Biorąc pod uwagę dodatnią liczbę całkowitą, znajdź jej najmniejszą wielokrotność dodatnią liczbę całkowitą, która jest przebiegiem 9, a następnie opcjonalnym przebiegiem 0. Innymi słowy, znajdź jego najmniejszą dodatnią liczbę całkowitą, która jest dopasowana przez wyrażenie regularne /^9+0*$/.

Na przykład, jeśli podaną dodatnią liczbą całkowitą jest 2, to zwróć 90, ponieważ 90 jest dodatnią liczbą całkowitą wielokrotności 2 i jest najmniejszą z wyrażeń regularnych /^9+0*$/.

Przypadki testowe:

n  f(n)
1  9
2  90
3  9
4  900
5  90
6  90
7  999999
8  9000
9  9
10 90
11 99
12 900
13 999999
14 9999990
15 90
16 90000

To jest golf golfowy . Najkrótsza odpowiedź w bajtach wygrywa. Obowiązują standardowe luki .

Galaretka , 13 11 bajtów

ṚḌ‘DS=ḍ@ð1#

Wypróbuj online!

Jak to działa

ṚḌ‘DS=ḍ@ð1#  Main link. Argument: n

        ð    Start a dyadic chain with arguments n and n.
         1#  Execute the chain to the left with left argument k = n, n+1, n+2, ...
             and right argument n until 1 match has been found. Return the match.
Ṛ                Get the decimal digits of k, reversed.
 Ḍ               Convert from base 10 to integer.
                 This essentially removes trailing zeroes. As a side effect, it
                 reverses the digits, which doesn't matter to us.
  ‘              Increment the resulting integer. If and only if it consisted
                 entirely of 9's, the result is a power of 10.
   DS            Compute the sum of the digits. The sum is 1 if and only if the
                 integer is a power of 10. Note that the sum cannot be 0.
      ḍ@         Test k for divisibility by n.
     =           Compare the results.

Python 2 , 55 54 bajtów

n=r=input()
while int(`10*r`.lstrip('9')):r+=n
print r

Wypróbuj online!

Python 2 , 51 bajtów

f=lambda n,r=9:r%n and f(n,10*r-10**n*r%-n/n*9)or r

Wypróbuj online!

JavaScript (ES6), 47 43 42 bajtów

-4 bajty dzięki @Arnauld
-1 bajtów dzięki @Luke

n=>eval('for(i=0;!/^9+0*$/.test(i);)i+=n')

Testy

let f=
n=>eval('for(i=0;!/^9+0*$/.test(i);)i+=n')

for(let i=1;i<=16;i++)console.log(`f(${i}) = `+f(i))

Rozwiązanie rekurencyjne (błąd 7, 13 i 14), 38 bajtów

n=>g=(i=0)=>/^9+0*$/.test(i+=n)?i:g(i)

Nazywa się jak f(5)(). Osiąga maksymalny rozmiar stosu wywołań w Chrome i Firefox dla n=7, n=13i n=14.

f=
n=>g=(i=0)=>/^9+0*$/.test(i+=n)?i:g(i)

for(let i=1;i<=16;i++){
	try { console.log(`f(${i}) = `+f(i)()) }
	catch(e) { console.log(`f(${i}) = FAILED`) }
}

Rubinowy , 36 bajtów

->x{r=0;1until"#{r+=x}"=~/^9+0*$/;r}

Brute-forcing - trwa wiecznie dla x = 17.

Wypróbuj online!

Java 8, 61 57 bajtów

n->{int r=0;for(;!(""+r).matches("9+0*");r+=n);return r;}

-4 bajty (i szybsze wykonanie) dzięki @JollyJoker .

Wyjaśnienie:

Wypróbuj tutaj.

n->{                              // Method with integer as parameter and return-type
  int r=0;                        //  Result-integer
  for(;!(""+r).matches("9+0*");   //  Loop as long as `r` doesn't match the regex
    r+=n                          //   And increase `r` by the input every iteration
  );                              //  End of loop
  return r;                       //  Return the result-integer
}                                 // End of method

Python 3 , 56 bajtów

f=lambda n,r=0:{*str(r).strip('0')}!={'9'}and n+f(n,n+r)

Wypróbuj online!

Brachylog , 16 bajtów

;I×≜.ẹḅhᵐc~a₀90∧

Wypróbuj online!

To jest dość powolne

Wyjaśnienie

;I×≜.              Output = Input × I
    .ẹḅ            Deconcatenate into runs of consecutive equal digits
       hᵐ          Take the head of each run
         c         Concatenate into a number
          ~a₀90∧   That number is a prefix of 90 (i.e. it's 9 or 90)

05AB1E , 10 bajtów

0[+D9Û0«_#

Wypróbuj online!

Po prostu dodaje wartość wejściową do 0, aż wynik minus wiodące 9 równa się 0.

RProgN 2 , 18 bajtów

x={x*'^9+0*$'E}éx*

Wyjaśnił

x={x*'^9+0*$'E}éx*
x=                  # Set the value of "x" to the input.
  {           }é    # Find the first positive integer in which passing it to the defined function returns truthy.
   x*               # Multiply the index by x, this essentially searches multiples now.
     '^9+0*$'       # A Regex defined by a literal string.
             E      # Does the multiple match the regex?
                x*  # Multiple the outputted index by x, giving the result.

Wypróbuj online!

Matematyka , 71 bajtów

(x=#;While[!StringMatchQ[ToString@x,RegularExpression@"9+0*"],x+=#];x)&

Wypróbuj online!

Niezbyt interesujące rozwiązanie brutalnej siły, ale przewyższa drugą odpowiedź Mathematiki, która wykorzystuje kilka sprytnych sztuczek.

The one redeeming quality Mathematica has in regards to this challenge is the fact that StringMatchQ requires a full match, so I can do 9+0* rather than ^9+0*$.

Batch, 175 bytes

@set/pn=
@set s=
:g
@set/ag=-~!(n%%2)*(!(n%%5)*4+1)
@if not %g%==1 set s=0%s%&set/an/=g&goto g
@set r=1
:r
@set s=9%s%
@set/ar=r*10%%n
@if %r% gtr 1 goto r
@echo %s%

Takes input on STDIN. Not a brute force solution but in fact based on my answer to Fraction to exact decimal so it will work for 17, 19, etc. which would otherwise exceed its integer limit anyway.

Mathematica, 127 bytes

Select[FromDigits/@Select[Tuples[{0,9},c=#],Count[#,9]==1||Union@Differences@Flatten@Position[#,9]=={1}&],IntegerQ[#/c]&][[1]]&

Input

[17]

Output

9999999999999999

here are the first 20 terms

{9, 90, 9, 900, 90, 90, 999999, 9000, 9, 90, 99, 900, 999999, 9999990, 90, 90000, 9999999999999999, 90, 999999999999999999, 900}

Haskell, 53 bytes

f takes and returns an integer.

f n=filter(all(<'1').snd.span(>'8').show)[n,n+n..]!!0

Try it online!

This times out for 17, which conveniently is just beyond the test cases. A faster version in 56 bytes:

f n=[x|a<-[1..],b<-[0..a-1],x<-[10^a-10^b],mod x n<1]!!0

Try it online!

How it works

• f generates all multiples of n, converts each to a string, filters out those with the right format, then takes the first one.

• The faster version instead uses that the required numbers are of the form 10^a-10^b, a>=1, a>b>=0. For golfing purposes it also uses the fact that for the minimal a, only one b can work, which allows it to generate the bs in the slightly shorter "wrong" order.

Ruby, 38 + 1 = 39 bytes

Uses -p flag.

$_=y=eval$_
1until"#{$_+=y}"=~/^9+0*$/

-p surrounds the program with:

while gets
    ...
end
puts $_

gets stores its result in $_. eval is used to convert it to a number, as it's shorter than .to_i, then brute force is used, incrementing $_ until it matches the regex. "#{}" is sttring interpolation, it's shorter than a .to_s call as that would require parantheses around $_+=y. Finally, $_ is printed.

Try it online!

Try all test cases!

Dyalog APL, 34 bytes

{∧/'0'=('^9+'⎕R'')⊢⍕⍵:⍵⋄∇⍵+r}n←r←⎕

Recursive dfns, based on Dennis' Python solution.

C++, 106 bytes

int main(){long N,T=9,j=10,M;cin>>N;while(T%N){if(T/j){T+=(M/j);j*=10;}else{T=(T+1)*9;j=10;M=T;}}cout<<T;}

Detailed Form:

int main()
{
    long N,T=9,j=10,M;
    cin >> N;

    while (T%N)
    {
        if (T/j)
        {
            T += (M/j);
            j *= 10;
        }
        else
        {
            T = (T+1)*9;
            j = 10;
            M = T;
        }
    } 

    cout << T;
}

TRY it online!

Python 2, 79 bytes

x=input();n=10;y=9
while y%x:
 b=n
 while(b-1)*(y%x):b/=10;y=n-b
 n*=10
print y

Try it online!

Some explanations It finds the smallest natural of form 10**n-10**b with n>b>=0 that divides the input.

Some IO

f(1) = 9
f(2) = 90
f(3) = 9
f(4) = 900
f(5) = 90
f(6) = 90
f(7) = 999999
f(8) = 9000
f(9) = 9
f(10) = 90
f(11) = 99
f(12) = 900
f(13) = 999999
f(14) = 9999990
f(15) = 90
f(16) = 90000
f(17) = 9999999999999999
f(18) = 90
f(19) = 999999999999999999

PHP, 39 bytes

for(;ltrim($r=$argn*++$i,9)>0;);echo$r;

Try it online!

PHP, 52 bytes

for(;!preg_match("#^9+0*$#",$r=$argn*++$i););echo$r;

Try it online!

Swift 3.0, Bytes:121

var i=2,m=1,n=""
while(i>0){n=String(i*m)
if let r=n.range(of:"^9+0*$",options:.regularExpression){print(n)
break};m=m+1}

Try it online!

Python 3, 62 bytes

This function takes an integer n and initializes m to zero. Then it removes all zeros from the ends of m and checks if the result only contains 9's, returning m if it does. If not, it adds n to m and checks again, etc.

def f(n,m=0):
 while{*str(m).strip('0')}!={'9'}:m+=n
 return m

Try it online!

Java (OpenJDK 8), 66 bytes, doesn't choke on 17

n->{long a=10,b=1;for(;(a-b)%n>0;b=(b<10?a*=10:b)/10);return a-b;}

Try it online!

Longer than @KevinCruijssen's solution but can handle slightly larger numbers. It calculates the candidate numbers like 10^6 - 10^3 = 999000. 64-bit longs are still the limit, breaking for n=23.

Can probably be golfed a bit but already took too long to make it work...

><>, 35 bytes

&a:v ;n-<
:,a/?(1:^!?%&:&-}:{
a*:\~

Try it online, or watch it at the fish playground!

Assumes the input is already on the stack. Works by looking for numbers of the form 10^a − 10^b, with a < b (yes, that's a less than sign — it takes fewer bytes!) until that's divisible by the input, then printing 10^b − 10^a. This is much faster than the brute force method (which would be difficult in ><> anyway).

V, 19 14 bytes

é0òÀ/[1-8]ü09

Try it online!

Explanation

é0              ' <m-i>nsert a 0
  ò             ' <m-r>ecursively
   À            ' <m-@>rgument times
               ' <C-A> increment the number (eventually gives all multiples)
     /[1-8]ü09  ' find ([1-8]|09) if this errors, the number is of the form
                ' (9+0*) (because there won't ever be just zeros)
                ' implicitly end the recursion which breaks on the above error

JavaScript (ES6), 51 49 bytes

let
f=(n,x=1,y=1)=>(x-y)%n?f(n,x,y*10):x-y||f(n,x*10)

<input type=number value=1 step=1 min=1 oninput=O.value=f(value)>
<input type=number value=9 id=O disabled>

Not the shortest approach, but it is wicked fast.

Mathematica, 82 bytes

Using the pattern of submission from @Jenny_mathy 's answer...

(d=x=1;y=0;f:=(10^x-1)10^y;n:=If[y>0,y--;x++,y=d;d++;x=1];While[Mod[f,#]!=0,n];f)&

Input:

[17]

Output:

9999999999999999

And relative to the argument in comments at @Jenny_mathy's answer with @Phoenix ... RepeatedTiming[] of application to the input [17] gives

{0.000518, 9999999999999999}

so half a millisecond. Going to a slightly larger input, [2003] :

{3.78, 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999}

a bit under 4 seconds.

Test table: On the first 30 positive integers, the results are

{9, 90, 9, 900, 90, 90, 999999, 9000, 9, 90, 99, 900, 999999, 
9999990, 90, 90000, 9999999999999999, 90, 999999999999999999, 900, 
999999, 990, 9999999999999999999999, 9000, 900, 9999990, 999, 
99999900, 9999999999999999999999999999, 90}

Explanation: The only magic here is the custom iterator ("iterator" in the CS sense, not the M'ma sense)

n := If[ y>0  ,  y-- ; x++  ,  y=d ; d++ ; x=1]

which acts on the global variables x, the number of leading "9"s, y, the number of trailing "0"s, and d, the total number of digits. We wish to iterate through the number of digits, and, for each choice of number of digits, start with the most "0"s and the least "9"s. Thus the first thing the code does is initialize d to 1, forcing x to 1 and y to 0. The custom iterator checks that the string of "0"s can be shortened. If so, it shortens the string of "0"s by one and increases the string of "1"s by one. If not, it increments the number of digits, sets the number of "0"s to one less than the number of digits, and sets the number of "9"s to 1. (This is actually done in a slightly different order to shave a few characters, since the old value of d is the desired value of y.)

Ti-Basic (TI-84 Plus CE), 48 41 bytes

Prompt X
For(K,1,0
For(M,-K+1,0
10^(K)-10^(-M
If 0=remainder(Ans,X
Return
End
End

Input is Prompt-ed during the program; output is stored in Ans.

Explanation:

Tries numbers of the form (10ⁿ)(10^m-1) = 10^k-10^m, where m+n=k starts at 1 and increases, and for each value of k, it tries m=1,n=k-1; m=2,n=k-2; ... m=k,n=0; until it finds a multiple of X.

This works up to 16; 17 gives a domain error because remainder( can only accept dividends up to 9999999999999 (13 nines), and 17 should output 9999999999999999 (16 nines).

Prompt X               # 3 bytes, input number
For(K,1,0              # 7 bytes, k in the description above; until a match is found
For(M,-K+1,0           # 10 bytes, start with n=1, m=(k-n)=k-1;
                           # then n=2, m=(k-n)=k-2, up to n=k, m=(k-n)=0
                           # (M=-m because it saved one byte)
10^(K)-10^(-M           # 8 bytes, n=(k-m) nines followed by m zeroes → Ans
If not(remainder(Ans,X # 8 bytes, If X is a factor of Ans (remainder = 0)
Return                 # 2 bytes, End program, with Ans still there
End                    # 2 bytes,
End                    # 1 byte (no newline)

QBIC, 53 bytes

{p=p+1┘o=p*9_F!o$|┘n=!A!~(_l!n$|=_l!n+1$|)-(o%:)|\_Xo

Explanation

{        infinitely DO
p=p+1    raise p (starts out as 0)
┘o=p*9   Get the mext multiple of 9 off of p
_F!o$|   Flip a string representation of p*9
┘n=!A!   and set 'n' to be an int version of the flipped p*9 
         (this effectively drops trailing 0's)
~        This IF will subtract two values: the first is either 0 for n=x^10, or -1
         and the second bit does (p*9) modulo 'a' (input number): also 0 for the numbers we want
(
 _l!n$|  the length of n's string representation
=        is equal to
_l!n+1$| the length of (n+1)'s string rep (81 + 1 = 82, both are 2 long; 99 + 1 = 100, there's a difference)
)        The above yields -1 (Qbasic's TRUE value) for non-9 runs, 0 for n=x^10
-        Subtract from that 
(o%:)    (p*9) modulo a     0 for p*9 = a*y
|       THEN (do nothing, since we want 0+0=0 in the conditionals above, execution of the right path jumps to ELSE
\_Xo    ELSE quit, printing (p*9)

C (gcc), 126 bytes

#include<stdio.h>
main(x,n,y,b){n=10;y=9;scanf("%d",&x);while(y%x){b=n;while((b-1)*(y%x)){b/=10;y=n-b;}n*=10;}printf("%d",y);}

Try it online!

Some explanations It finds the smallest natural of form 10**n-10**b with n>b>=0 that divides the input.

Some IO

f(1) = 9
f(2) = 90
f(3) = 9
f(4) = 900
f(5) = 90
f(6) = 90
f(7) = 999999
f(8) = 9000
f(9) = 9
f(10) = 90
f(11) = 99
f(12) = 900
f(13) = 999999
f(14) = 9999990
f(15) = 90
f(16) = 90000

Perl 5, 23 + 2 (-pa) = 25 bytes

Brute Force Method

$_+=$F[0]while!/^9+0*$/

Try it online!

It's slow, but it's tiny.

More Efficient Method:

41 + 2 (-pa) = 43 bytes

$_=9;s/0/9/||($_=9 .y/9/0/r)while$_%$F[0]

Try it online!

It works well for any input, but it's longer code.