Outputting ordinal numbers (1st, 2nd, 3rd)

Question

I would like to generate (as a return result of a function, or simply as the output of a program) the ordinal suffix of a positive integer concatenated to the number.

Samples:

1st 
2nd 
3rd 
4th 
... 
11th 
12th 
13th 
... 
20th 
21st
22nd
23rd
24th

And so on, with the suffix repeating the initial 1-10 subpattern every 10 until 100, where the pattern ultimately starts over.

The input would be the number and the output the ordinal string as shown above.

What is the smallest algorithm for this?

• Ilmari Karonen: Hi, NickC, and welcome to codegolf.SE! Just to clarify, do you mean that we should read a number like 11 as input, and output e.g. 11th? Is each number in the input on a separate line, and should the output numbers be on separate lines too? And do we need to handle more than one line of input?
• Toto: Are you looking for smallest algorithm or smallest code?
• Nicole: @Ilmari I am looking for 11 as input and 11th as output. I don't mind if it processes multiple lines but what I had in mind was processing just a single number.
• Nicole: @M42 You know, I'm not really sure. I don't have a strict requirement - but I was probably thinking smallest algorithm.
• sergiol: The German version of this challenge would be to add a dot after the number...

Answer 1

Python 2, 49 bytes

lambda n:`n`+'tsnrhtdd'[n%5*(n%100^15>4>n%10)::4]

An anonymous function. A full program would be counted at 55 bytes.

'tsnrhtdd'[i::4] encodes the suffixes th st nd rd for values of i from 0 to 3. Given this, all we need is a way to map the values of n to the index of the corresponding suffix, i. A straightforward expression that works is (n%10)*(n%10<4 and not 10<n%100<14). We can easily shorten this by dropping the first set of parentheses and observing that n%5 gives the same results as n%10 for the values of n with the special suffixes. With a bit of trial and error, one may also shorten not 10<n%100<14 to n%100^15>4, which can be chained with the other conditional to save even more bytes.

• cat: this is some black magic right here c:
• Morgan Thrapp: terminally-incoherent.com/blog/wp-content/uploads/2012/08/…
• je je: the straightforward expression looks to have a mistake: the index should be non-zero if n%100 is not between 10 and 14.
• xsot: @jeje Good eye! I fixed up the explanation. Funny how this error has gone unnoticed for so many years.

Answer 2

Perl, 37 + 1 characters

s/1?\d\b/$&.((0,st,nd,rd)[$&]||th)/eg

This is a regexp substitution that appends the appropriate ordinal suffix to any numbers in $_ that are not already followed by a letter. To apply it to file input, use the p command line switch, like this:

perl -pe 's/1?\d\b/$&.((0,st,nd,rd)[$&]||th)/eg'

This is a complete Perl program that reads input from stdin and writes the processed output to stdout. The actual code is 37 chars long, but the p switch counts as one extra character.

Sample input:

This is the 1 line of the sample input...
...and this is the 2.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
101 102 103 104 105 106 107 108 109 110

Output:

This is the 1st line of the sample input...
...and this is the 2nd.
1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th
11th 12th 13th 14th 15th 16th 17th 18th 19th 20th
21st 22nd 23rd 24th 25th 26th 27th 28th 29th 30th
101st 102nd 103rd 104th 105th 106th 107th 108th 109th 110th

Numbers already followed by letters will be ignored, so feeding the output again through the filter won't change it. Spaces, commas and periods between numbers are not treated specially, so they're assumed to separate numbers like any other punctuation. Thus, e.g. 3.14159 becomes 3rd.14159th.

How does it work?

• First, this is a global regexp replacement (s///g). The regexp being matched is 1?\d\b, where \d matches any digit and \b is a zero-width assertion matching the boundary between an alphanumeric and a non-alphanumeric character. Thus, 1?\d\b matches the last digit of any number, plus the previous digit if it happens to be 1.

• In the substitution, which is evaluated as Perl code due to the /e switch, we take the matched string segment ($&) and append (.) to it the suffix obtained by using $& itself as an integer index to the list (0,st,nd,rd); if this suffix is zero or undefined (i.e. when $& is zero or greater than three), the || operator replaces it with th.

---

Edit: If the input is restricted to a single integer, then this 35 character solution will suffice:

s/1?\d$/$&.((0,st,nd,rd)[$&]||th)/e

• Mr. Llama: Should be able to drop the g off the substitution if you specify that each number must be on it's own line. Also, that'd let you change the word boundary to be $. But overall, +1, damn clever solution.
• Ilmari Karonen: @GigaWatt: Indeed, I wrote the code before NickC answered my question about the input format, so I made it as generic as possible.
• Dom Hastings: I know this is pretty old and rules were different, but using $\ instead of s/// shaves a few bytes off: Try it online!
• Dom Hastings: In fact, even shorter using /.../g syntax: Try it online!

Answer 3

Python, 68 characters

i=input()
k=i%10
print"%d%s"%(i,"tsnrhtdd"[(i/10%10!=1)*(k<4)*k::4])

• DLosc: This is really late, but you can take off 7 bytes by doing `i`+"tsnrhtdd". Otherwise, this is the exact solution I just got.

Answer 4

Mathematica (削除) 39 (削除ここまで) 45 bytes

Note: In recent versions of Mathematica, asking for the nth part of p, where p is undefined, generates an error message, but returns the correct answer anyway. I've added Quiet to prevent the error message from printing.

Quiet@StringSplit[SpokenString[p[[#]]]][[2]]&

Usage

Quiet@StringSplit[SpokenString[p[[#]]]][[2]] &[31]

31st

Quiet@StringSplit[SpokenString[p[[#]]]][[2]] &/@Range[21]

{"1st", "2nd", "3rd", "4th", "5th", "6th", "7th", "8th", "9th", "10th", "11th", "12th", "13th", "14th", "15th", "16th", "17th", "18th", "19th", "20th", "21st"}

---

How it works

SpokenString writes out an any valid Mathematica expression as it might be spoken. Below are two examples from the documentation for SpokenString,

SpokenString[Sqrt[x/(y + z)]]

"square root of the quantity x over the quantity y plus z" *)

SpokenString[Graphics3D[Sphere[{{0, 0, 0}, {1, 1, 1}}]], "DetailedGraphics" -> True]

"a three-dimensional graphic consisting of unit spheres centered at 0, 0, 0 and 1, 1, 1"

---

Now, for the example at hand,

Quiet@SpokenString[p[[#]]] &[31]

"the 31st element of p"

Let's represent the above string as a list of words:

StringSplit[%]

{"the", "31st", "element", "of", "p"}

and take the second element...

%[[2]]

31st

• Mr.Wizard: I'm confused; where is p defined? EDIT: never mind, I see how you're using this; unfortunately it does not work on my system. :-/
• DavidC: It works on v.9.0.1. (I seem to recall that you are using 7.0.) Yes, p does not need to be defined.
• Mr.Wizard: I understand that now. However, in v7 I get from SpokenString @ p[[117]] the output " part 117 of p".
• DavidC: So SpokenString gets revised from time to time. I wouldn't be surprised it this code (codegolf.stackexchange.com/questions/8859/…) also doesn't work on v. 7. BTW, it wasn't meant to be an enduring solution.
• Mr.Wizard: Hah, I hadn't seen that answer before.

Answer 5

Javascript (ES6) (削除) 50 (削除ここまで) 44 Bytes

a=>a+=[,"st","nd","rd"][a.match`1?.$`]||"th"

Notes

• takes imput as a string
• Removed 6 bytes, thanks @user81655

• user81655: a+ -> a+=, remove parentheses, \d -> ., remove [0], and if you take the number as a string: a.match`1?.$` instead of /1?.$/.exec(a).

Answer 6

Ruby, 60

It's not as good as the Perl entry, but I figured I'd work on my Ruby skills.

def o(n)n.to_s+%w{th st nd rd}[n/10%10==1||n%10>3?0:n%10]end

Function takes one integer argument, n, and returns a string as the ordinal form.

Works according to the following logic:
If the tens digit is a 1 or the ones digit is greater than 3 use the suffix 'th'; otherwise find the suffix from the array ['th', 'st', 'nd', 'rd'] using the final digit as the index.

• hammar: o(113) is "113rd", should be "113th". The tens digit check doesn't account for numbers with more than two digits.
• Mr. Llama: Ok, pitched in another %10 to compensate. Added 3 characters. (I feel like %10 appears enough where it should be shortened somehow, but I can't think of a solution)
• jamylak: You can never beat Perl in writing horrible incomprehensible code that's as short as possible :)
• wizzwizz4: Set a variable to 10?
• CalculatorFeline: I think setting a variable to n%10 is better.

Answer 7

Javascript, (削除) 68 (削除ここまで) 71

function o(n){return n+([,'st','nd','rd'][~~(n/10%10)-1?n%10:0]||'th')}

Joint effort with ItsCosmo.

EDIT: Wasn't working properly with numbers > 100

• Eliseo D'Annunzio: You could bring it down to 62 with: function o(n)n+([,'st','nd','rd'][~~(n/10%10)-1?n%10:0]||'th') and you can bring it down further to 54 if you are happy to use fat arrow notation: o=n=>n+([,'st','nd','rd'][~~(n/10%10)-1?n%10:0]||'th')
• aebabis: @WallyWest For some reason, I really like that my solution works in modern browsers. Feel free to post your own answer; I think its sufficiently different.
• Eliseo D'Annunzio: No, that's fine! Just providing a few alternatives!
• DaCuteRaccoon: I made it work with ARGV: for(n of arguments)print(n+([,'st','nd','rd'][~~(n/10%10)-1?n%10:0]||'th'))

Answer 8

JavaScript, 64 characters (ES3) or 47 characters (ES6)

ES3 (64 characters):

function(n){return n+=[,'st','nd','rd'][n%100>>3^1&&n%10]||'th'}

ES6 (47 characters):

n=>n+=[,'st','nd','rd'][n%100>>3^1&&n%10]||'th'

Explanation

The expression n % 100 >> 3 ^ 1 evaluates to 0 for any positive n ending with digits 08–15. Thus, for any n mod 100 ending in 11, 12, or 13, the array lookup returns undefined, leading to a suffix of th.

For any positive n ending in other digits than 08–15, the expression n % 100 >> 3 ^ 1 evaluates to a positive integer, invoking the expression n % 10 for array lookup, returning st,nd, or rd for n which ends with 1, 2, or 3. Otherwise, th.

• Shaggy: Save a bytes with n+=[,'st','nd','rd'][n%100>>3^1&&n%10]||'th'.
• guest271314: FYI github.com/guest271314/SpeechSynthesisSSMLParser/blob/master/…, github.com/guest271314/SpeechSynthesisSSMLParser/blob/master/…
• Tomas Langkaas: @guest271314, fun to see this code being used. Note that it only works for positive numbers, another alternative is n+=[,"st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10]||"th", adapted from this post.

Answer 9

Haskell, (削除) 95 (削除ここまで) 100 chars

h=foldr g"th".show
g '1'"th"="1st"
g '2'"th"="2nd"
g '3'"th"="3rd"
g '1'[x,_,_]='1':x:"th"
g a s=a:s

Testing:

*Main> map h [1..40]
["1st","2nd","3rd","4th","5th","6th","7th","8th","9th","10th","11th","12th","13t
h","14th","15th","16th","17th","18th","19th","20th","21st","22nd","23rd","24th",
"25th","26th","27th","28th","29th","30th","31st","32nd","33rd","34th","35th","36
th","37th","38th","39th","40th"]

Must be loaded with -XNoMonomorphismRestriction.

• Khuldraeseth na'Barya: I'm seeing 100 here, but 95 using an infix operator.
• Will Ness: thanks, fixed. probably didn't know we have to count the newlines as well, back then.

Answer 10

Golfscript, 34 characters

~.10/10%1=!110ドル%*.4<*'thstndrd'2/=

Answer 11

J - (削除) 44 (削除ここまで) 41 char

Nothing in J? This is an outrage!

(":,th`st`nd`rd{::~4|4<.10(|*|~:-~)100&|)

Explained (note that 1 is boolean true in J and 0 is false):

• 100&| - First, reduce the input modulo 100.
• 10(|*|~:-~) - In this subexpression, | is the input mod 10 and -~ is the input (mod 100) minus 10. You don't get many trains as clean as this one in J, so it's a treat to see here!
  • ~: is not-equals, so |~:-~ is false if the tens digit is 1 and true otherwise.
  • * is just multiplication, so we're taking the input mod 10 and multiplying by a boolean, which will zero it out if false. The result of this is "input mod 10, except 0 on the tens".
• 4|4<. - Min (<.) the above with 4, to clamp down larger values, and then reduce modulo 4 so that they all go to zero.
• th`st`nd`rd{::~ - Use that as an index into the list of suiffixes. Numbers ending in X1 or X2 or X3, but not in 1X, will find their suffix, and everything else will take the 0th suffix th.
• ":, - Finally, take the original input, convert it to a string (":), and append the suffix.

Usage is obvious, though as-is the verb can only take one ordinal, not a list.

 (":,th`st`nd`rd{::~4|4<.10(|*|~:-~)100&|) 112 NB. single use
112th
 (":,th`st`nd`rd{::~4|4<.10(|*|~:-~)100&|) 1 2 3 4 5 NB. doing it wrong
|length error
| (":,th`st`nd`rd{::~4|4<.10(|*|~:-~)100&|)1 2 3 4 5
 NB. i.5 10 makes a 5x10 grid of increasing integers
 NB. &.> to operate on each integer separately, and box the result after
 (":,th`st`nd`rd{::~4|4<.10(|*|~:-~)100&|)&.> i.5 10 NB. all better
+----+----+----+----+----+----+----+----+----+----+
|0th |1st |2nd |3rd |4th |5th |6th |7th |8th |9th |
+----+----+----+----+----+----+----+----+----+----+
|10th|11th|12th|13th|14th|15th|16th|17th|18th|19th|
+----+----+----+----+----+----+----+----+----+----+
|20th|21st|22nd|23rd|24th|25th|26th|27th|28th|29th|
+----+----+----+----+----+----+----+----+----+----+
|30th|31st|32nd|33rd|34th|35th|36th|37th|38th|39th|
+----+----+----+----+----+----+----+----+----+----+
|40th|41st|42nd|43rd|44th|45th|46th|47th|48th|49th|
+----+----+----+----+----+----+----+----+----+----+

Alternative 41 char solution showing off a couple logical variants:

(":;@;st`nd`rd`th{~3<.10(|+3*|=-~)100|<:)

Previously I had an overlong explanation of this 44 char hunk of junk. 10 10#: takes the last two decimal digits and /@ puts logic between them.

(":,th`st`nd`rd{::~10 10(]*[(~:*])4>])/@#:])

Answer 12

PowerShell, 92

process{"$_$(switch -r($_){"(?<!1)1$"{'st'}"(?<!1)2$"{'nd'}"(?<!1)3$"{'rd'}default{'th'}})"}

Works with one number per line of input. Input is given through the pipeline. Making it work for only a single number doesn't reduce the size.

Answer 13

R, (削除) 79 (削除ここまで) 76 bytes

Since there is no R solution yet... no tricks here, basic vector indexing, golfed down 3 chars thanks to Giuseppe. Previously tried index: [1+(x%%10)-(x%%100==11)] and [1+(x%%10)*(x%%100!=11)].

function(x)paste0(x,c("th","st","nd","rd",rep("th",6))[1+x%%10*!x%%100==11])

Try it online!

With substr, 79 bytes:

function(x,y=1+2*min(x%%10-(x%%100==11),4))paste0(x,substr("thstndrdth",y,y+1))

Try it online!

• Giuseppe: 1+x%%10*!x%%100==11 for the index?
• JayCe: @Giuseppe I need to calm down on the parenthesis :). Clever use of ! in front of expression instead of !=.
• Giuseppe: Yeah, the operators have weird precedence; ^ is really high, then %%-type operators, then */ and +- and I think == and &| comes next. ! has quite low precedence so you can use it as a separator between operations.

Answer 14

APL (Dyalog Unicode), (削除) 38 (削除ここまで) 36 bytes

Thanks to ngn for fixing a bug while maintaining byte count.

Anonymous tacit prefix function. Requires ⎕IO (Index Origin) set to 0, which is default on many systems. Even works for 0!

⍕,{2↑'thstndrd×ばつ⊃⍵⌽∊1 0 8\⊂10↑⍳4}

Try it online!

{...} anonymous lambda; ⍵ is argument:

⍳4 first four ɩndices; [0,1,2,3]

10↑ take first ten elements from that, padding with zeros: [0,1,2,3,0,0,0,0,0,0]

⊂ enclose to treat as single element; [[0,1,2,3,0,0,0,0,0,0]]

1 0 8\ expand to one copy, a prototypical copy (all-zero), eight copies;
[[0,1,2,3,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0],
[0,1,2,3,0,0,0,0,0,0],
[0,1,2,3,0,0,0,0,0,0],
⋮ (5 more)
[0,1,2,3,0,0,0,0,0,0]]

∊ εnlist (flatten);
[0,1,2,3,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,
0,1,2,3,0,0,0,0,0,0,
0,1,2,3,0,0,0,0,0,0,
⋮ (50 more)
0,1,2,3,0,0,0,0,0,0]

⍵⌽ cyclically rotate left as many steps as indicated by the argument

⊃ pick the first number (i.e. the argument-mod-100'th number)

×ばつ multiply two by that (gives 0, 2, 4, or 6)

'thstndrd'↓⍨drop that many characters from this string

2↑ take the first two of the remaining characters

⍕, concatenate the stringified argument to that

• ngn: "31th"?­­­­­­­­
• ngn: ⍕,{2↑'thstndrd'↓⍨2×⊃⍵⌽∊1 0 8\⊂10↑⍳4}
• ngn: sorry, I forgot to mention ⎕io←0. I can see you guessed that, but there are a few 1,2,3,4,0,0... that should be 0,1,2,3,0,0...
• Adám: @ngn Fixed. And happens to work for 0 too!

Answer 15

Javascript, (削除) 75 (削除ここまで) 60

With the new arrow notation:

o=(s)=>s+((0|s/10%10)==1?"th":[,"st","nd","rd"][s%10]||"th")

Old version, without arrow notation (75 chars):

function o(s){return s+((0|s/10%10)==1?"th":[,"st","nd","rd"][s%10]||"th")}

• user100411: 52 bytes (V8)

Answer 16

Bash (Cardinal to ordinal) 91 Bytes:

function c(){ case "1ドル" in *1[0-9]|*[04-9])"1ドル"th;;*1)"1ドル"st;;*2)"1ドル"nd;;*3)"1ドル"rd;;esac }

• RGS: You can save at least 2 bytes in your solution, just rename your c2o function something with a 1-byte name.
• user100411: 58 bytes.
• stephanmg: Wow @tailsparkrabbitear I think you should write an answer. Or do you want me to edit my answer?
• user100411: If you say so; posted mine.

Answer 17

PHP, 151

I know that this program is not comparable to the others. Just felt like giving a solution.

<?$s=explode(' ',trim(fgets(STDIN)));foreach($s as$n){echo$n;echo(int)(($n%100)/10)==1?'th':($n%10==1?'st':($n%10==2?'nd':($n%10==3?'rd':'th')))."\n";}

• karth: you can save a few characters by using foreach($s as $n){echo$n;

Answer 18

K - 44 char

It so happens that this is exactly as long as the J, and works in almost the same way.

{x,$`th`st`nd`rd@{y*(y<4)*~1=x}.-2#0,.:'x$:}

Explained:

• x$: - First, we convert the operand x into a string, and then assign that back to x. We will need its string rep again later, so doing it now saves characters.
• .:' - Convert (.:) each (') digit back into a number.
• -2#0, - Append a 0 to the front of the list of digits (in case of single-digit numbers), and then take the last two.
• {y*(y<4)*~1=x}. - Use the two digits as arguments x and y to this inner function, which returns y if y is less than 4 and x is not equal to 1, otherwise 0.
• `th`st`nd`rd@ - Index the list of suffixes by this result.
• x,$ - Convert the suffix from symbol to string, and append it to the original number.

Usage:

 {x,$`th`st`nd`rd@{y*(y<4)*~1=x}.-2#0,.:'x$:} 3021
"3021st"
 {x,$`th`st`nd`rd@{y*(y<4)*~1=x}.-2#0,.:'x$:}' 1 2 3 4 11 12 13 14 /for each in list
("1st"
 "2nd"
 "3rd"
 "4th"
 "11th"
 "12th"
 "13th"
 "14th")

Answer 19

PHP, 98 bytes

function c($n){$c=$n%100;$s=['','st','nd','rd'];echo$c>9&&$c<14||!$s[$c%10]?$n.'th':$n.$s[$c%10];}

The 11-13 bit is killing me here. Works for any integer $n >= 0.

For any integer $n:

PHP, 103 bytes

function c($n){$c=abs($n%100);$s=['','st','nd','rd'];echo$c>9&&$c<14||!$s[$c%10]?$n.'th':$n.$s[$c%10];}

Answer 20

Javascript ES6, 52 chars

n=>n+(!/1.$/.test(--n)&&'snr'[n%=10]+'tdd'[n]||'th')

Answer 21

C#, 62 bytes

n=>n+(n/10%10==1||(n%=10)<1||n>3?"th":n<2?"st":n<3?"nd":"rd");

Full program and verification:

using System;
namespace OutputOrdinalNumbers
{
 class Program
 {
 static void Main(string[] args)
 {
 Func<int,string>f= n=>n+(n/10%10==1||(n%=10)<1||n>3?"th":n<2?"st":n<3?"nd":"rd");
 
 for (int i=1; i<=124; i++)
 Console.WriteLine(f(i));
 }
 }
}

• Kevin Cruijssen: You can golf it by two bytes by changing the || to |.

Answer 22

Mathematica 29 + 5 = 34 bytes

SpokenStringDump`SpeakOrdinal

+5 bytes because Speak function must be called before using this built-in.

Usage

SpokenStringDump`SpeakOrdinal[1]

"1st "

SpokenStringDump`SpeakOrdinal[4707]

"4,707th "

Answer 23

Java 7, (削除) 91 (削除ここまで) 81 bytes

String d(int n){return n+(n/10%10==1|(n%=10)<1|n>3?"th":n<2?"st":n<3?"nd":"rd");}

Port from @adrianmp's C# answer.

Old answer (91 bytes):

String c(int n){return n+((n%=100)>10&n<14?"th":(n%=10)==1?"st":n==2?"nd":n==3?"rd":"th");}

Ungolfed & test code:

Try it here.

class M{
 static String c(int n){
 return n + ((n%=100) > 10 & n < 14
 ?"th"
 : (n%=10) == 1
 ? "st"
 : n == 2
 ? "nd"
 : n == 3
 ? "rd"
 :"th");
 }
 static String d(int n){
 return n + (n/10%10 == 1 | (n%=10) < 1 | n > 3
 ? "th"
 : n < 2
 ? "st"
 : n < 3
 ? "nd"
 :"rd");
 }
 public static void main(String[] a){
 for(int i = 1; i < 201; i++){
 System.out.print(c(i) + ", ");
 }
 System.out.println();
 for(int i = 1; i < 201; i++){
 System.out.print(d(i) + ", ");
 }
 }
}

Output:

1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, 12th, 13th, 14th, 15th, 16th, 17th, 18th, 19th, 20th, 21st, 22nd, 23rd, 24th, 25th, 26th, 27th, 28th, 29th, 30th, 31st, 32nd, 33rd, 34th, 35th, 36th, 37th, 38th, 39th, 40th, 41st, 42nd, 43rd, 44th, 45th, 46th, 47th, 48th, 49th, 50th, 51st, 52nd, 53rd, 54th, 55th, 56th, 57th, 58th, 59th, 60th, 61st, 62nd, 63rd, 64th, 65th, 66th, 67th, 68th, 69th, 70th, 71st, 72nd, 73rd, 74th, 75th, 76th, 77th, 78th, 79th, 80th, 81st, 82nd, 83rd, 84th, 85th, 86th, 87th, 88th, 89th, 90th, 91st, 92nd, 93rd, 94th, 95th, 96th, 97th, 98th, 99th, 100th, 101st, 102nd, 103rd, 104th, 105th, 106th, 107th, 108th, 109th, 110th, 111th, 112th, 113th, 114th, 115th, 116th, 117th, 118th, 119th, 120th, 121st, 122nd, 123rd, 124th, 125th, 126th, 127th, 128th, 129th, 130th, 131st, 132nd, 133rd, 134th, 135th, 136th, 137th, 138th, 139th, 140th, 141st, 142nd, 143rd, 144th, 145th, 146th, 147th, 148th, 149th, 150th, 151st, 152nd, 153rd, 154th, 155th, 156th, 157th, 158th, 159th, 160th, 161st, 162nd, 163rd, 164th, 165th, 166th, 167th, 168th, 169th, 170th, 171st, 172nd, 173rd, 174th, 175th, 176th, 177th, 178th, 179th, 180th, 181st, 182nd, 183rd, 184th, 185th, 186th, 187th, 188th, 189th, 190th, 191st, 192nd, 193rd, 194th, 195th, 196th, 197th, 198th, 199th, 200th, 
1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, 12th, 13th, 14th, 15th, 16th, 17th, 18th, 19th, 20th, 21st, 22nd, 23rd, 24th, 25th, 26th, 27th, 28th, 29th, 30th, 31st, 32nd, 33rd, 34th, 35th, 36th, 37th, 38th, 39th, 40th, 41st, 42nd, 43rd, 44th, 45th, 46th, 47th, 48th, 49th, 50th, 51st, 52nd, 53rd, 54th, 55th, 56th, 57th, 58th, 59th, 60th, 61st, 62nd, 63rd, 64th, 65th, 66th, 67th, 68th, 69th, 70th, 71st, 72nd, 73rd, 74th, 75th, 76th, 77th, 78th, 79th, 80th, 81st, 82nd, 83rd, 84th, 85th, 86th, 87th, 88th, 89th, 90th, 91st, 92nd, 93rd, 94th, 95th, 96th, 97th, 98th, 99th, 100th, 101st, 102nd, 103rd, 104th, 105th, 106th, 107th, 108th, 109th, 110th, 111th, 112th, 113th, 114th, 115th, 116th, 117th, 118th, 119th, 120th, 121st, 122nd, 123rd, 124th, 125th, 126th, 127th, 128th, 129th, 130th, 131st, 132nd, 133rd, 134th, 135th, 136th, 137th, 138th, 139th, 140th, 141st, 142nd, 143rd, 144th, 145th, 146th, 147th, 148th, 149th, 150th, 151st, 152nd, 153rd, 154th, 155th, 156th, 157th, 158th, 159th, 160th, 161st, 162nd, 163rd, 164th, 165th, 166th, 167th, 168th, 169th, 170th, 171st, 172nd, 173rd, 174th, 175th, 176th, 177th, 178th, 179th, 180th, 181st, 182nd, 183rd, 184th, 185th, 186th, 187th, 188th, 189th, 190th, 191st, 192nd, 193rd, 194th, 195th, 196th, 197th, 198th, 199th, 200th, 

Answer 24

Vyxal, 6 bytes

∆o2NȯJ

Try it Online!

Answer 25

Python 2.7, 137 chars

def b(n):
 for e,o in[[i,'th']for i in['11','12','13']+list('4567890')]+[['2','nd'],['1','st'],['3','rd']]:
 if n.endswith(e):return n+o

n should be a string

I know I'm beaten by the competition here already, but I thought I'd provide my idea anyways

this just basically generates a list of key, value pairs with number (as a string) ending e and the ordinal o. It attempts to match 'th' first (hence why I didn't use a dictionary), so that it won't accidentally return 'st', for example, when it should be 'th'. This will work for any positive integer

• hlt: n[-1]==e is 5 chars shorter than n.endswith(e)

Answer 26

Scala 86

def x(n:Int)=n+{if(n%100/10==1)"th"else(("thstndrd"+"th"*6).sliding(2,2).toSeq(n%10))}

Scala 102:

def x(n:Int)=if((n%100)/10==1)n+"th"else if(n%10<4)n+("thstndrd".take(n+1)%5*2.drop(n%5*2))else n+"th"

102 as well:

def x(n:Int)=if((n%100)/10==1)n+"th"else if(n%10<4)n+("thstndrd".sliding(2,2).toSeq(n%10))else n+"th"

ungolfed:

def x (n: Int) =
 n + { if (((n % 100) / 10) == 1) "th" 
 else (("thstndrd" + ("th" * 6)).sliding (2, 2).toSeq (n % 10))
 }

Answer 27

C: 95 characters

A ridiculously long solution:

n;main(){scanf("%d",&n);printf("%d%s",n,n/10%10-1&&(n=n%10)<4&&n?n>2?"rd":n<2?"st":"nd":"th");}

It needs to be mangled more.

• dmckee --- ex-moderator kitten: A user flagged this saying "is incorrect: doesn't loop at all, i.e. it works just for a given number, not _all_ numbers below it. see ideone.com/pO6UwV ." That's not a reason to flag as moderators do not judge correctness, but it may be an issue.
• Fors: Based upon this comment from the question creator: "@Ilmari I am looking for 11 as input and 11th as output. I don't mind if it processes multiple lines but what I had in mind was processing just a single number. – NickC Jan 20 '12 at 21:00" it is doing what it should. I.e. it only should process a single number.
• Will Ness: sorry for that flag; I misread the requirement, and didn't have enough rep at the time, to comment directly. Sorry again. :)

Answer 28

OCaml

I'm pretty new to OCaml, but this is the shortest i could get.

let n x =
 let v = x mod 100 and string_v = string_of_int x in
 let z = v mod 10 in
 if v=11 || v=12 || v=13 then string_v^"th" 
 else if v = 1 || z = 1 then string_v^"st" else if v = 2 || z = 2 then string_v^"nd" else if v = 3 || z = 3 then string_v^"rd" else string_v^"th";;

I created a function n that takes a number as a parameter and does the work. Its long but thought it'd be great to have a functional example.

• user3614: input: 11 would yield output: 11st
• Joseph Elcid: yeah, you're right.. I've made the correction. Thanks
• Gareth: I've just tidied up the formatting on your code. Each line needs to be preceded by four spaces in order to be recognized as a code block.
• Gaffi: Wouldn't your first if check be shorter as if v>10 && v<14? I'm not familiar with ocaml, but is it necessary to have the string_v variable be so long?
• Joseph Elcid: No its not necessary, I could have chosen w or x. Just wanted something meaningful. But you're right, it would have made the code a little shorter.

Answer 29

C - (削除) 95 (削除ここまで) 83 characters

main(n,k){scanf("%d",&n);k=(n+9)%10;printf("%d%s\n",n,k<3?"st0円nd0円rd"+3*k:"th");}

Degolfed:

main(n,k)
{
 scanf("%d",&n);
 k=(n+9)%10; // xx1->0, xx2->1, xx3->2
 printf("%d%s\n",n,k<3?"st0円nd0円rd"+3*k:"th");
}

We could do k=(n-1)%10 instead of adding 9, but for n=0 we would get an incorrect behaviour, because in C (-1)%10 evaluates to -1, not 9.

Answer 30

Javascript, 75

function s(i){return i+'thstndrd'.substr(~~(i/10%10)-1&&i%10<4?i%10*2:0,2)}