Qualche tempo fa c'era una sfida sulla moltiplicazione delle stringhe. Ci ha mostrato come possiamo moltiplicare non solo i numeri, ma anche le stringhe. Tuttavia, non possiamo ancora moltiplicare un numero per una stringa correttamente. C'è stato un tentativo di farlo, ma questo è ovviamente sbagliato. Dobbiamo sistemarlo!

Il tuo compito:

Scrivi una funzione o un programma che moltiplica due input, una stringa e un numero intero. Per moltiplicare (correttamente) una stringa per un numero intero, dividere la stringa in caratteri, ripetere ogni carattere un numero di volte uguale all'intero, quindi incollare nuovamente i caratteri. Se il numero intero è negativo, utilizziamo il suo valore assoluto nel primo passaggio, quindi invertiamo la stringa. Se l'ingresso è 0, non viene emesso nulla (qualsiasi cosa moltiplicata per 0 è uguale a nulla).

Ingresso:

Una stringa composta esclusivamente da caratteri ASCII e newline stampabili e un numero intero (possibile negativo).

Produzione:

La stringa moltiplicata per l'intero.

Esempi:

Hello World!, 3            --> HHHeeellllllooo   WWWooorrrlllddd!!!
foo, 12                    --> ffffffffffffoooooooooooooooooooooooo
String, -3                 --> gggnnniiirrrtttSSS
This is a fun challenge, 0 --> 
Hello
World!, 2                  --> HHeelllloo

                               WWoorrlldd!!

punteggio:

Questo è code-golf , vince il conteggio di byte più basso!

Gelatina , 6 5 4 byte

²Ɠxm

Provalo online!

Come funziona

²Ɠxm  Main link. Argument: n (integer)

²     Yield n².
 Ɠ    Read and eval one line of input. This yields a string s.
  x   Repeat the characters of s in-place, each one n² times.
   m  Takes each |n|-th character of the result, starting with the first if n > 0, 
      the last if n < 0.

JavaScript (ES6), 63 byte

Accetta input nella sintassi del curry (s)(n) .

s=>n=>[...s].reduce((s,c)=>n<0?c.repeat(-n)+s:s+c.repeat(n),'')

Casi test

let f =

s=>n=>[...s].reduce((s,c)=>n<0?c.repeat(-n)+s:s+c.repeat(n),'')

console.log(f(`Hello World!`)(3))
console.log(f(`foo`)(12))
console.log(f(`String`)(-3))
console.log(f(`This is a fun challenge`)(0))
console.log(f(`Hello
World!`)(2))

Python 3 , 44 byte

f=lambda s,n:s and s[0]*n+f(s[1:],n)+s[0]*-n

Provalo online!

Python 2 , 59 57 50 46 byte

-2 byte grazie ad Anders Kaseorg. -4 byte grazie a Dennis.

lambda s,n:''.join(i*n**2for i in s)[::n or 1]

Provalo online!

05AB1E , 10 byte

S²Ä×J²0‹iR

Provalo online!

S          # Split the string into characters
 ²Ä×       # Repeat each character abs(integer) times
    J      # Join into a string
     ²0‹i  # If the integer is less than 0...
         R #   Reverse the string

MATL , 9 byte

y|Y"w0<?P

Gli input sono: numero, quindi stringa.

Stringhe con a capo sono inseriti utilizzando char 10come segue: ['first line' 10 'second line'].

Provalo online! Oppure verifica tutti i casi di test .

Spiegazione

Considera gli input -3 e 'String'.

y      % Implicitly take two inputs. Duplicate from below
       % STACK: -3, 'String', -3
|      % Absolute value
       % STACK: -3, 'String', 3
Y"     % Run-length decoding
       % STACK: -3, 'SSStttrrriiinnnggg'
w      % Swap
       % STACK: 'SSStttrrriiinnnggg', -3
0<     % Less than 0?
       % STACK: 'SSStttrrriiinnnggg', 1
?      % If so
  P    %   Flip
       %   STACK: 'gggnnniiirrrtttSSS'
       % End (implicit). Display (implicit)

Haskell , 41 36 byte

f n|n<0=reverse.f(-n)|1<3=(<*[1..n])

Provalo online!

Esempio di utilizzo: f (-3) "abc" rese "cccbbbaaa".

Modifica: -5 byte grazie a xnor!

V , 29, 23, 18 , 17 byte

æ_ñÀuñÓ./&ò
ÀäëÍî

Provalo online!

hexdump:

00000000: e65f f1c0 75f1 d32e 2f26 f20a c0e4 ebcd  ._..u.../&......
00000010: ee                                       .

Grazie a @ nmjcman101 per aver salvato 6 byte, il che mi ha incoraggiato a salvarne altri 5!

La revisione originale è stata piuttosto terribile, ma ora sono davvero orgoglioso di questa risposta perché gestisce numeri negativi sorprendentemente bene. (V ha quasi nessun supporto numerico e nessun supporto per i numeri negativi)

Spiegazione:

æ_          " Reverse the input
  ñ  ñ      " In a macro:
   À        "   Run the arg input. If it's positive it'll give a count. If it's negative
            "   running the '-' will cause V to go up a line which will fail since we're
            "   on the first line, which will break out of this macro
    u       "   (if arg is positive) Undo the last command (un-reverse the line)
      Ó./&ò " Put every character on it's own line

A questo punto, il buffer si presenta così:

H
e
l
l
o

w
o
r
l
d
!
<cursor>

È importante non la nuova riga finale e che il cursore sia su di essa.

À           " Run arg again. If it's negative, we will move up a line, and then give the 
            " absolute value of the count. If it's positive (or 0) it'll just give the
            " count directly (staying on the last line)
 ä          " Duplicate... (count times)
  ë         "   This column. 
   Íî       " Remove all newlines.

R, 83 78 76 byte

function(s,i)cat('if'(i<0,rev,`(`)(rep(el(strsplit(s,'')),e=abs(i))),sep='')

Funzione anonima.

Federico ha salvato 3 byte, Giuseppe ha salvato 2 4.

Spiegazione:

     el(strsplit(s,''))                      # split string into list characters
 rep(                  ,e=abs(i)))           # repeat each character abs(i) times


    'if'(i<0,rev,   ){...}                 # if i>0, reverse character list
                 `(`                       # otherwise leave it alone: `(` is the identity function
cat(                      ,sep='')         # print the result

test:

> f('Hello World!', 3 )
HHHeeellllllooo   WWWooorrrlllddd!!!
> f('foo', 12)
ffffffffffffoooooooooooooooooooooooo
> f('String', -3)
gggnnniiirrrtttSSS
> f('This is a fun challenge', 0)
> f('Hello
+ World!', 2)
HHeelllloo

WWoorrlldd!!

05AB1E, 10 bytes

0‹FR}ʒ¹Ä×?

Try it online!

Explanation

0‹F         # input_1 < 0 times do:
   R        # reverse input_2
    }       # end loop
     ʒ      # filter
      ¹Ä×   # repeat current char abs(input_1) times
         ?  # print without newline

PHP>=7.1, 65 bytes

for([,$s,$n]=$argv;$i<strlen($s)*abs($n);)echo$s[$i++/$n-($n<0)];

PHP Sandbox Online

Brain-Flak (BrainHack), 154 152 bytes

([(({})(<()>))]<>)<>{({}()<([{}]()<([{}])>)<>({}<>)<>>)<>}{}<>{}<>({}<([][()]){{}({<({}<(({}<>)<>)>())>[()]}<{}{}>)([][()])}{}{}<>>){{}{({}<>)<>}(<>)}{}

Try it online!

Just here to give DJMcMayhem some competition. ;)

Explanation

Here's a modified version of DJMcMayhem's explanation

#Compute the sign and negative absolute value 
([(({})<(())>)]<>)<>{({}()<([{}]()<([{}])>)<>({}<>)<>>)<>}{}<>{}<>

#Keep track of the sign
({}<

    #For each char in the input string:
    ([][()])
    {
        {}

        #Push n copies to the alternate stack
        ({<({}<(({}<>)<>)>())>[()]}<{}{}>)

        #Endwhile
        ([][()])
    }{}{}<>

#Push the sign back on
>)

#If so...
{{}

    #Reverse the whole stack
    {({}<>)<>}

    #And toggle over, ending the loop
    (<>)
}

#Pop the counter off
{}

J, 19 15 13 bytes

(#~|)A.~0-@>]

Try it online!

Explanation

        0-@>]      NB. first or last index depending on sign of right arg
     A.~           NB. get first or last Anagram of left arg
(#~|)              NB. copy left arg, absolute-value-of-right-arg times

Dyalog APL, 15 bytes

{⌽⍣(⍵<0)⊢⍺/⍨|⍵}

String as a left argument, number as a right argument.

Try it online!

How?

⍺/⍨ - repeat the string

|⍵ - abs(number) times

⌽⍣ - reverse if

(⍵<0) - the number is below 0

MATLAB, 37 bytes

@(s,n)flip(repelem(s,abs(n)),(n<0)+1)

This defines and anonymous function with inputs s: string and n: number.

Example runs:

>> @(s,n)flip(repelem(s,abs(n)),(n<0)+1)
ans = 
    @(s,n)flip(repelem(s,abs(n)),(n<0)+1)

>> f = ans;

>> f('String', 3)
ans =
SSStttrrriiinnnggg

>> f('String', -3)
ans =
gggnnniiirrrtttSSS

>> f('String', 0)
ans =
   Empty matrix: 1-by-0

Brain-Flak (Haskell), 202 192 bytes

(({})<(([({})]<>)){({}()<([{}])<>({}<>)<>>)<>}{}([{}]<><{}>)([][()]){{}({<({}<(({}<>)<>)>[()])>()}<{}{}>)([][()])}{}{}<>>)([({}<(())>)](<>)){({}())<>}{}{((<{}>))<>{}}{}<>{}{{}{({}<>)<>}(<>)}{}

Try it online!

This is probably the worst possible language to do it in, but it's done. Thanks to @Wheatwizard for providing the Haskell interpreter, which allows mixed input formats. This would be about 150 bytes longer without it.

Explanation:

#Keep track of the first input (n)
(({})<

    #Push abs(n) (thanks WheatWizard!)
    (([({})]<>)){({}()<([{}])<>({}<>)<>>)<>}{}([{}]<><{}>)

    #For each char in the input string:
    ([][()])
    {
        {}

        #Push n copies to the alternate stack
        ({<({}<(({}<>)<>)>[()])>()}<{}{}>)

        #Endwhile
        ([][()])
    }{}{}<>

#Push the original n back on
>)

#Push n >= 0
([({}<(())>)](<>)){({}())<>}{}{((<{}>))<>{}}{}<>{}

#If so...
{{}

    #Reverse the whole stack
    {({}<>)<>}

    #And toggle over, ending the loop
    (<>)
}

#Pop the counter off
{}

Java (OpenJDK 8), 99 98 89 87 85 bytes

s->n->{for(int i=s.length*(n<0?n:-n),r=n<0?0:~i;i++<0;)System.out.print(s[(i+r)/n]);}

Try it online!

• -2 bytes thanks to @Xanderhall
• -2 bytes thanks to @Nevay

Octave, 49 bytes

@(s,n){t=repmat(s,abs(n),1)(:)',flip(t)}{2-(n>0)}

Try it online!

I will provide an explanation tomorrow.

Ruby, 59 +1 = 60 bytes

Uses -n flag.

n=eval$_
a=$<.read
a.reverse!if n<0
a.chars{|i|$><<i*n.abs}

Try it online!

Charcoal, 16 bytes

Ｆθ¿‹η0Ｆ±Ｉη←ιＦＩηι

Try it online! Link is to verbose version of code. Explanation:

Ｆθ              For each character in the input string
  ¿‹η0          If the input number is less than zero
      Ｆ±Ｉη      Repeat the negation of the input number times
          ←ι    Print the character leftwards (i.e. reversed)
      ＦＩη       Otherwise repeat the input number times
         ι      Print the character

CJam, 9 bytes

q~__*@e*%

Try it online!

Japt, 12 bytes

®pVaìr!+sVg

Try it online!

Explanation

Implicit input of string U and integer V.

®pVaÃ

Map (®) each letter of U (implicitly) to itself repeated (p) abs(V) (Va) times.

¬r

Turn the string into an array of chars (¬) and reduce (r) that with...

!+sVg

"!+".slice(sign(V)) - this either reduces with + → a + b, or with !+ → b + a.
Thanks @Arnauld for the backwards-reduce idea!

WendyScript, 46 bytes

<<f=>(s,x){<<n=""#i:s#j:0->x?x>0n+=i:n=i+n/>n}

f("Hello World", -2) // returns ddllrrooWW  oolllleeHH

Try it online!

Explanation (Ungolfed):

let f => (s, x) {
  let n = ""
  for i : s
    for j : 0->x
      if x > 0 n += i
      else n = i + n
  ret n
}

C89 bytes

main(int c,char**v){for(;*v[1];v[1]++)for(c=atoi(v[2]+(*v[2]=='-'));c--;)putchar(*v[1]);}

I saw Ben Perlin's version and wondered if you couldn't be shorter still and also have a full program; surely, atoi() and putchar() aren't that expensive in terms of bytes? Seems I was right!

Pyth, 13 11 bytes

*sm*.aQdz._

Try it!

-2 bytes thanks to @jacoblaw

explanation

*sm*.aQdz._   
  m     z     # map onto the input string (lambda var: d)
   *.aQd      # repeat the char d as often as the absolute value of the input number 
 s            # sum the list of strings into a single string
*        ._Q   # Multiply with the sign of the implicit input value: reverse for negative Q 

old approach, 13 bytes

_W<Q0sm*.aQdz

Try it!

Python 3, 68 bytes

h=lambda s,n:h(s[::-1],-n)if n<0 else s[0]*n+h(s[1:],n)if s else s*n

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QBIC, 32 bytes

g=sgn(c)[_l;||[:*g|?_sA,b*g,1|';

Explanation

            Takes inputs A$ ('Hello'), and c (-3) from the cmd line
g=sgn(c)    Save the sign of c          -1
[_l;||      FOR each char in A$
[:*g|       FOR the number of repetitions wanted    (ie: -3 * -1)
            Note that : reads a number from the cmd line, and c is the first 
            available variable to save it in after a and b got used as FOR counters.
            Also note that a negative value times the sign becomes positive.
?_s         PRINT a substring
  A         of A$
 ,b*g       startng at char n, where n is the first FOR loop counter times the sign
                That means that when c is negative, so is this. A negative starting index
                on Substring instructs QBIC to take from the right.
 ,1|        taking 1 char.
';          This bit injects a literal ; in the output QBasic, to suppress newlines om PRINT

Mathematica, 89 bytes

(T=Table;t=""<>T[s[[i]]~T~Abs@#2,{i,Length[s=Characters@#]}];If[#2>0,t,StringReverse@t])&

input

["Hello World!", 3]

Braingolf, 22 bytes

1-v{R.[v.R]v}R[v>R]v&@

Try it online!

Eeh, not bad.

Takes input as an integer and an array of characters.

Alternatively:

Braingolf, 31 bytes

l1->[M]1-v&,{R.[v.R]v}R[v>R]v&@

Try it online!

Takes input as an integer and a string

C, 109 bytes

char *f(int n, char *s){char *o=calloc(n,strlen(s)+1),*t=o;while(*s){for(int i=n;i--;)*t++=*s;s++;}return o;}

Starting with a function declaration that takes an int and a string and produces a string (it seems implied that memory is not preallocated and must be created) it seems that the straight-forward approach is shorter than any attempts at being cleaver that I had tried.

char *f(int n, char *s){
  char *o=calloc(n, strlen(s)+1),
    *t=o;

  while (*s) {
    for(int i=n; i--; )
      *t++=*s;
    s++;
  }

 return o;

}