Twoim zadaniem jest napisanie programu, który, biorąc pod uwagę liczbę i ciąg, dzieli ciąg na fragmenty tego rozmiaru i odwraca je.

Zasady

Twój program otrzyma dodatnią liczbę całkowitą n, a także ciąg so długości co najmniej jeden składający się tylko z drukowalnego ASCII (bez białych znaków). Łańcuch powinien być następnie podzielony na kawałki długości n, jeśli długość łańcucha nie jest podzielna przezn resztki na końcu, należy uznać za własny kawałek. Następnie odwróć kolejność kawałków i połącz je ponownie.

Przypadki testowe

n   s           Output

2   abcdefgh    ghefcdab
3   foobarbaz   bazbarfoo
3   abcdefgh    ghdefabc
2   a           a
1   abcdefgh    hgfedcba
2   aaaaaa      aaaaaa
2   baaaab      abaaba
50  abcdefgh    abcdefgh
6   abcdefghi   ghiabcdef

To jest golf golfowy , więc powinieneś dążyć do jak najmniejszej liczby bajtów.

Galaretka , 2 bajty

sṚ

Pełny program, który drukuje wynik.

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W jaki sposób?

sṚ - Main link: string, number                                   e.g. 'abcdefg', 3
s  - split string into chunks of length number (keeping any overflow) ["abc","def","g"]
 Ṛ - reverse the resulting list                                       ["g","def","abc"]
   - implicit print                                                   gdefabc

Python 3 , 35 bajtów

f=lambda s,n:s and f(s[n:],n)+s[:n]

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05AB1E , 5 4 3 bajty

-1 dzięki Dennis
-1 dzięki carusocomputing

ôRJ

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     # Implicit: push input
 ô   # Split in pieces of length b
  RJ # Reverse and join

JavaScript (ES6), 37 bajtów

n=>F=s=>s&&F(s.slice(n))+s.slice(0,n)

Pobiera dane wejściowe przez curry: najpierw liczba, a następnie ciąg znaków, jak f(2)("abcdefgh").

let f =
n=>F=s=>s&&F(s.slice(n))+s.slice(0,n)

let g = (n, s) => console.log(`f(${n})("${s}"): ${f(n)(s)}`)

g(2, "abcdefgh")
g(3, "foobarbaz")
g(3, "abcdefgh")
g(2, "a")
g(1, "abcdefgh")
g(2, "aaaaaa")
g(2, "baaaab")
g(50, "abcdefgh")

Perl 6 ,  28  20 bajtów

{$^b.comb($^a).reverse.join}

Spróbuj

{[R~] $^b.comb($^a)}

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Rozszerzony:

{  # bare block lambda with placeholder parameters ｢$a｣ and ｢$b｣
  [R[~]] # reduce the following using the reverse meta operator `R`
         # combined with the string concatenation operator

    # `comb` with a number splits the invocant into chunks of that size
    $^b.comb($^a)
}

Bash + coreutils, 22

fold -$1|tac|tr -d \\n

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Haskell , 32 bajty

n#""=""
n#s=n#drop n s++take n s

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PHP, 53 bajtów

<?=join(array_reverse(str_split($argv[2],$argv[1])));

Röda , 36 bajtów

f n{[[_]..[try head n-1]]|reverse|_}

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Jest to funkcja, która wymaga jednego argumentu. Znaki ciągu muszą znajdować się w strumieniu.

trysłuży do odrzucania błędów w przypadku, gdy headfunkcja nie może odczytać n-1wartości.

Wyjaśnienie:

f n{[[_]..[try head n-1]]|reverse|_}
f n{                               } /* Function declaration */
                                     /* In a loop: */
      _                              /*   Pull one value */
           try head n-1              /*   Pull n-1 values (or less) */
     [ ]..[            ]             /*   Make an array */
    [                   ]            /*   Push it to the stream */
                         |reverse    /* Reverse all values in the stream */
                                 |_  /* Flat all arrays in the stream */
                                     /* Characters in the stream are printed */

Nie tak zaciemnione jak zwykle. Myślę, że to całkiem piękne. :)

CJam , 5 bajtów

q~/W%

Dane wejściowe to liczba i ciąg zamknięty w podwójnych cudzysłowach, oddzielone spacjami.

Wypróbuj online! Lub sprawdź wszystkie przypadki testowe .

Wyjaśnienie

q~   e# Read all input and evaluate: pushes a number and a string
/    e# Split string into chunks of that size. Last chunk may be
     e# smaller. Gives an array of strings
W%   e# Reverse the array. Implicitly display

Partia, 74 bajty

@if %2=="" (echo %~3)else set s=%~2&call %0 %1 "%%s:~%1%%" "%%s:~,%1%%%~3"

Raczej denerwująco kończy się to raczej rekurencyjnym niż rekurencyjnym.

V , 13 10 bajtów

òÀ|lDÏpòÍî

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ò      ò    ' Recursively
 À|         ' Go to the "nth" column
   l        ' Move one character right (breaks loop when no more chunks)
    D       ' Delete from here to the end of the line
     Ï      ' Add a line above the current line (now contains one chunk)
      p     ' Paste the remainder of the line that was deleted
        Íî  ' Remove all newlines

W akcji:

abcdefghijkl

przemienia się w

efghijkl
abcd

która staje się

ijkl
efgh
abcd

zanim wszystkie nowe wiersze zostaną usunięte

pieprzenie mózgu , 78 bajtów

,<<<+[[>]>+>[[>],<[<]>+>-]<-[->>[>]>>+<<<[<]<]>>]<<<<[[<]>[-[+.[-]]+>]<[<]<<<]

Pierwszy bajt danych wejściowych to wielkość porcji podana przez wartość bajtu. Resztę bajtów uważa się za ciąg.

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Rozszerzony i skomentowany

Read the chunk size byte
This cell will become a counter cell
,

Move left a few cells an increment; 
this is to make the starting position 
line up with the relative positioning
needed to fit in with the loop
<<<+

While the current cell is nonzero:
[

 Move right to the first zero cell
 [>]

 Move right once and increment and then move right to the counter cell
 The increment is required because of "move to zero cell" loops
 >+>

 This loop will store one chunk of the input in consecutive memory cells
 [
  [>]   Move right until a zero cell is hit
  ,     Store 1 byte of input there
  <[<]  Move back left until a zero cell (other than the current one) is hit
  >+>-  Increment the temporary cell by 1 and decrement the counter
 ] (end loop once the counter hits zero)

 Decrement the temp cell (because we needed to have 1 there initially to make the cell location work)
 <-

 Move the temp cell to three cells after the end of the chunk
 This is the new counter cell for the next chunk
 [->>[>]>>+<<<[<]<]

 Move two cells right from where the temp cell was
 This is the first cell of the chunk; if it's 0
 then the input is finished and the loop should end
 >>
]

Due to the way the counter is kept track of the tape head
will always be four cells to the right of the last input cell
when the loops breaks
<<<<

Now the chunks are printed one by one
At the start of an iteration the tape head is at the end of a chunk
[
 Locate the start of the last chunk
 [<]>

 Print the chunk:
 [
  Print the byte held in the current cell if it isn't 1
  This is necessary because we left a stray 1 in a cell at
  the start which shouldn't be printed
  -[+.[-]]+

  Move to the next cell
  >
 ]

 Move to just left of the chunk
 <[<]

 Move three cells over to the end of the next chunk
 <<<
]

PowerShell, 56 49 bajtów

-7 bajtów dzięki mazzy

param($n,$s)$s-split"(.{$n})"-ne''|%{$r=$_+$r};$r

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Mathematica, 46 bytes

""<>Reverse@Partition[Characters@#2,#,#,1,{}]&

Anonymous function. Takes a number and a string as input and returns a string as output. Not much to see here.

Javascript - 54 47 46 bytes

Remade:

(s,n)=>s.match(eval(`/.{1,${n}}/g`)).reverse()

Used as

f=(s,n)=>s.match(eval(`/.{1,${n}}/g`)).reverse()
alert(f("abcdefgh",2));

Thank you to @ETHproductions for some RegEx quickenning Thank you to @Shaggy for an extra byte in the eval!

Original:

(s,n)=>s.match(new RegExp('.{1,'+n+'}','g')).reverse()

Pyth, 5 bytes

s_c.*

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Explanation

   .*  splat implicit input
  c    split into chunks length n
 _     reverse
s      join

Retina, 38 bytes

1 byte saved thanks to @LeakyNun

^

+`(.* (1)+¶)((?<-2>.)+)
$3$1
 1+¶

(Note the space on the second line, and the trailing space)

This program takes input as unary on the first line, and the string on the second.

Try it online!

Test Suite! (slightly modified)

Explanation

The first step is to prepend a space (will become important later on).

^
 

Now we reverse. This uses .NET's balancing groups. It is important to note that groups here act as stacks, so every match is essentially pushed onto the stack. Here we capture every digit in the unary number into group 2. Now each time a character in the string is found, a match is popped from group 2. This ensures the the number of characters does not exceed that of the unary number.

+`(.* (1)+¶)                       Capture the unary number in group 2
             ((?<-2>.)+)           Balancing group for substrings
$3$1                               Reverse

And finally remove the unary number and the newline.

 1+¶

Java, 147 138 Bytes

String r(String s,int n){String r="";int l=s.length();for(int i=l/n*n;i>=0;i-=n)if(!(i>=l))r+=(i+n)>=l?s.substring(i):s.substring(i,i+n);return r;}

Saved 9 Bytes thanks to Kevin Cruijssen!

String r(String s,int n){String r="";int l=s.length(),i=l/n*n;for(;i>=0;i-=n)if(i<l)r+=i+n>=l?s.substring(i):s.substring(i,i+n);return r;}

In expanded form:

String r(String s,int n){
    String r="";
    int l=s.length(),i=l/n*n;
    for(;i>=0;i-=n)
        if(i<l)
            r+=i+n>=l?s.substring(i):s.substring(i,i+n);
    return r;
}

This is actually my first try to codegolf ever, so any feedback is welcome!

Perl 5, 25 bytes

Uses the -lnM5.010 flags.

say reverse<>=~/.{1,$_}/g

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Shoutout to Grinnz for telling me about =~ m/.{1,$n}/g

-M5.010 enables the use of the say function, which for our purposes is print with a shorter name.

-n puts the first line of input into $_, and -l chomps off the trailing newline.

We then get the second line of input using <>, and apply it to the regex .{1,$_}: any character, between 1 and $_ (the first input) times. Since this is greedy by default, it tries to always match $_ characters. The 1, is needed for the possible leftover chunk at the end.

The /g modifier gives us every match of that regex in the input string as a list, which is then reversed and printed. In Perl, passing a list to say joins it without any delimiter by default.

Dyalog APL Extended, 16 15 bytes

{∊⌽⍵⊂⍨(≢⍵)⍴=⍳⍺}

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Python, 62 bytes

lambda n,s:''.join([s[i:i+n]for i in range(0,len(s),n)][::-1])

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Stacked, 9 bytes

#<rev''#`

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#< chunks, rev reverses, and ''#` joins by empty string. Quite simple.

QBIC, 24 bytes

:;[1,_lA|,a|Z=_sA,b,a|+Z

This makes excellent use of the new substring-function I recently added to QBIC:

:;          Read in the cmd line params a (number) and A$ (text)
[1,_lA|,a|  Set up a FOR loop: FOR b = 1; b <= A$.length; b += a
Z=          Modify Z$; Z$ is autoprinted at the end of QBIC code
_sA,b,a|    SUBSTRING: _s is the function followed by the string 
               to take from, the starting pos and the # of chars
+Z          Take chunks from further into A$, put them before Z$

Pyth, 4 bytes

s_cF

Takes input as "s",n: Try it for yourself!

Convex, 2 bytes

/¥

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C, 69 bytes

i;f(s,n)char*s;{i=strlen(s);for(i-=i%n;printf("%.*s",n,s+i),i;i-=n);}

Result is printed out to the standard output.

Scala, 57 55 bytes

(n:Int,s:String)=>(""/:s.grouped(n).toSeq.reverse)(_+_)

Thanks Jacob! Try it here.

Note: By using the symbol form of foldLeft ("/:"), I was able to take off a couple more bytes.

Ohm, 5 bytes

σ]QWJ

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Explanation

σ]QWJ
σ         # Split input1 into input2 pieces
 ]        # Flatten array
  Q       # Reverses stack
   W      # Wraps stack to array
    J     # Joins stack
          # Implicit print

R, 69 60 bytes

function(s,n)cat(substring(s,(x=nchar(s):0*n)+1,x+n),sep="")

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Thanks to Kirill L. for the suggestion to remove seq.