absolute value and reorders them, and uses one of the 2 calculations above, based on whether a*b>0 initially.

Python 2, 54 bytes

i,j=input();k=0
while i<=j:k+=not"5"in`i`;i+=1
print k

Try it online!

Not the shortest Python answer Uses same algorithm but a different way of implementing with a while loop and is not a lambda function.

Java 7, 77 bytes

This is an improvement of Kevins Answer, but since I don't have the reputation to comment yet, this new answer will have to do.

So what I did was:

• Replace the indexOf statements with contains (-1 byte)
• Move the incrementing part of the for-loop into the conditional statement (-2 bytes)

for-loop (77 bytes):

int c(int a,int b){int r=1;for(;a++<b;)r+=(""+a).contains("5")?0:1;return r;}

recursive (79 bytes):

int d(int r,int a,int b){r+=(""+a).contains("5")?0:1;return a!=b?d(r,a+1,b):r;}

Output:

8
12

8
12

Test it here !

C#, 67 bytes

a=>b=>{int c=0;for(;a<=b;)c+=(a+++"").Contains("5")?0:1;return c;};

JavaScript (ES6), 58 56 49 bytes

let f =

(s,e)=>{for(c=0;s<=e;)c+=!/5/.test(s++);return c}

console.log(f(1, 9));
console.log(f(4, 17));
console.log(f(-9, -1));

Golfed 7 bytes thanks to ETHproductions.

MATL, 10 bytes

&:!V53-!As

Try it online!

Explanation

        % Implicitly grab two input arguments
&:      % Create an array from [input1....input2]
!V      % Convert to a string where each number is it's own row
53-     % Subtract ASCII '5' from each character.
!A      % Detect which rows have no false values (no 5's). Returns a logical array
s       % Sum the logical array to get the # numbers without 5's
        % Implicitly display the result

C#, 77 bytes

(n,m)=>{var g=0;for(var i=n;i<m+1;i++)g+=(i+"").Contains("5")?0:1;return g;};

Anonymous lambda call.

Uses n (first number) and m (last number) as input, then checks via string containment ("".Contains("")).

Actually, 13 bytes

u@x`$'5íuY`░l

Try it online!

Explanation:

u@x`$'5íuY`░l
u@x            range(a, b+1)
   `$'5íuY`░   take where:
    $            string representation
     '5íuY       does not contain "5"
            l  length