6
Python 3, 149 \$\cdots\$ 145 140 bytes
from datetime import*
def f(d):
n,s=(date(*d)-date(1,1,1)).days+1137143,""
for i in[144000,7200,360,20,1]:s+=f".{n//i}";n%=i
return s[1:]
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Takes Gregorian date as [year, month, day] list and returns Mayan date as f"{B'ak'tun}.{K'atun}.{Tun}.{Winal}.{Day}"(loosely as a python f-string).
5
Mathematica, 220 218 bytes
For[n=1,1>0,n++,For[m=1,1>0,m++,x=Array[a,{m,n}];d=Table[(Norm[x[[i]]-x[[j]]]^2-1)*(Norm[x[[i]]-x[[j]]]^2-y)==0,{i,1,m-1},{j,i+1,m}];If[Not[CylindricalDecomposition[Flatten@{d,y>0},Flatten@{y,x}]],Break[]]];Print[m-1]]
Try it online! Un-golfed version:
For[ n = 1, 1>0, n++,
For[ m = 1, 1>0, m++,
x = Array[a, {...
4
Jelly, 14 bytes
ŒṖḌD$ƑƇḌQƑƇLÞṪ
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A monadic link taking a list of base-10 digits and returning a list of integers.
Added a byte (and switched to numbers) because of the zero padding issue.
Added 5 bytes to fix a further issue with zeros pointed out by @KevinCruijssen - thanks!
Explanation
ŒṖ | Partitions of list
ḌD$ƑƇ ...
4
JavaScript (ES6), 96 bytes
Assumes that the system is set up to UTC time (as the TIO servers are)
Takes input as 3 distinct parameters (year, month, day).
(y,m,d)=>[144e3,7200,360,20,1].map(k=>n/(n%=k,k)|0,n=new Date(y+4e3,m-1,d+395335)/864e5).join`.`
Try it online!
How?
We can't safely pass a year \$y<100\$ to the Date constructor, as it gets ...
3
APL (Dyalog Unicode), 2 bytesSBCS
Anonymous tacit prefix function. Requires ⎕IO←0
⊃⍒
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⍒ gives the "grade". The indices into the argument that would stably sort it in descending order
If the first element is less than the second, it gives [1,0]
If the the elements are equal, it gives [0,1]
If the second element is less than the second, it ...
3
Microsoft PowerPoint (33 Slides, 512 shapes to meet minimum requirements)
Storage representation is in hexadecimal.
Directions
Click the blue square to advance the counter (or propagate the carry), or make an AutoHotKey Script to click it for you.
You can view the number's digits at any time using the green buttons. (It causes unintended behavior if you ...
3
APL (Dyalog Extended), 16 bytesSBCS
⎕0+⌈3÷⍨1 2×⌊⎕×√3
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A full program that takes y then x from standard input and prints a 2-element vector of integers.
How it works: the math
First of all, note that any Gaussian integer will be placed on the vertical diagonal of a diamond, with the point \$Z\$ placed at \$(x,\sqrt3y)\$ for some integer \$x,...
3
05AB1E, 54 52 bytes
365*Š3‹¹α4т‚DPª÷®β•ë˜¿•ºS₂+²£`•H`Ø•OŽQív₂y-‰R`})R'.ý
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First step: compute the day number.
365 # literal 365
* # multiply
# => 365*year is left on the stack for now
Š # get the other two inputs
# => day is left on the stack
3‹ ...
3
JavaScript (Node.js), 74 bytes
d=>(i=5,g=(a,m=20-8%--i)=>i?g(a/m|0)+'.'+a%m:a)(new Date(d)/864e5+1856305)
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Input as yyyy-MM-dd format string.
Thanks Arnauld, save 4 bytes.
3
Jelly, 7 bytes
ṁ€æl/}Z
A dyadic Link accepting a list of the two lists on the left and a list of their respective lengths on the right which yields a lists of lists.
(A monadic Link accepting only the left argument costs 8 bytes, ṁ€Ẉæl/$Z)
Try it online! Or see the test-suite.
How?
ṁ€æl/}Z - Link: list of lists [a, b], list [length(a), length(b)]
} ...
3
R, 69 60 54 51 bytes
function(a,b,x,y)cbind(rep(a,y/sum(!(1:x*y)%%x)),b)
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Loosely based on my answer to this question. Takes a,b, and the lengths as x and y, respectively.
Returns a two-column matrix.
Thanks to Robin Ryder for saving 4 bytes, and then another 3.
Relies on the identity \$xy=GCD(x,y)\cdot LCM(x,y)\$; repeats a by \$y/GCD(x,...
3
J, 14 13 bytes
,./@($&>~*./)
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-1 byte thanks to Bubbler
This answer takes the two lists as its left arg, and the two list lengths as its right arg.
($&>~*./) is a dyadic hook which applies gcd between the two right arg lengths *./ and the reshapes the two left arg lists $ to that length after opening them &>. At this ...
2
naz, 50 bytes
1x1f1a1o2x1v0m9a1a1o1v0x1f1f1f1f1f1f1f1f1f8s1o1s1o
Explanation (with 0x commands removed)
1x1f # Function 1
1a1o # Add 1 to the register and output
2x1v # Store the new value in variable 1
0m9a1a1o # Output a newline
1v # Read variable 1 into the register
...
2
BrainFuck, 41 Bytes
-[----->+>+<<]>---<++++[->+.+.<]>+.>--.-.
Output
12345678910
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2
APL, 9 chars/bytes
2≡≢∪∨∘⍳⍨⎕
get the input
⎕
compute least common divisor between n and all integers from 1 to n
∨∘⍳⍨
count of unique values
≢∪
if it's exactly 2, then it's prime
2≡
2
Intcode, 22 bytes
3,0,3,1,2,0,1,2,4,2,99
Fairly basic. IO boilerplate takes up most of the bytecount, given that the language has multiplication as a builtin (2,0,1,2 would be a valid answer, if snippets were allowed)
2
Keg, 5 bytes
¿:0=+
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Explanation
¿: # Take 2 inputs
0= # Check whether the input is 0
# (Yields 1 if the input is 0)
# (Yields 0 otherwise)
+ # Add the input with this result
# i.e. 0->1, 5->5
Keg, 4 bytes
Keg is not yet updated
:0=+
2
Perl 6, 49 bytes
{m:ex/(0|<![0]>.+)+<!{$0>set ~<<$0}>/.max(+*[0])}
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Outputs a regex match with a list of submatches as the integers.
Explanation:
{ } # Anonymous code block
m:ex/ / # Match all
( )+ # ...
2
05AB1E, 10 bytes
.œʒÙïJQ}éθ
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2
C# (Visual C# Interactive Compiler), 186 163 bytes
string f(int[]t){int x,n=(int)(new DateTime(t[0],t[1],t[2])-new DateTime(1,1,1)).TotalDays+1137143;return string.Join(".",new[]{144000,7200,360,20,1}.Select(e=>{x=n/e;n%=e;return x;}));}
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Edit: Use int[] as input instead of DateTime to meet the specification better. Thanks @keizerharm for ...
2
Perl 6, 35 bytes
{$^a <<,<<flat $^b xx($a lcm$b)/$b}
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2
C (clang), 70 68 bytes
i;f(*a,*b,x,y){for(i=0;printf("(%d,%d)",a[i%x],b[i%y]),++i%x+i%y;);}
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Takes input as pointer to a and b and the lengths, prints q to std out.
We use the modulo of an iterator by the data lengths e.g. i % x and i % y to access data, when both modules are equal to 0 , i is LCM and the function terminates.
1
JavaScript (ES6), 100 bytes
Takes input as a string.
f=([v,...a],p=o=[],c='')=>(v&&f(a,p,c+v),c?p.includes(c)|[+c]!=c||f(a,[...p,c],v):p[o.length]?o=p:o)
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Commented
f = ( // f is a recursive function taking:
[v, // v = next digit
...a], // a[] = array of remaining ...
1
Ruby, 115 112 114 bytes
Takes input as a string. A recursive function that takes a string and a list of previously-used numbers; it gets all substrings that start with the first character, and for each gets the result of the recursive call on the rest of the string. It then decides what the longest string that still matches is.
f=->s,*e{(1..l=s.size)....
1
TI-BASIC, 3 bytes
Ans-0
Very simple and works the same way as the top J answer.
Input is an integer in Ans.
Examples:
4:Ans-0
4
4:0-Ans
-4
1
W j, 7 / 2 = 3.5 points
I updated the ord to support W's own code page.
0“„√*"C
Explanation
0“„√*" % Define a string
C % Convert to codepoints [48,15,16,23,42]
Flag:j % Join the output list without a separator
```
1
Burlesque, 39 bytes
riXX:nz{J2dv{{9r@}{ro}}che!<-}m[tp)im++
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ri #Read as int
XX #Return list of digits
:nz #Filter for non-zero
{
J2dv #Divisible by 2?
{
{9r@} #Range a, 9
{ro} #Range 1, a
}che! #Run based on if divisible
<- #Reverse the range
}m[ #Apply to each digit
tp #Transpose digits
)im #...
1
05AB1E, 16 bytes
0KεDÈi9ŸëL]íõζJO
Try it online or verify all test cases.
Or alternatively:
0KεDÈ8*>Ÿ{R}õζJO
0Kε9Ÿ¬L‚yèR}õζJO
Explanation:
0K # Remove all 0s from the (implicit) input-integer
ε # Map each digit to:
D # Duplicate the digit
Èi # If it's even:
9Ÿ # Pop and push a list in the range [digit, 9]
...
1
C (gcc), 19 bytes
#define f(n)n*-~n/2
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C (gcc), 15 bytes
f(n){n*=-~n/2;}
-5 bytes thanks to ceilingcat!
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1
Clojure - 19 bytes
#(apply +(range %))
Boring answer is boring...
Only top voted, non community-wiki answers of a minimum length are eligible
