Étant donné un entier non négatif, renvoyez la différence absolue entre la somme de ses chiffres pairs et la somme de ses chiffres impairs.

Règles par défaut

• Les échappatoires standard s'appliquent.

• Vous pouvez prendre des entrées et fournir des sorties par n'importe quelle méthode d'entrée / sortie standard.

• Vous pouvez prendre l'entrée comme une chaîne, comme un entier ou comme une liste de chiffres.

• C'est le code-golf , donc le code le plus court en octets dans toutes les langues gagne!

Cas de test

Entrée ~> Sortie

0 ~> 0 (| 0-0 | = 0)
1 ~> 1 (| 1-0 | = 1)
12 ~> 1 (| 2-1 | = 1)
333 ~> 9 (| 0- (3 + 3 + 3) | = 9)
459 ~> 10 (| 4- (5 + 9) | = 10)
2469 ~> 3 (| (2 + 4 + 6) -9 | = 3)
1234 ~> 2 (| (2 + 4) - (1 + 3) | = 2)

Gelée , 6 octets

-*æ.¹A

Essayez-le en ligne!

Comment ça fonctionne

-*æ.¹A  Main link. Argument: A (digit array)

-*      Raise -1 to A's digits, yielding 1 for even digits and -1 for odd ones.
    ¹   Identity; yield A.
  æ.    Take the dot product of the units and the digits.
     A  Apply absolute value.

Scripts SHENZHEN I / O MCxxxx, 197 (126 + 71) octets

Puce 1 (MC6000):

• x0: entrée sous forme de liste
• x2: puce 2 x1
• x3: MC4010

mov 0 dat
j:
mov x0 acc
mov 70 x3
mov -1 x3
mov acc x3
mul x3
mov acc x2
mov dat acc
add 1
tlt acc 3
mov acc dat
+jmp j
slx x0

Puce 2 (MC4000):

• p0: Sortie
• x0: MC4010
• x1: Puce 1 x2

mov x1 acc
add x1
add x1
mov 30 x0
mov 0 x0
mov acc x0
mov x0 p0
slx x1

Python 2, 39 octets

Prend un entier comme liste. Essayez-le en ligne

lambda A:abs(sum((-1)**i*i for i in A))

-3 octets grâce à @ Mr.Xcoder
-1 octet grâce à @ovs

TI-Basic, 18 9 octets

abs(sum((-1)^AnsAns

Explication

Multiplie chaque chiffre de la liste par -1 à sa puissance, annulant chaque chiffre impair, avant de les additionner.

C (gcc) , 59 58 57 octets

i;f(char*v){for(i=0;*v;v++)i+=*v%2?*v-48:48-*v;v=abs(i);}

Essayez-le en ligne!

R, 30 29 octets

abs(sum((d=scan())-2*d*d%%2))

d = scan() prend le numéro saisi d'un chiffre après l'autre.

-1 octet grâce à @Giuseppe!

C #, 57 octets

namespace System.Linq{i=>Math.Abs(i.Sum(n=>n%2<1?n:-n))}

Prend l'entrée au fur iet à mesure et résume les nombres entiers en tournant les cotes à négatif.

Mathematica, 20 octets

Abs@Tr[(-1)^(g=#)g]&

prend en entrée une liste de chiffres

merci spécial à @LLlAMnYP pour m'avoir informé des "nouvelles règles"

Japt , 8 octets

x_*JpZÃa

Testez-le en ligne!

Explication

 x_  *JpZÃ a
UxZ{Z*JpZ} a
                Implicit: U = list of digits
UxZ{     }      Take the sum of each item Z in U mapped through the following function:
      JpZ         Return (-1) ** Z
    Z*            times Z. This gives Z if even, -Z if odd.
           a    Take the absolute value of the result.
                Implicit: output result of last expression

Neim , 7 octets

ΓDᛃΞ𝐍}𝐬

Explication:

Γ        Apply the following to each element in the input array
 D         Duplicate
  ᛃ        Modulo 2, then perform logical NOT
   Ξ       If truthy, then:
    𝐍        Multiply by -1
      }  Close all currently running loops/conditionals etc
       𝐬 Sum the resulting array

APL, 8 octets

|⊢+.ׯ1*⊢

Essayez-le en ligne!

Comment?

¯1*⊢- -1 ⁿ pourn dans⍵

_{[ 4 5 9→1 ¯1 ¯1 ]}

⊢+.× - multiplication vérifiée avec o , puis somme

_{[ +/ 4 5 9 × 1 ¯1 ¯1→ +/ 4 ¯5 ¯9→¯10 ]}

| - valeur absolue

JavaScript (ES6), 43 38 octets

Prend l'entrée sous forme de chaîne d' un tableau de chiffres.

a=>a.map(d=>s+=d&1?d:-d,s=0)&&s>0?s:-s

Cas de test

let f =

a=>a.map(d=>s+=d&1?d:-d,s=0)&&s>0?s:-s

console.log(f([1]))     // ~> 1 ( |1-0| = 1)
console.log(f([0]))     // ~> 0 ( |0-0| = 0)
console.log(f([1,2]))    // ~> 1 ( |1-2| = 1)
console.log(f([4,5,9]))   // ~> 10 ( |4-(5+9)| = 10)
console.log(f([2,4,6,9]))  // ~> 3 ( |(2+4+6)-9| = 3)
console.log(f([3,3,3]))   // ~> 9 ( |(3+3+3)-0| = 9)

EDIT: Une approche plus centrée sur le golf:

EXCEL, 42 36 29 octets

Enregistré 6 octets grâce à Magic Octopus Urn Enregistré 7 octets en utilisant l'approche Dennis '-1 ^ (qui, je viens de l'apprendre, fonctionne sur des tableaux dans Excel)

=ABS(SUMPRODUCT(A:A,-1^A:A))

Prend une liste d'entiers dans une colonne pour l'entrée. Peut probablement être joué plus loin, ou en utilisant la version chaîne, en prenant une chaîne en A1 pour entrée.

EXCEL, 256 octets

=ABS(LEN(SUBSTITUTE(A1,1,""))-2*LEN(SUBSTITUTE(A1,2,""))+3*LEN(SUBSTITUTE(A1,3,""))-4*LEN(SUBSTITUTE(A1,4,""))+5*LEN(SUBSTITUTE(A1,5,""))-6*LEN(SUBSTITUTE(A1,6,""))+7*LEN(SUBSTITUTE(A1,7,""))-8*LEN(SUBSTITUTE(A1,8,""))+9*LEN(SUBSTITUTE(A1,9,""))-5*LEN(A1))

Julia 0,5 , 19 octets

!x=(-1).^x⋅x|>abs

Essayez-le en ligne!

Husk , 7 octets

≠0ṁṠ!¡_

Essayez-le en ligne!

Prend une liste de chiffres en entrée.

Il manque toujours un "abs" intégré, mais un bon résultat tout de même :)

Explication

Ṡ!¡_est une fonction qui prend un nombre net applique ensuite n-1la fonction _(négation) à n. Cela se traduit npar impair nou -npairn .

ṁ applique une fonction à chaque élément d'une liste et additionne les résultats.

≠0 renvoie la différence absolue entre un nombre et 0.

05AB1E , 6 octets

Merci à Dennis pour le power-trick -1. Prend la saisie sous forme de liste de chiffres

®sm*OÄ

Essayez-le en ligne!

Explication

®sm*OÄ                                               Example [4, 5, 9]
®      # Push -1                                   % STACK: -1
 sm    # Take -1 to the power of every number      % STACK: [1, -1, -1]
         in the input list 
   *   # Multiply with original input              % Multiply [1, -1, -1] with [4, 5, 9] results in STACK: [4, -5, -9]
    O  # Sum them all together                     % STACK: -10
     Ä # Absolute value                            % STACK: 10
       # Implicit print

PHP, 51 octets

while(~$n=$argn[$i++])$s+=$n&1?$n:-$n;echo abs($s);

ajoute un chiffre $ssi impair, soustrait s'il est pair. Exécuter en tant que tuyau avec -nR.

ou

while(~$n=$argn[$i++])$s+=(-1)**$n*$n;echo abs($s);

en utilisant le tour de -1force de Dennis .

Mathematica, 67 octets

(s=#;Abs[Subtract@@(Tr@Select[IntegerDigits@s,#]&/@{EvenQ,OddQ})])&

PHP , 54 octets

for(;~$n=$argn[$i++];)${eo[$n&1]}+=$n;echo abs($e-$o);

Essayez-le en ligne!

PHP , 57 octets

stocker les sommes paires et impaires dans un tableau

for(;~$n=$argn[$i++];)$r[$n&1]+=$n;echo abs($r[0]-$r[1]);

Essayez-le en ligne!

PHP , 57 octets

stocker les sommes paires et impaires dans deux variables

for(;~$n=$argn[$i++];)$n&1?$o+=$n:$e+=$n;echo abs($e-$o);

Essayez-le en ligne!

Haskell, 47 42 39 38 26 25 bytes

-1 thanks to nimi

-12 thanks to Bruce

-1 thanks to xnor

abs.sum.map(\x->x*(-1)^x)

Try it online!

Perl 6, 28 bytes

{abs sum $_ Z*.map(*%2*2-1)}

Try it online!

Takes a list of digits as input.

• $_ is the input argument.
• .map(* % 2 * 2 - 1) maps each digit to either 1 or -1 depending on whether the digit is odd or even, respectively.
• Z* zips the original list of digits with the even/odd list using multiplication.

Braingolf, 18 bytes

{.2%?M|}&+v&+c-!s*

Try it online!

Takes input as a list of digits

Explanation

{.2%?M|}&+v&+c-!s*  Implicit input from commandline args
{......}            Foreach loop, runs on each item in the stack..
 .2%                ..Parity check, push 1 if odd, 0 if even
    ?               ..If last item != 0 (pops last item)..
     M              ....Move last item to next stack
      |             ..Endif
        &+          Sum entire stack
          v&+       Switch to next stack and sum entire stack
             c-     Collapse into stack1 and subtract
               !s   Sign check, push 1 if last item is positive, -1 if last item is
                    negative, 0 if last item is 0
                 *  Multiply last item by sign, gets absolute value
                    Implicit output

R, 72 43 bytes

b=(d=scan())%%2<1;abs(sum(d[b])-sum(d[!b]))

First, d = scan() takes the number as input, one digit after the other (thanks to @Giuseppe comment !)
Then, b = d %% 2 <1 associates to b a TRUE or FALSE value at each index depending on the digits' parity. Therefore, b values are TRUE for even numbers, and !b are TRUE for odd values.

Finaly, abs(sum(d[b]) - sum(d[!b])) does the job.

Bash 141 139 99 Bytes

while read -n1 a; do
[ $[a%2] = 0 ]&&e=$[e+a]||o=$[o+a]
done
(($[e-o]>0))&&echo $[e-o]||echo $[o-e]

Try it online!

Java (OpenJDK 8), 55 bytes

a->{int s=0;for(int d:a)s+=d%2<1?-d:d;return s<0?-s:s;}

Try it online!

Naive implementation.

C#, 67 bytes

namespace System.Linq{a=>Math.Abs(a.Sum(n=>n%2<1)-a.Sum(n=>n%2>1))}

05AB1E, 7 bytes

È2*<*OÄ

Try it online!

        # Input              | 1234567
È       # Is Even?           | [0,1,0,1,0,1,0]
 2*<    # (a * 2) - 1        | [-1,1,-1,1,-1,1,-1]
    *   # Multiply w/ input. | [-1,2,-3,4,-5,6,-7]
     O  # Sum.               | -10
      Ä # Absolute value of. | 10

x86-64 Machine Code, 30 bytes

31 C0 99 8B 4C B7 FC F6 C1 01 74 04 01 CA EB 02 01 C8 FF CE 75 ED 29 D0 99 31 D0 29 D0 C3

The above code defines a function that accepts a list/array of integer digits and returns the absolute difference between the sum of its even digits and the sum of its odd digits.

As in C, assembly language doesn't implement lists or arrays as first-class types, but rather represents them as a combination of a pointer and a length. Therefore, I have arranged for this function to accept two parameters: the first is a pointer to the beginning of the list of digits, and the second is an integer specifying the total length of the list (total number of digits, one-indexed).

The function conforms to the System V AMD64 calling convention, which is standard on Gnu/UNIX systems. In particular, the first parameter (pointer to the beginning of the list) is passed in RDI (as this is 64-bit code, it is a 64-bit pointer), and the second parameter (length of the list) is passed in ESI (this is only a 32-bit value, because that's more than enough digits to play with, and naturally it is assumed to be non-zero). The result is returned in the EAX register.

If it's any clearer, this would be the C prototype (and you can use this to call the function from C):

int OddsAndEvens(int *ptrDigits, int length);

Ungolfed assembly mnemonics:

; parameter 1 (RDI) == pointer to list of integer digits
; parameter 2 (ESI) == number of integer digits in list (assumes non-zero, of course)
OddsAndEvens:
   xor  eax, eax              ; EAX = 0 (accumulator for evens)
   cdq                        ; EDX = 0 (accumulator for odds)
.IterateDigits:
   mov  ecx, [rdi+rsi*4-4]    ; load next digit from list
   test cl, 1                 ; test last bit to see if even or odd
   jz   .IsEven               ; jump if last bit == 0 (even)
.IsOdd:                       ; fall through if last bit != 0 (odd)
   add  edx, ecx              ; add value to odds accumulator
   jmp  .Continue             ; keep looping
.IsEven:
   add  eax, ecx              ; add value to evens accumulator
.Continue:                    ; fall through
   dec  esi                   ; decrement count of digits in list
   jnz  .IterateDigits        ; keep looping as long as there are digits left

   sub  eax, edx              ; subtract odds accumulator from evens accumulator

   ; abs
   cdq                        ; sign-extend EAX into EDX
   xor  eax, edx              ; XOR sign bit in with the number
   sub  eax, edx              ; subtract sign bit

   ret                        ; return with final result in EAX

Here's a brief walk-through of the code:

• First, we zero out the EAX and EDX registers, which will be used to hold the sum totals of even and odd digits. The EAX register is cleared by XORing it with itself (2 bytes), and then the EDX register is cleared by sign-extending the EAX into it (CDQ, 1 byte).

• Then, we go into the loop that iterates through all of the digits passed in the array. It retrieves a digit, tests to see if it is even or odd (by testing the least-significant bit, which will be 0 if the value is even or 1 if it is odd), and then jumps or falls through accordingly, adding that value to the appropriate accumulator. At the bottom of the loop, we decrement the digit counter (ESI) and continue looping as long as it is non-zero (i.e., as long as there are more digits left in the list to be retrieved).

  The only thing tricky here is the initial MOV instruction, which uses the most complex addressing mode possible on x86.^* It takes RDI as the base register (the pointer to the beginning of the list), scales RSI (the length counter, which serves as the index) by 4 (the size of an integer, in bytes) and adds that to the base, and then subtracts 4 from the total (because the length counter is one-based and we need the offset to be zero-based). This gives the address of the digit in the array, which is then loaded into the ECX register.

• After the loop has finished, we do the subtraction of the odds from the evens (EAX -= EDX).

• Finally, we compute the absolute value using a common trick—the same one used by most C compilers for the abs function. I won't go into details about how this trick works here; see code comments for hints, or do a web search.

__
_{* The code can be re-written to use simpler addressing modes, but it doesn't make it any shorter. I was able to come up with an alternative implementation that dereferenced RDI and incremented it by 8 each time through the loop, but because you still have to decrement the counter in ESI, this turned out to be the same 30 bytes. What had initially given me hope is that add eax, DWORD PTR [rdi] is only 2 bytes, the same as adding two enregistered values. Here is that implementation, if only to save anyone attempting to outgolf me some effort :-)}

                    OddsAndEvens_Alt:
31 C0                   xor    eax, eax
99                      cdq
                    .IterateDigits:
F6 07 01                test   BYTE PTR [rdi], 1
74 04                   je     .IsEven
                    .IsOdd:
03 17                   add    edx, DWORD PTR [rdi]
EB 02                   jmp    .Continue
                    .IsEven:
03 07                   add    eax, DWORD PTR [rdi]
                    .Continue:
48 83 C7 08             add    rdi, 8
FF CE                   dec    esi
75 ED                   jne    .IterateDigits
29 D0                   sub    eax, edx
99                      cdq
31 D0                   xor    eax, edx
29 D0                   sub    eax, edx
C3                      ret

TI-BASIC, 11 6 bytes

abs(sum(Anscos(πAns

Takes input as a list. i²^Ans saves two bytes over (-1)^Ans because we don't need the parentheses.

abs(sum(Anscos(πAns
           cos(πAns                  1 for evens, -1 for odds
        Ans                          Multiply by original list
abs(sum(                             Sum the list and take absolute value, which also
                                     fixes rounding errors from cos(.

J, 14 bytes

|-/(2&|+//.[),

Try it online!

explanation

|                absolute value of
 -/              the difference between
                 the items on the list returned by this fork
   (2&|          is an item odd? (1 for yes, 0 for no)
       +//.      the "key" verb /. which partitions on above (even / odd) then sums
           [)    identity, ie, the list of digits passed
             ,   turn it into a list (to handle 1 element case)