22
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Given a number 1≤n≤365, output the nth day of the year in "Day-numberth Month" format. For example, given 1, you should output "1st January", without "of".

The Gregorian calendar will be used and the program should not account for leap years, so your program should never output "29th February" in any circumstance. Any method can be used, as long as it follows the "Day-numberth Month" format mentioned before. Your program should also output ordinals correctly, meaning it should always output 1st, 2nd, 3rd, should 1, 2 or 3 respectively be the day numbers for any input. Leading spaces or other indentation are allowed.

This is code golf, so the shortest solution by characters wins.

Test cases:

1 gives 1st January
2 gives 2nd January
3 gives 3rd January
365 gives 31st December
60 gives 1st March
11 gives 11th January
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  • 4
    \$\begingroup\$ Also, do you need to force an error message on numbers > 365? Can the program just assume that's invalid input and it won't need to handle that? \$\endgroup\$ – Rɪᴋᴇʀ Mar 13 at 15:29
  • 5
    \$\begingroup\$ As not everyone is a native English speaker, you may want to add that day numbers 11, 12, and 13 get "th", numbers ending in "1" get "st", "2" get "nd", "3" get "rd", and all other get "th". \$\endgroup\$ – Adám Mar 13 at 15:38
  • 9
    \$\begingroup\$ Whoa, don't accept answers so quickly. Especially not wrong answers! \$\endgroup\$ – Adám Mar 13 at 16:10
  • 6
    \$\begingroup\$ You should add at least 11 (11th January) and 21 (21st January) to the test cases. \$\endgroup\$ – Arnauld Mar 13 at 16:30
  • 1
    \$\begingroup\$ And while you're editing test cases, maybe specify what exactly your test case format is. A couple of answerers have thought that 123= was part of the required output. Or simply edit your test cases to read something like: 365 gives 31st December \$\endgroup\$ – Adám Mar 13 at 16:36

20 Answers 20

9
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PHP, 38 40 30 28 bytes

<?=date("jS F",86399*$argn);

Try it online!

Run with php -nF input is from STDIN. Example (above script named y.php):

$ echo 1|php -nF y.php
1st January
$ echo 2| php -nF y.php
2nd January
$ echo 3| php -nF y.php
3rd January
$ echo 11|php -nF y.php
11th January
$ echo 21|php -nF y.php
21st January
$ echo 60|php -nF y.php
1st March
$ echo 365|php -nF y.php
31st December

Explanation

Construct an epoch timestamp for the desired day in 1970 (conveniently not a leap year) by multiplying the day number * number of seconds per day (86400). However, this would yield one day higher so instead multiply by number of seconds in a day - 1 (86399) which for the range of input numbers (1≤n≤365) will result with the timestamp of the end of each correct day. Then just use PHP's built-in date formatting for output.

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  • \$\begingroup\$ why's the -n necessary? \$\endgroup\$ – Ven Mar 13 at 16:23
  • \$\begingroup\$ @Ven it might not be in all cases, but just disables any settings in local php.ini that might create inconsistent behavior. \$\endgroup\$ – 640KB Mar 13 at 16:31
6
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Jelly,  79 78  77 bytes

-1 fixing a bug :) (shouldn't pre-transpose to find index, should post-reverse, but then we can tail rather than head)
-1 using reflection (⁽©ṅB+30_2¦2 -> ⁽0ṗb4+28m0)

⁽0ṗb4+28m0SRṁRƲœiµṪȮ%30%20«4ị“nḄƲf⁷»s3¤Ṗ,ị“£ṢtẒ⁽ẹ½MḊxɲȧėAṅ ɓaṾ¥D¹ṀẏD8÷ṬØ»Ḳ¤$K

A full program which prints the result

Try it online!

How?

will update this later...

⁽©ṅB+30_2¦2SRṁRƲZœiµḢȮ%30%20«4ị“nḄƲf⁷»s3¤Ṗ,ị“...»Ḳ¤$K - Main Link: integer, n
⁽©ṅB+30_2¦2SRṁRƲZœi - f(n) to get list of integers, [day, month]
⁽©ṅ                 - compressed literal 2741
   B                - to a list of binary digits -> [ 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1]
    +30             - add thirty                    [31,30,31,30,31,30,31,31,30,31,30,31]
         ¦          - sparse application...
        2           - ...to indices: [2]
       _  2         - ...action: subtract two       [31,28,31,30,31,30,31,31,30,31,30,31]
               Ʋ    - last four links as a monad - i.e. f(x):
           S        -   sum x                       365
            R       -   range                       [1..365]
              R     -   range x (vectorises)        [[1..31],[1..28],...]
             ṁ      -   mould like                  [[1..31],[32..59],...]
                Z   - transpose                     [[1,32,...],[2,33,...],...]
                 œi - 1st multi-dimensional index of n  -> [day, month]

µḢȮ%30%20«4ị“nḄƲf⁷»s3¤Ṗ,ị“...»Ḳ¤$K - given [day, month] format and print
µ                                  - start a new monadic chain - i.e. f(x=[day, month])
 Ḣ                                 - head -- get the day leaving x as [month])
  Ȯ                                - print it (with no newline) and yield it
   %30                             - modulo by thirty
      %20                          - modulo by twenty
         «4                        - minimum of that and four
                     ¤             - nilad followed by link(s) as a nilad:
            “nḄƲf⁷»                -   dictionary words "standard"+" the" = "standard the"
                   s3              -   split into threes = ["sta","nda","rd ","the"]
           ị                       - index into
                      Ṗ            - remove rightmost character
                               ¤   - nilad followed by link(s) as a nilad:
                         “...»     -   dictionary words "January"+" February"+...
                              Ḳ    -   split at spaces = ["January","February",...]
                        ị          - index into (vectorises across [month])
                       ,           - pair                  e.g. ["th", ["February"]]
                                K  - join with spaces           ["th ", "February"]
                                   - print (implicitly smashes)   th February
\$\endgroup\$
  • 4
    \$\begingroup\$ The "standard the" trick is amazing. \$\endgroup\$ – Ven Mar 15 at 9:17
  • \$\begingroup\$ I agree with @Ven, great trick! It also saved a byte in my 05AB1E answer in comparison to the compressed string "thstndrd" split into parts of size 2 (.•oθ2(w•2ô), so thanks. :) \$\endgroup\$ – Kevin Cruijssen Mar 15 at 11:03
  • 1
    \$\begingroup\$ This has to be one of the longest Jelly programs I have ever seen. \$\endgroup\$ – JAD Mar 15 at 12:41
6
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C# (Visual C# Interactive Compiler), 115 113 109 98 bytes

g=>$"{f=(g=p.AddDays(g-1)).Day}{"tsnr"[f=f%30%20<4?f%10:0]}{"htdd"[f]} {g:MMMM}";DateTime p;int f;

Thanks to @someone for saving 9 bytes

Try it online!

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  • 1
    \$\begingroup\$ @KevinCruijssen I got the modulos out of order, should be fixed now. \$\endgroup\$ – Embodiment of Ignorance Mar 14 at 14:59
  • \$\begingroup\$ .code.tio(2,22): error CS0165: Use of unassigned local variable 'p' It appears that the struct thing doesn't work. \$\endgroup\$ – JAD Mar 15 at 12:45
  • \$\begingroup\$ var g=new DateTime().AddDays(n-1) works though \$\endgroup\$ – JAD Mar 15 at 12:50
  • \$\begingroup\$ @JAD mistake on my part, fixed \$\endgroup\$ – Embodiment of Ignorance Mar 15 at 15:00
  • \$\begingroup\$ 110 bytes I think \$\endgroup\$ – someone Jun 15 at 15:15
5
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Python 3.8 (pre-release), 112 bytes

lambda x:str(d:=(t:=gmtime(x*86399)).tm_mday)+'tsnrhtdd'[d%5*(d%30%20<4)::4]+strftime(' %B',t)
from time import*

Try it online!

Weirdly enough, I don't have to parenthesize d:=(t:=gmtime(~-x*86400), probably because the interpreter only checks if there are () characters around the assignment expression and not that the expression itself is parenthesized.

-2 thanks to gwaugh.
-5 thanks to xnor.

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5
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Perl 6, 166 161 bytes

{~(.day~(<th st nd rd>[.day%30%20]||'th'),<January February March April May June July August September October November December>[.month-1])}o*+Date.new(1,1,1)-1

Try it online!

Hardcodes all the month names, which takes up most of the space. Man, Perl 6 really needs a proper date formatter.

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4
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Hack, 115 59 39 bytes

$x==>date("jS F",mktime(0,0,0,1,$x));

Since @gwaugh got to the same solution as mine while I was golfing, I'm posting this in Hack instead :).

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  • \$\begingroup\$ Wow, great minds think alike. :) +1 to you sir! \$\endgroup\$ – 640KB Mar 13 at 16:17
  • \$\begingroup\$ @gwaugh haha, I didn't know I could just have a top-level program. I'll edit mine to make it top-level too, and find a way to get a better scor e;-) \$\endgroup\$ – Ven Mar 13 at 16:18
  • 1
    \$\begingroup\$ @gwaugh Made mine Hack instead. \$\endgroup\$ – Ven Mar 13 at 16:26
  • 1
    \$\begingroup\$ You'll probably want to specify a non-leap year parameter to your mktime() call otherwise it will return the wrong output if run on a leap year. (had to do to my answer). \$\endgroup\$ – 640KB Mar 13 at 16:42
4
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JavaScript (ES6),  117  113 bytes

Saved 4 bytes thanks to @tsh

d=>(n=(d=new Date(1,0,d)).getDate())+([,'st','nd','rd'][n%30%20]||'th')+' '+d.toLocaleString('en',{month:'long'})

Try it online!

Commented

d =>                     // d = input day
  ( n =                  //
    ( d =                // convert d to
      new Date(1, 0, d)  //   a Date object for the non leap year 1901
    ).getDate()          // save the corresponding day of month into n
  ) + (                  //
    [, 'st', 'nd', 'rd'] // ordinal suffixes
    [n % 30 % 20]        // map { 1, 2, 3, 21, 22, 23, 31 } to { 'st', 'nd', 'rd' }
    || 'th'              // or use 'th' for everything else
  ) + ' ' +              // append a space
  d.toLocaleString(      // convert d to ...
    'en',                // ... the English ...
    { month: 'long' }    // ... month name
  )                      //

Without date built-ins, 188 bytes

f=(d,m=0)=>d>(k=31-(1115212>>m*2&3))?f(d-k,m+1):d+([,'st','nd','rd'][d%30%20]||'th')+' '+`JanuaryFebruaryMarchAprilMayJuneJulyAugustSeptemberOctoberNovemberDecember`.match(/.[a-z]*/g)[m]

Try it online!

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  • \$\begingroup\$ Fails for 11th,12th,13th of each month \$\endgroup\$ – Expired Data Mar 13 at 16:04
  • 1
    \$\begingroup\$ @ExpiredData Thanks for reporting this. Fixed now. \$\endgroup\$ – Arnauld Mar 13 at 16:16
  • \$\begingroup\$ Ignore my comment, I made an ID10T error. \$\endgroup\$ – asgallant Mar 13 at 21:10
  • \$\begingroup\$ I'm not sure how nodejs handle language tags, but it seems using 0 would work as using "en". And changing to toLocaleString would save 4 bytes. 110 bytes \$\endgroup\$ – tsh Mar 14 at 2:39
  • \$\begingroup\$ @tsh It seems that toLocaleString is using the system default settings when it's passed an unrecognized string or a numeric value. So, it can be anything. This parameter is basically ineffective on a TIO instance, because only English locales are installed anyway. \$\endgroup\$ – Arnauld Mar 14 at 8:17
4
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Smalltalk, 126 bytes

d:=Date year:1day:n.k:=m:=d dayOfMonth.10<k&(k<14)and:[k:=0].o:={#st.#nd.#rd}at:k\\10ifAbsent:#th.m asString,o,' ',d monthName
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  • 1
    \$\begingroup\$ I don't know Smalltalk, but is this correct for 11th,12th,13th? If I read correctly you integer-divide the day by 10, but that would mean it would result in 11st,12nd,13rd, unless something else in the code fixes this while I'm unaware of it. \$\endgroup\$ – Kevin Cruijssen Mar 14 at 10:58
  • \$\begingroup\$ @KevinCruijssen You are right. Thanks for calling my attention on this. I'll need to spend some more bytes to fix this. \$\endgroup\$ – Leandro Caniglia Mar 14 at 16:19
  • 1
    \$\begingroup\$ @KevinCruijssen, Done. Thanks again. \$\endgroup\$ – Leandro Caniglia Mar 15 at 1:58
3
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C# (Visual C# Interactive Compiler), 141 139 133 124 122 bytes

a=>{var d=s.AddDays(a-1);int x=d.Day,m=x%30%20;return x+"thstndrd".Substring(m<4?m*2:0,2)+d.ToString(" MMMM");};DateTime s

Thanks to Arnauld for faster method of removing 11,12,13th saving 4 bytes

Try it online!

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  • \$\begingroup\$ Using C# 8, this can be reduced to: a=>{var d=s.AddDays(a-1);int x=d.Day,m=x%30%20;return x+"thstndrd"[(m<4?m*2:0)..2]+$" {d:MMMM}";};DateTime s The interactive compiler doesn't seem to support changing its language level to "preview" at this time, though. \$\endgroup\$ – Arcanox Mar 13 at 19:07
  • \$\begingroup\$ 116 bytes \$\endgroup\$ – Embodiment of Ignorance Mar 13 at 20:02
  • \$\begingroup\$ I'm pretty sure you have to add a semi-colon after the DataTime s \$\endgroup\$ – Embodiment of Ignorance Mar 14 at 1:31
3
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R, 158 134 bytes

-24 bytes @Nick Kennedy for golfing the 'st', 'nd', 'rd', & 'th'. Thanks!

f=format;paste0(a<-as.double(f(d<-as.Date(scan(,''),'%j'),'%e')),`if`((a-1)%%10>2|a%/%10==1,'th',c("st","nd","rd")[a%%10]),f(d,' %B'))

Try it online!

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3
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MySQL, 47 45 42 bytes

SELECT DATE_FORMAT(MAKEDATE(1,n),"%D %M")

1901 can be replaced with any year that was/is not a leap year.

Edit: saved two bytes by removing spaces and another three bytes by changing the year to 1, thanks to @Embodyment of Ignorance.

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  • \$\begingroup\$ Can you remove the spaces between 1901, n and the string? \$\endgroup\$ – Embodiment of Ignorance Mar 14 at 15:31
  • \$\begingroup\$ @EmbodimentofIgnorance yes I can, thanks! \$\endgroup\$ – NicolasB Mar 14 at 15:32
  • \$\begingroup\$ Also, why not replace 1901 with a year like 1? 1 isn't a leap year, and it's 3 bytes shorter \$\endgroup\$ – Embodiment of Ignorance Mar 14 at 15:32
  • \$\begingroup\$ @EmbodimentofIgnorance done and done :-) \$\endgroup\$ – NicolasB Mar 14 at 15:35
3
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05AB1E, 81 79 78 76 75 74 73 71 70 69 bytes

•ΘÏF•ºS₂+.¥-D0›©ÏθDT‰ć≠*4šß„—ÊØ3ôsè¨ð”……‚應…ä†ï€¿…Ë…ê†Ä…æ…Ì…Í”#®OèJ

-9 bytes thanks to @Grimy.
-1 byte thanks to @JonathanAllan's standard the trick for th,st,nd,rd, which he used in his Jelly answer.

Try it online or verify all possible test cases.

Explanation:

•ΘÏF•        # Push compressed integer 5254545
     º       # Mirror it vertically: 52545455454525
      S      # Converted to a list of digits: [5,2,5,4,5,4,5,5,4,5,4,5,2,5]
       ₂+    # And 26 to each: [31,28,31,30,31,30,31,31,30,31,30,31,28,31]
             # (the additional trailing 28,31 won't cause any issues)
.¥           # Undelta this list (with automatic leading 0):
             #  [0,31,59,90,120,151,181,212,243,273,304,334,365,393,424]
  -          # Subtract each from the (implicit) input-integer
   D0›       # Duplicate the list, and check for each if it's positive (> 0)
      ©      # Store the resulting list in the register (without popping)
       Ï     # Only leave the values at those truthy indices
        θ    # And get the last value from the list, which is our day
D            # Duplicate this day
 T‰          # Take the divmod-10 of this day: [day//10, day%10]
   ć         # Extract the head; pop and push the remainder-list and head: [day%10], day//10
    ≠        # Check whether the day//10 is NOT 1 (0 if day//10 == 1; 1 otherwise)
     *       # Multiply that by the [day%10] value
      4š     # Prepend a 4 to this list
        ß    # Pop and push the minimum of the two (so the result is one of [0,1,2,3,4],
             # where the values are mapped like this: 1..3→1..3; 4..9→4; 10..19→0; 20..23→0..3; 24..29→4; 30,31→0,1)
 …thŠØ       # Push dictionary string "th standards"
      3ô     # Split it into parts of size 3: ["th ","sta","nda","rds"]
        sè   # Swap and index the integer into this list (4 wraps around to index 0)
          ¨  # And remove the trailing character from this string
ð            # Push a space " "
”……‚應…ä†ï€¿…Ë…ê†Ä…æ…Ì…Í”
             # Push dictionary string "December January February March April May June July August September October November"
 #           # Split on spaces
  ®          # Push the list of truthy/falsey values from the register again
   O         # Get the amount of truthy values by taking the sum
    è        # Use that to index into the string-list of months (12 wraps around to index 0)
J            # Join everything on the stack together to a single string
             # (and output the result implicitly)

See this 05AB1E tip of mine to understand why:

  • (section How to use the dictionary?) ”……‚應…ä†ï€¿…Ë…ê†Ä…æ…Ì…Í” is "December January February March April May June July August September October November"
  • (section How to use the dictionary?) …thŠØ is "th standards"
  • (section How to compress large integers?) •ΘÏF• is 5254545
\$\endgroup\$
  • 1
    \$\begingroup\$ -2 bytes by using 5в28+ for compression: TIO \$\endgroup\$ – Grimmy Mar 14 at 14:35
  • 1
    \$\begingroup\$ Using S is a good idea, -1 byte again: TIO \$\endgroup\$ – Grimmy Mar 14 at 15:45
  • 1
    \$\begingroup\$ @Grimy Thanks for the -1 byte for •EË7Óæ•S₂+, but your -3 golf doesn't work unfortunately. Indexing automatically wraps around in 05AB1E, so the 5st,6nd,7rd,25st,26nd,27rd,29st will be wrong. PS: if it would have worked, could have been for an additional -1. :) \$\endgroup\$ – Kevin Cruijssen Mar 14 at 18:48
  • 1
    \$\begingroup\$ -1 again (using "th standards" instead of "standard the" removes the need for Á). \$\endgroup\$ – Grimmy Mar 19 at 16:52
  • 1
    \$\begingroup\$ -1 (•C.ñÒā• to •ΘÏF•º, the extra digits don't matter) \$\endgroup\$ – Grimmy Mar 20 at 12:51
2
\$\begingroup\$

bash, 82 80 bytes

-2 bytes thanks to @ASCII-only

a=(th st nd rd);set `printf "%(%e %B)T" $[$1*86399]`;echo $1${a[$1%30%20]-th} $2

TIO

bash +GNU date, 77 bytes

a=(th st nd rd);set `date -d@$[$1*86399] +%e\ %B`;echo $1${a[$1%30%20]-th} $2
\$\endgroup\$
  • \$\begingroup\$ 80? \$\endgroup\$ – ASCII-only Mar 14 at 11:35
  • \$\begingroup\$ @ASCII-only, yes subtracting 100s for each day, 100*365 = 36500s which is less than one day (86400), works also with 86399 (subtract 1s by day) \$\endgroup\$ – Nahuel Fouilleul Mar 14 at 11:42
  • \$\begingroup\$ :/ still looks really long but haven't found a better way yet \$\endgroup\$ – ASCII-only Mar 14 at 11:52
2
\$\begingroup\$

Shell + coreutils, 112 90 bytes

date -d0-12-31\ $1day +%-dth\ %B|sed 's/1th/1st/;s/2th/2nd/;s/3th/3rd/;s/\(1.\).. /\1th /'

Try it online! Link includes test cases. Edit: Saved 22 bytes thanks to @NahuelFouilleul. Explanation:

date -d0-12-31\ $1day

Calculate the number of day(s) after the first day preceding a non-leap year. (Sadly you can't do relative date calculations from @-1.)

+%-dth\ %B|sed

Output the day of month (without leading zero), th, and the full month name.

's/1th/1st/;s/2th/2nd/;s/3th/3rd/;

Fix up 1st, 2nd, 3rd, 21st, 22nd, 23rd and 31st.

s/\(1.\).. /\1th /'

Restore 11th to 13th.

\$\endgroup\$
  • \$\begingroup\$ i saw this answer after mine, could save 18bytes using one sed command, also s in days can be removed, and 19 in 1969 \$\endgroup\$ – Nahuel Fouilleul Mar 14 at 11:09
  • \$\begingroup\$ @NahuelFouilleul That last one uses a Bash-ism so should be posted as a separate answer, but thanks for the other tips! \$\endgroup\$ – Neil Mar 14 at 11:49
2
\$\begingroup\$

Jelly, 115 114 101 97 bytes

%30%20¹0<?4Ḥ+ؽị“thstndrd”ṭ
“5<Ḟ’b4+28ÄŻ_@µ>0T,>0$ƇZṪµ1ịị“£ṢtẒ⁽ẹ½MḊxɲȧėAṅ ɓaṾ¥D¹ṀẏD8÷ṬØ»Ḳ¤,2ịÇƊṚK

Try it online!

Long by Jelly standards, but done from first principles.

Thanks to @JonathanAllan for saving 13 bytes through better understanding of string compression.

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  • \$\begingroup\$ “£ṢtẒ⁽ẹ½MḊxɲȧėAṅ ɓaṾ¥D¹ṀẏD8÷ṬØ»Ḳ¤ would save 13 (Compress.dictionary looks for a leading space and has special handling for it). \$\endgroup\$ – Jonathan Allan Mar 14 at 12:13
2
\$\begingroup\$

Google Sheets, 118 103 86 bytes

=day(A1+1)&mid("stndrdth",min(7,1+2*mod(mod(day(A1+1)-1,30),20)),2)&text(A1+1," mmmm")

I can't edit my comment so, here's a working version of the Google Sheets code.

Try it Online!

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1
\$\begingroup\$

Red, 124 bytes

func[n][d: 1-1-1 + n - 1[rejoin[d/4 either 5 > t: d/4 % 30 % 20[pick[th st nd rd]t + 1]['th]]pick system/locale/months d/3]]

Try it online!

Adds n - 1 days to 1-1-1 (1-Jan-2001) to form a date, than uses Arnauld's method to index into month suffixes. Too bad Red is 1-indexed, this requires additional tweaking. The good thing is that Red knows the names of the months :)

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1
\$\begingroup\$

APL(NARS), 235 chars, 470 bytes

{k←↑⍸0<w←+\v←(1-⍵),(12⍴28)+13561787⊤⍨12⍴4⋄k<2:¯1⋄d←1+v[k]-w[k]⋄(⍕d),({d∊11..13:'th'⋄1=10∣d:'st'⋄2=10∣d:'nd'⋄3=10∣d:'rd'⋄'th'}),' ',(k-1)⊃(m≠' ')⊂m←'January February March April May June July August September October November December'}

13561787 is the number that in base 4 can be summed to (12⍴28) for obtain the lenght of each month... test:

  f←{k←↑⍸0<w←+\v←(1-⍵),(12⍴28)+13561787⊤⍨12⍴4⋄k<2:¯1⋄d←1+v[k]-w[k]⋄(⍕d),({d∊11..13:'th'⋄1=10∣d:'st'⋄2=10∣d:'nd'⋄3=10∣d:'rd'⋄'th'}),' ',(k-1)⊃(m≠' ')⊂m←'January February March April May June July August September October November December'}     
  ⊃f¨1 2 3 365 60 11
1st January  
2nd January  
3rd January  
31st December
1st March    
11th January 
\$\endgroup\$
0
\$\begingroup\$

C (gcc), 174 155 bytes

i;char a[99],*b="thstndrd";f(long x){x--;x*=86400;strftime(a,98,"%d   %B\0",gmtime(&x));i=*a==49?0:a[1]-48;a[2]=b[i=i>3?0:i*2];a[3]=b[++i];x=*a==48?a+1:a;}

Try it online!

\$\endgroup\$
-2
\$\begingroup\$

Python 3, 95 Bytes

Datetimed it :P

from datetime import *;f=lambda s:(datetime(2019,1,1)+timedelta(days=s-1)).strftime("%d of %B")

Try it online!

\$\endgroup\$
  • 2
    \$\begingroup\$ This doesn't produce the ordinal suffixes, and has leading zeroes in the day number. The of is also unnecessary \$\endgroup\$ – Jo King Mar 19 at 12:10

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