12
Python 2, 76 bytes
f=lambda n,k=1000:n>19and(f(n/k,k/10)+`k-k%4*5`)*(n>=k)+f(n%k,k/10)or`n`[:n]
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Test suite from Chas Brown.
A fully recursive solution that splits on place values k=1000, k=100, and k=10. The separator for the place values is computed as k-k%4*5, which maps
1000 -> 1000
100 -> 100
10 -> 0
By stopping ...
7
Haskell, 82 73 bytes
r=read.reverse.show
f 1=0
f a=1+(minimum$f(a-1):[f$r a|r a<a,mod a 10>0])
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Simplest recursion method.
-9 bytes thanks to Christian Sievers
6
Wren, 126 124 bytes
As always Wren is the longest. I hate Wren's verbosity.
Fn.new{|a|a.map{|i|i>0?"":"-"+(i.abs.toString[-1..-(i.abs.toString.count)])}.map{|i|Num.fromString(i)}.reduce{|x,y|x<y?x:y}}
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Explanation
Fn.new{ // ...
6
Python 2, 61 60 59 bytes
lambda n:[a*(1+.03*n)**i*75for a in 4,8for i in range(n/2)]
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-1 byte, thanks to Joseph Sible
5
C++, 140 159 147 145 bytes
Edit: new solution without standard library, using a recursive function and pointer magic (constant 20 experimentally determined and not the same on other compilers)
Edit 2: -2 bytes thanks to ceilingcat
int*s,S,*G,x;int F(int Z,int*p=&x){int W=1,a=(*p)++,r=0;for(;r=r*10+a%10,a/=10;);G?W=r,p<=G?G=&W,S++,p=s:0:p=s=G=&...
5
Perl 5, 51 bytes:
sub f{($x,$y)=map{/-?/;$&.reverse$'}@_;$x>$y?$x:$y}
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...or if using a function from a core module is allowed (max in List::Util) it's 35 bytes:
35 bytes:
sub f{max map{/-?/;$&.reverse$'}@_}
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4
Gaia, 44 43 bytes
:(!>
ℍZ:₸Z¤1>∧'0פṙ$×:dẏ×,↑ℍṙ×
@3eZ,↑¦↓3eṙ×
Longer than my original answer, but it works now.
fixed my code again but now it's twice as sad
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@3eZ,↑¦↓3eṙ× Main function.
@3eZ Input divmod 1000, pushes input div 1000 then input mod 1000
, Pair those two value into a list
↑¦ Run the ...
4
Zsh, 63 bytes
for x;a+=(${(M)x#-}${(j::)${(Oas::)x#-}})
<<<$a[1+$[a[1]<a[2]]]
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${x#-} removes the leading - if it exists. Adding the (M) flag causes the - to be substituted instead of what remains.
Then this construct reverses the remaining string: ${(j::)${(Oas::)var}}.
For each element, we append to an array, then use a ...
4
05AB1E, 32 29 25 bytes
žDL.Xʒ•BÌË"•4в2ôsÇ87%èøOQ
-7 bytes thanks to @Grimy.
Try it online or verify all test cases.
Explanation:
žD # Push builtin 4096 (the maximum Roman number is 3999)
L # Create a list in the range [1,4096]
.X # Convert each integer to a Roman number
ʒ # Filter this list ...
4
Python 3, 89 86 bytes
Disclaimer: This answer does not work for the second example #4 due to floating point accuracy constraints in Python. Feel free to disqualify.
Update: This answer now works for all examples thanks to @JonathanAllan
f=float
r=lambda a:-f(a[:1:-1])if eval(a)<0else f(a[::-1])
m=lambda a,b:max(r(a),r(b))
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4
JavaScript (ES6), 100 bytes
Takes input as (a)(b).
a=>b=>Math.max((g=n=>(s=n>0||-1)*[...(s*n).toFixed(9).replace(/\.0+$/,0)].reverse().join``)(a),g(b))
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4
05AB1E, 6 7 6 bytes
+1 byte to fix cases when only one of the inputs is negative.
-1 byte by Grimmy
í'-δ†Z
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Test all inputs
The input is taken as a list of strings [x, y].
Explanation:
í Reverse each input
'- Push a minus sign
δ† Apply "filter out the minus sign to the front" to each input
Z Take the maximum
4
Python 3.8, 69 bytes
lambda*v:max(map(lambda x:(-1)**(z:=eval(x)<0)*float(x[z:][::-1]),v))
An unnamed function accepting 2 (actually, any number of) strings, which yields the maximal reversed version as a floating point number.
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4
Wolfram Language (Mathematica), 36 bytes
{#,2#}&[300(1+.03#)^Range[0,#/2-1]]&
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3
My first try on Code Golf, please point out some optimizations!
Python 3.7, 465 449 364 363 284 Bytes
Edit: Thanks for the improvements!
Jo King managed to squeeze this logic from 449 to just 364 bytes,
I took a single Byte off it so I felt justified to update it. :D
Description of the old function, which remains can still be found in the code:
To ...
3
Jelly, 16 bytes
Recursion might well end up being less bytes.
ṚḌ;‘))Fṭ
1Ç¡ċ€ċ0
A monadic Link accepting a positive integer which yields a non-negative integer
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Or try a faster, 17 byte version
How?
ṚḌ;‘))Fṭ - Helper Link: next(achievable lists)
) - for each (list so far):
) - for each (value, V, in that list):
Ṛ - ...
3
APL (Dyalog Unicode), 5 bytes
⍳⊥⊢÷!
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Using the mixed base trick found in my answer of another challenge. Uses ⎕IO←0.
How it works
⍳⊥⊢÷! Right argument: n, the number of terms
⊢÷! v: 1÷(n-1)!
⍳ B: The array of 0 .. n-1
⊥ Expand v to length-n array V,
then mixed base conversion of V in base B
Base | Digit | Value
-------...
3
RUBY, 24 bytes 18 bytes
n=0;10.times{puts n+=1}
thanks to @A_
10.times{|n|p n+1}
3
PHP, 137 136 143 bytes
function f($n){return$n>99?$n>999?f(substr($n,0,-3)).(1e3).f(substr($n,-3)):$n[0].'100'.f($n[1].$n[2]):($n>20&&$n%10?(0^$n/10).'0'.$n[1]:+$n);}
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3
Charcoal, 34 32 31 29 bytes
⪫Ｅ↨Ｎφ⪫Ｅ↨ι¹⁰⁰⪫↨λχ…0›Πλ﹪λχ100Ｉφ
Try it online! Link is to verbose version of code. Edit: Saved 2 bytes thanks to @Grimmy. Saved a further byte by finding a better way to choose between 0 or the empty string from a boolean. Saved a further 2 bytes because Base(l, 10) conveniently returns an empty array when l is zero. Explanation:
...
3
Python 3+inflect, 193 bytes
import inflect,re
exec("for i in re.sub('(and)|\W',' ',%sinput())).split():\n for j in range(0,6**8):\n if i in %sj):\n print(j,end='')\n break"%(('inflect.engine().number_to_words(',)*2))
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Very naive solution which uses Python's inflect library. Here is the unfurled code:
import inflect,re
for i in re.sub('(...
3
JavaScript, 69 bytes
Recursively splits and joins based on place values stored in k. Inspired by xnor's Python answer.
f=(n,k=1e3)=>n<20?+n||'':(n<k?'':f(n/k|0,k/10)+[k-10&&k])+f(n%k,k/10)
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3
Red, 74 bytes
func[a][forall a[a/1:(sign? a/1)* do reverse to""absolute a/1]max a/1 a/2]
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Takes the input as a list of two numbers.
The 5th test case (0.0000001 0) doesn't work in TIO, but works fine in the Red console:
3
JavaScript (Node.js), 64 bytes
a=>b=>(R=([h,...t])=>h?h+1<0?h+R(t):R(t)+h:t)(a)>+R(b)?R(a):R(b)
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Input / output as strings.
3
05AB1E, 15 bytes
₆v;F.03*>NmyтP,
Port of @TFeld's Python 2 answer, so make sure to upvote him.
Try it online or verify all test cases.
Explanation:
₆ # Push builtin integer 36
v # Loop `y` over both digits:
; # Halve the (implicit) input-integer
F # Inner loop `N` in the range [0, input/2):...
3
Jelly, 16 bytes
×.03‘*HḶ$×300;Ḥ$
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-2 bytes thanks to Nick Kennedy
Explanation
×.03‘*HḶ$×300;Ḥ$ Main Link
×.03 Multiply the number of bottles by 3%
‘ Increment (add 100%)
* (Vectorized) exponent to the power of:
HḶ$ [0, 1, ..., n-1]
×300 Multiply each factor by 300
...
3
Julia 1.0, 42 38 bytes
f(n,z=[75*(1+.03n).^(0:n÷2-1)])=4z,8z
-4 bytes using ideas from John and Joseph Sible.
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3
Haskell, 47 bytes
f n=[75*a*(1+0.03*n)**i|a<-[4,8],i<-[0..n/2-1]]
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3
Python 2, 55 bytes
n=input()/2
for c in 4,8:exec"print c*75;c*=1+.06*n;"*n
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The *75 trick is taken from other answers.
2
JavaScript, 95 93 92 bytes
f=n=>n>999?f(n/1e3|0)+[1e3]+f(n%1e3):n>99?(n/100|0)+'100'+f(n%100):n>20&&(x=n%10)?n-x+''+x:n
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-2 bytes (@Shaggy): store n%10 in a variable
-1 byte (@Arnauld): replace '1000' with [1e3]
Only top voted, non community-wiki answers of a minimum length are eligible
