给定数字1≤n≤365，以“天数月”格式输出一年中的第n天。例如，给定1，您应输出“ 1st January”，不带“ of”。

将使用公历，该程序不应计入leap年，因此您的程序在任何情况下均不应输出“ 2月29日”。只要遵循前面提到的“日数月”格式，就可以使用任何方法。您的程序还应该正确输出序号，这意味着它应该始终输出1st，2nd，3rd，应该分别将1、2或3作为任何输入的日期。允许前导空格或其他缩进。

这是代码高尔夫，所以最短的解决方案就是胜出。

测试用例：

1 gives 1st January
2 gives 2nd January
3 gives 3rd January
365 gives 31st December
60 gives 1st March
11 gives 11th January

PHP，38 40 30 28字节

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

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php -nF输入来自的运行STDIN。示例（脚本名为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

说明

通过将day number * number of seconds per day（86400）乘以构造1970年所需日期（通常不是a年）的纪元时间戳。但是，这会产生高出一天的结果，因此乘以number of seconds in a day - 1（86399），对于输入数字范围（1≤n≤365），将得到每正确一天结束时的时间戳记。然后，只需使用PHP的内置日期格式进行输出即可。

果冻， 79 78  77 字节

-1修复错误：）_{（不应预先转置以找到索引，而应反向后转，但我们可以尾部而不是头部）}
-1使用反射_{（⁽©ṅB+30_2¦2-> ⁽0ṗb4+28m0）}

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

完整的程序可以打印结果

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怎么样？

稍后会更新...

⁽©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

C＃（Visual C＃中交互式编译器），115个 113 109 98字节

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;

感谢@someone节省了9个字节

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Python 3.8（预发布），112字节

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

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奇怪的是，我不必加上括号d:=(t:=gmtime(~-x*86400)，这可能是因为解释器仅检查()赋值表达式周围是否存在字符，而不是对表达式本身进行括号括起来。

-2感谢gwaugh。
-5感谢xnor。

Perl 6的，166个 161字节

{~(.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

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对所有月份名称进行硬编码，这将占用大部分空间。伙计，Perl 6确实需要适当的日期格式。

哈克115 59 39字节

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

由于@gwaugh在打高尔夫球时获得了与我相同的解决方案，因此我将其发布在Hack中:)。

的JavaScript（ES6）， 117个  113字节

@tsh节省了4个字节

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

在线尝试！

已评论

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
  )                      //

没有日期内置，188字节

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]

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Smalltalk，126个字节

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

C＃（Visual C＃中交互式编译器），141个 139 133 124 122字节

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

感谢Arnauld提供了更快的删除11,12,13th的方法，节省了4个字节

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R，158134字节

-24字节@Nick Kennedy，用于打高尔夫球“ st”，“ nd”，“ rd”和“ th”。谢谢！

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'))

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MySQL，47 45 42字节

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

1901年可以用任何曾经/不是a年的年份代替。

编辑：感谢@Embodyment of Ignorance，通过删除空格节省了两个字节，将年份更改为1则节省了另外三个字节。

05AB1E，81 79 78 76 75 74 73 71 70 69 字节

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

-9个字节，感谢@Grimy。
-1字节感谢@JonathanAllan的standard thetrick俩th,st,nd,rd，他在Jelly答案中使用了trick 俩。

在线尝试或验证所有可能的测试用例。

说明：

•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)

请参阅我的05AB1E技巧以了解原因：

• （部分如何使用字典？）”……‚應…ä†ï€¿…Ë…ê†Ä…æ…Ì…Í”是"December January February March April May June July August September October November"
• （部分如何使用字典？）…thŠØ是"th standards"
• （部分如何压缩大整数？）•ΘÏF•是5254545

bash，82个 80字节

-2个字节，仅@ASCII

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

蒂奥

bash + GNU日期，77个字节

a=(th st nd rd);set `date -d@$[$1*86399] +%e\ %B`;echo $1${a[$1%30%20]-th} $2

Shell + coreutils，112个 90字节

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

在线尝试！链接包括测试用例。编辑：由于@NahuelFouilleul，节省了22个字节。说明：

date -d0-12-31\ $1day

计算非-年之前的第一天之后的天数。（遗憾的是，您无法从中进行相对日期计算@-1。）

+%-dth\ %B|sed

输出月份中的日期（不带前导零）th，和月份的全名。

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

修复了1st，2nd，3rd，21st，22nd，23rd和31st。

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

恢复11th到13th。

果冻，115个 114 101 97字节

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

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长期以果冻为标准，但是从第一原则开始的。

感谢@JonathanAllan通过更好地理解字符串压缩节省了13个字节。

谷歌表格118个 103 86字节

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

我无法编辑我的评论，因此，这是Google表格代码的有效版本。

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红色，124字节

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]]

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n在1-1-1（2001年1月1日）上加上-1天以形成日期，然后使用Arnauld的方法索引月份后缀。太糟糕了，红色是1索引的，这需要额外的调整。好消息是Red知道月份的名称:)

APL（NARS），235个字符，470个字节

{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是基数4可以累加为（12⍴28）的数字以获得每月的长度...测试：

  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 

C（gcc），174155字节

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;}

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Python 3，95字节

日期：P

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

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