澳大利亚有8个州和地区，每个州和地区的字母缩写为2或3：

• ACT：澳大利亚首都领地
• 新南威尔士州：新南威尔士州
• NT：北领地
• 昆士兰州：昆士兰州
• SA：南澳大利亚
• 塔斯马尼亚州（TAS）：塔斯马尼亚州
• VIC：维多利亚
• 西澳大利亚州：西澳大利亚州

您的任务是编写一个函数/程序，该函数/程序必须使用有效的澳大利亚州大写字母缩写，并返回正确的全名（如上所述，区分大小写）。

除单个尾随换行符外，不允许有多余的空格，这是该语言中每个程序不可避免的副作用。

例如：

f("TAS") => "Tasmania"

标准 代码高尔夫球 规则（以字节为单位的最短代码为准！），并且存在漏洞。

视网膜，135121字节

@Neil通过巧妙的空间插入策略打高尔夫球了13个字节，这启发了再打一个字节

AS
asmania
^W
Western 
NT
NorthernT
T$
 Territory
A
Australia
aC
an Capital
IC
ictoria
LD
ueensland
SW
ew SWales
S
South 

在线尝试！

注意第四行和最后一行的尾随空格。

果冻，56 字节

OS%15ị“ÞÑṿ“ıÐṁṾṗ“¡ʋẹḢ⁾ḅ“¢T¡ż¶““msẒw²ẉṪP“ØÑĊ“©$Ḅ3ẊḳƓ“ÇH°»

用于获取和返回字符列表的单子链接。

在“ 在线试用”中查看所有案例！

注意：重用单词“ Australia”和“ Territory”无助于减少字节数。

怎么样？

将八个状态缩写的序数的总和取模15得出8个唯一值。取那些模9的结果可保持唯一性。

OS%15ị“ ... » - Main link: list of characters a
O             - cast to ordinals
 S            - sum
  %15         - modulo 15
      “ ... » - list of compressed strings:
              -   "Victoria"
              -   "Western Australia"
              -   "Northern Territory"
              -   "South Australia"
              -   ""
              -   "Australian Capital Territory"
              -   "Tasmania"
              -   "New South Wales"
              -   "Queensland"
     ị        - index into (1-indexed and modular - hence the single empty entry)

Haskell，193192字节

f"NSW"="New South Wales"
f"QLD"="Queensland"
f"ACT"=a++"n Capital"++t
f"VIC"="Victoria"
f"TAS"="Tasmania"
f"SA"="South "++a
f"NT"="Northern"++t
f"WA"="Western "++a
a="Australia"
t=" Territory"

有点天真的解决方案，但是我想第一次尝试一个相当简单的挑战。我什至不确定Haskell是否可以更好。

通过将空间移到t中节省了一个字节

Mathematica，177个字节

明显的解决方案：（199字节）

<|"ACT"->"Australian Capital Territory","NSW"->"New South Wales","NT"->"Northern Territory","QLD"->"Queensland","SA"->"South Australia","TAS"->"Tasmania","VIC"->"Victoria","WA"->"Western Australia"|>

改进的解决方案：（182字节，只能在我的Wolfram Mathematica 10.2计算机上运行）

{,"Northern Territory","Queensland",,,,,"Australian Capital Territory","New South Wales",,,"Tasmania",,,"Western Australia","Victoria",,,,,,,,,,,,"South Australia"}[[Hash@#~Mod~29]]&

最佳解决方案：

{,"Northern Territory","Queensland",a="Australia",,,,a<>"n Capital Territory","New South Wales",,,"Tasmania",,,"Western "<>a,"Victoria",,,,,,,,,,,,"South "<>a}[[Hash@#~Mod~29]]&

不幸的是，重复" Territory"只能保存0个字节。

似乎Mathematica的默认哈希函数会在版本之间进行更改。但是添加方法会使代码更长。现在，使用Mathematica沙箱的默认哈希函数可以使代码缩短约9个字节。

Mathematica（沙盒-现在-版本11），168字节

{a="Australia","Western "<>a,,,,"South "<>a,,a<>"n Capital Territory",,,,"Tasmania","Northern Territory",,,,"New South Wales","Victoria","Queensland"}[[Hash@#~Mod~20]]&

Python，181字节

@Jonathan Allan节省了1个字节

lambda a,s='Australia',t=' Territory':[s+'n Capital'+t,'New South Wales','Northern'+t,'Queensland','South '+s,'Tasmania','Victoria','Western '+s]['ACNSNTQLSATAVIWA'.index(a[:2])//2]

在线尝试！

桑达，167个 157字节

s/T$/ Territory/
s/N /Northern /
s/AS/asmania/
s/IC/ictoria/
s/LD/ueensland/
s/W$/Wales/
s/A/Australia/
s/C/n Capital/
s/NS/New S/
s/^W/Western /
s/S/South /

PHP，148143字节

<?=strtr(["0n Capital 3",Victoria,Queensland,West20,10,Tasmania,North23,"New 1 Wales"][md5(hj.$argn)%8],[Australia,"South ","ern ",Territory]);

由于JörgHülsermann节省了5个字节。

C＃，289字节

数据

• 输入 String s状态的缩写。
• 输出 String扩展状态名称

打高尔夫球

(string s)=>{return new System.Collections.Generic.Dictionary<string,string>(){{"ACT","Australian Capital Territory"},{"NSW","New South Wales"},{"NT","Northern Territory"},{"QLD","Queensland"},{"SA","South Australia"},{"TAS","Tasmania"},{"VIC","Victoria"},{"WA","Western Australia"}}[s];};

不打高尔夫球

( string s ) => {
    return new System.Collections.Generic.Dictionary<string, string>() {
        { "ACT", "Australian Capital Territory" },
        { "NSW", "New South Wales" },
        { "QLD", "Queensland" },
        { "TAS", "Tasmania" },
        { "VIC", "Victoria" },
        { "NT", "Northern Territory" },
        { "SA", "South Australia" },
        { "WA", "Western Australia" }
    }[ s ];
};

非高尔夫可读

// Takes a state name abbreviated 
( string s ) => {

    // Creates a dictionary with the state name abbreviated and the full state name and returns the one that match
    return new System.Collections.Generic.Dictionary<string, string>() {
        { "ACT", "Australian Capital Territory" },
        { "NSW", "New South Wales" },
        { "QLD", "Queensland" },
        { "TAS", "Tasmania" },
        { "VIC", "Victoria" },
        { "NT", "Northern Territory" },
        { "SA", "South Australia" },
        { "WA", "Western Australia" }
    }[ s ];
};

完整代码

using System;
using System.Collections.Generic;

namespace Namespace {
   class Program {
      static void Main( String[] args ) {
         Func<String, String> f = ( string s ) => {
             return new System.Collections.Generic.Dictionary<string, string>() {
                 { "ACT", "Australian Capital Territory" },
                 { "NSW", "New South Wales" },
                 { "QLD", "Queensland" },
                 { "TAS", "Tasmania" },
                 { "VIC", "Victoria" },
                 { "NT", "Northern Territory" },
                 { "SA", "South Australia" },
                 { "WA", "Western Australia" }
             }[ s ];
         };

         List<String>
             testCases = new List<String>() {
                 "ACT",
                 "NSW",
                 "QLD",
                 "TAS",
                 "VIC",
                 "NT",
                 "SA",
                 "WA",
             };

         foreach( String testCase in testCases ) {
             Console.WriteLine( $"Input: {testCase}\nOutput: {f( testCase )}\n");
         }

         Console.ReadLine();
      }
   }
}

发布

• 1.0 - 289 bytes-初始溶液。

笔记

• 没有

的Rexx（伊利），148个 147字节

a=Australia
t=Territory
s=South
ac=a'n' Capital t
ns=New s Wales
nt=Northern t
qs=Queensland
sa=s a
ta=Tasmania
vi=Victoria
wa=Western a
arg x 3
say value(x)

在线尝试！

JavaScript（ES6），167字节

s=>({A:`${A='Australia'}n Capital ${T='Territory'}`,N:s[2]?'New South Wales':'Northern '+T,T:'Tasmania',Q:'Queensland',S:'South '+A,V:'Victoria',W:'Western '+A}[s[0]])

f=
s=>({A:`${A='Australia'}n Capital ${T='Territory'}`,N:s[2]?'New South Wales':'Northern '+T,T:'Tasmania',Q:'Queensland',S:'South '+A,V:'Victoria',W:'Western '+A}[s[0]])

console.log('ACT NSW NT QLD SA TAS VIC WA'.split(' ').map(f))