Die Herausforderung

Bei einer Ganzzahleingabe von xwhere 1 <= x <= 255werden die Ergebnisse von Zweierpotenzen zurückgegeben, die bei Summierung ergeben x.

Beispiele

Angesichts der Eingabe:

86

Ihr Programm sollte folgendes ausgeben:

64 16 4 2

Eingang:

240

Ausgabe:

128 64 32 16

Eingang:

1

Ausgabe:

1

Eingang:

64

Ausgabe:

64

Der Ausgang kann Nullen enthalten, wenn die bestimmte Zweierpotenz nicht in der Summe vorhanden ist.

Zum Beispiel Eingabe 65 ausgegeben werden 0 64 0 0 0 0 0 1.

Wertung

Das ist Code-Golf , also gewinnt die kürzeste Antwort in jeder Sprache.

JavaScript (ES6), 28 Byte

f=n=>n?[...f(n&~-n),n&-n]:[]

Probieren Sie es online!

Pure Bash, 20

echo $[2**{7..0}&$1]

Try it online!

Erläuterung

          {7..0}     # Brace expansion: 7 6 5 4 3 2 1 0
       2**{7..0}     # Brace expansion: 128 64 32 16 8 4 2 1
       2**{7..0}&$1  # Brace expansion: 128&n 64&n 32&n 16&n 8&n 4&n 2&n 1&n (Bitwise AND)
     $[2**{7..0}&$1] # Arithmetic expansion
echo $[2**{7..0}&$1] # and output

Jelly, 4 bytes

-2 since we may output zeros in place of unused powers of 2 :)

Ḷ2*&

Try it online!

How?

Ḷ2*& - Link: integer, n         e.g. 10
Ḷ    - lowered range of n            [  0,  1,  2,  3,  4,  5,  6,  7,  8,  9]
 2*  - two to the power of           [  1,  2,  4,  8, 16, 32, 64,128,256,512]
   & - bit-wise and                  [  0,  2,  0,  8,  0,  0,  0,  0,  0,  0]

Jelly, 6 bytes

BUT’2*

Try it online!

Explanation

BUT here is an explanation (note: I had assumed that we may only output the powers of 2 themselves and nothing else):

BUT’2* – Monadic link. Takes a number N as input. Example: 86
B      – Convert N to binary.                              [1, 0, 1, 0, 1, 1, 0]
 U     – Reverse.                                          [0, 1, 1, 0, 1, 0, 1]
  T    – Truthy indices.                                   [2, 3, 5, 7]
   ’   – Decrement.                                        [1, 2, 4, 6]
    2* – Raise 2 to that power.                            [2, 4, 16, 64]

$X$$\{x_1, x_2, x_3,\cdots, x_n\}$$x_i\in\{0,1\},\:\forall\:\: i\in\overline{1,n}$

Python, 35 bytes

lambda n:[n&2**i for i in range(8)]

Little-endian with zeros at unused powers of 2.

Try it online!

APL (Dyalog Extended), 7 bytes^SBCS

Anonymous tacit prefix function. Requires 0-based indexing (⎕IO←0).

2*⍸⍢⌽⍤⊤

Try it online!

2 two
* raised to the power of
⍸ the ɩndices where true
⍢ while
⌽ reversed
⍤ of
⊤ the binary representation

Sledgehammer 0.2, 3 bytes

⡔⡸⢣

Decompresses into {intLiteral[2],call[NumberExpand,2]}.

Sledgehammer is a compressor for Wolfram Language code using Braille as a code page. The actual size of the above is 2.75 bytes, but due to current rules on meta, padding to the nearest byte is counted in code size.

05AB1E, 3 bytes

Ýo&

Port of @JonathanAllan's Jelly answer, so make sure to upvote him!

Contains zeros (including -loads of- trailing zeros).

Try it online or verify all test cases.

Explanation:

Ý      # Create a list in the range [0, (implicit) input]
       #  i.e. 15 → [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
       #  i.e. 16 → [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
 o     # Take 2 to the power of each value
       #  → [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768]
       #  → [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536]
  &    # Bitwise-AND each value with the (implicit) input
       # 15 → [1,2,4,8,0,0,0,0,0,0,0,0,0,0,0,0]
       # 16 → [0,0,0,0,16,0,0,0,0,0,0,0,0,0,0,0,0]
       # (and output the result implicitly)

Catholicon, 3 bytes

ṫĊŻ

Try it online!

Explanation:

ṫ       Decompose         into the largest values where:
 Ċ               the input
  Ż       the bit count is truthy (equal to one)

Wolfram Language (Mathematica), 17 bytes

#~NumberExpand~2&

Try it online!

Mathematica strikes again.

R, 27 23 bytes

bitwAnd(scan(),2^(7:0))

Try it online!

Unrolled code and explanation :

A = scan()         # get input number A from stdin
                   # e.g. A = 65

bitwAnd( A , 2^(7:0))  # bitwise AND between all powers of 2 : 2^7 ... 2^0 and A
                       # and implicitly print the result
                       # e.g. B = bitwAnd(65, c(128,64,32,16,8,4,2,1)) = c(0,64,0,0,0,0,0,1)

• 4 bytes thanks to @Kirill L.

C# (Visual C# Interactive Compiler), 29 bytes

Contains 5 unprintable characters.

n=>"€@ ".Select(a=>a&n)

Explanation

//Lambda taking one parameter 'n'
n=>
//String with ASCII characters 128, 64, 32, 16, 8, 4, 2, and 1
"€@ "
//Iterate through all the chars of the above string and transform them to
.Select(a=>
//A bitwise AND operation between the integer value of the current char and the input value
a&n)

Try it online!

C# (Visual C# Interactive Compiler), 38 bytes

x=>{for(int y=8;y-->0;Print(x&1<<y));}

Try it online!

C (gcc), 39 bytes

f(n){for(;n;n&=n-1)printf("%d ",n&-n);}

Try it online!

05AB1E, 7 bytes

2вRƶ<oò

explanation:

2в        convert input to binary array
R         reverse array
ƶ<        multiply each item by it's index and subtract 1
oò        2^item then round down

Try it online!

Haskell, 29 bytes

(mapM(\n->[0,2^n])[7,6..0]!!)

Try it online!

Ruby, 25 bytes

->n{8.times{|i|p n&2**i}}

Try it online!

C (clang), 133 110 63 58 bytes

58-byte solution thanks to @ceilingcat.

x=256;main(y){for(scanf("%d",&y);x/=2;)printf("%d ",y&x);}

Try it online!

MATL, 5 bytes

BPfqW

Try it online!

Explanation

Consider input 86 as an example.

B    % Implicit input. Convert to binary (highest to lowest digits)
     % STACK: [1 0 1 0 1 1 0]
P    % Flip
     % STACK: [0 1 1 0 1 0 1]
f    % Find: indices of nonzeros (1-based)
     % STACK: [2 3 5 7]
q    % Subtract 1, element-wise
     % STACK: [1 2 4 6]
W    % Exponential with base 2, element-wise. Implicit display
     % STACK: [2 4 16 64]

Perl 6, 16 12 bytes

-4 bytes thanks to Jonathan Allan

*+&2**all ^8

Try it online!

Returns an All Junction with 8 elements. This is a rather non-standard way of returning, but generally, Junctions can act as ordered (at least until autothreading is implemented) lists and it is possible to extract the values from one.

Explanation:

*+&              # Bitwise AND the input with
   2**           # 2 raised to the power of
      all ^8     # All of the range 0 to 7

Japt, 8 5 bytes

Æ&2pX

Try it

Æ&2pX     :Implicit input of integer U
Æ         :Map each X in the range [0,U)
 &        :  Bitwise AND of U with
  2pX     :  2 to the power of X

Alternative

Suggested by Oliver to avoid the 0s in the output using the -mf flag.

N&2pU

Try it

N&2pU     :Implicitly map each U in the range [0,input)
N         :The (singleton) array of inputs
 &        :Bitwise AND with
  2pX     :2 to the power of U
          :Implicitly filter and output

05AB1E, 9 bytes

Ýoʒ›}æʒOQ

Try it online!

This is also correct for 6-bytes, but it doesn't complete in time on TIO for 86:

05AB1E, 6 bytes

ÝoæʒOQ

Try it online!

Julia 0.6, 13 bytes

n->n&2.^(0:7)

Try it online!

CJam, 12 bytes

{:T{2\#T&}%}

Try it online!

K (oK), 19 16 bytes

-3 bytes thanks to ngn!

{*/x#2}'&|(8#2)\

Try it online!

oK does not have power operator, that's why I need a helper function {*/x#2} (copy 2 x times and reduce the resulting list by multiplication)

Alchemist, 125 bytes

_->In_x+128a+m
m+x+a->m+b
m+0x+a->n+a
m+0a->o+Out_b+Out_" "
n+b->n+x+c
n+0b+a->n+c
n+0a->p
o+b->o+c
o+0b->p
p+2c->p+a
p+0c->m

Try it online! or Test every input!

Explanation

_->In_x+128a+m           # Initialize universe with input, 128a (value to compare to) and m (state)
m+x+a->m+b               # If c has been halved, subtract min(a, x) from a and x and put its value into b
m+0x+a->n+a              # If x < a, continue to state n
m+0a->o+Out_b+Out_" "    # Else print and continue to state o
n+b->n+x+c               # Add min(a, x) (i.e. x) back to x, and add it to c (we're collecting a back into c)
n+0b+a->n+c              # Then, add the rest of a to c
n+0a->p                  # Then, go to state p
o+b->o+c                 # Add min(a, x) (i.e. a) to c - x _is_ greater than a and so contains it in its binary representation, so we're not adding back to x
o+0b->p                  # Then, go to state p
p+2c->p+a                # Halve c into a
p+0c->m                  # Then go to state m

PHP, 41 39 bytes

for($c=256;$c>>=1;)echo$argv[1]&$c,' ';

Try it online!

Or 38 with no fun >>= operator and PHP 5.6+:

for($x=8;$x--;)echo$argv[1]&2**$x,' ';

Or 36 with little-endian ("0 2 4 0 16 0 64 0") output:

while($x<8)echo$argv[1]&2**$x++,' ';

Really I just wanted to use the >>= operator, so I'm sticking with the 39.

Tests:

$php pow2.php 86
0 64 0 16 0 4 2 0

$php pow2.php 240
128 64 32 16 0 0 0 0

$php pow2.php 1
0 0 0 0 0 0 0 1

$php pow2.php 64
0 64 0 0 0 0 0 0

$php pow2.php 65
0 64 0 0 0 0 0 1

TSQL, 43 39 bytes

Can't find a shorter fancy solution, so here is a standard loop. -4 bytes thanks to MickyT and KirillL

DECLARE @y int=255

,@ int=128s:PRINT @y&@ SET @/=2IF @>0GOTO s

Try it out

Python 2, 43 40 bytes

f=lambda n,p=1:n/p*[1]and f(n,p*2)+[p&n]

Try it online!

C# (Visual C# Interactive Compiler), 33 bytes

n=>{for(;n>0;n&=n-1)Print(n&-n);}

Port of @Arnauld's JavaScript (ES6) answer, so make sure to upvote him!

Try it online.

Explanation:

n=>{            // Method with integer parameter and no return-type
  for(;n>0      //  Continue looping as long as `n` is larger than 0:
      ;         //    After every iteration:
       n&=n-1)  //     Bitwise-AND `n` by `n-1`
    Print(      //   Print with trailing newline:
      n&-n);}   //    `n` bitwise-AND `-n`