这是哪张卡？

Question

Introduction

很久以前，当我用通常的扑克牌编码纸牌游戏时，我通常会为每一张牌指定一个号码，然后调用一个函数来得到一张卡片。这在某种程度上激励了我提出这个挑战。

因此，对于那些不知道扑克牌的人来说，一副牌由52张牌组成(四套牌中每套13张，即红心牌、钻石牌、黑桃牌、棍棒)。在每套西装中，有13张牌--首先是2-10的牌号，然后是杰克(J)、皇后(Q)、国王(K)和Ace(A)。这是命令

挑战

挑战是以1-52之间的整数作为输入，并在该位置显示卡片。但是，你的输出必须是文字。此外，必须维持秩序，即前13张牌是红心牌，然后是钻石牌，然后是黑桃牌，最后是棍棒牌。

例如，如果有人选择数字30.The卡，则属于第三套，即黑桃。此外，这将是西服中的第四张卡，这意味着数字5。因此，你的文字输出必须是：five of spades，它应该始终遵循这种格式，即首先是卡片，最后是of和西装的名称，中间需要空格。

输入输出

输入将是介于1-52之间的整数(都包括在内)。请注意，这里的计数从1开始，您可以选择从0开始。然而，你必须维持上面提到的卡片的秩序。你的输出应该是用文字写在那个位置的卡片。您不需要处理无效的输入。此外，您的输出可能是小写的或大写的.

下文列出了所有可能的投入及其产出：

1 -> two of hearts
2 -> three of hearts
3 -> four of hearts
4 -> five of hearts
5 -> six of hearts
6 -> seven of hearts
7 -> eight of hearts
8 -> nine of hearts
9 -> ten of hearts
10 -> jack of hearts
11 -> queen of hearts
12 -> king of hearts
13 -> ace of hearts
14 -> two of diamonds
15 -> three of diamonds
16 -> four of diamonds
17 -> five of diamonds
18 -> six of diamonds
19 -> seven of diamonds
20 -> eight of diamonds
21 -> nine of diamonds
22 -> ten of diamonds
23 -> jack of diamonds
24 -> queen of diamonds
25 -> king of diamonds
26 -> ace of diamonds
27 -> two of spades
28 -> three of spades
29 -> four of spades
30 -> five of spades
31 -> six of spades
32 -> seven of spades
33 -> eight of spades
34 -> nine of spades
35 -> ten of spades
36 -> jack of spades
37 -> queen of spades
38 -> king of spades
39 -> ace of spades
40 -> two of clubs
41 -> three of clubs
42 -> four of clubs
43 -> five of clubs
44 -> six of clubs
45 -> seven of clubs
46 -> eight of clubs
47 -> nine of clubs
48 -> ten of clubs
49 -> jack of clubs
50 -> queen of clubs
51 -> king of clubs
52 -> ace of clubs

评分

这是密码-高尔夫，所以最短的代码获胜。

Answer 1

05AB1E，54字节

0-索引

“»€Å‹¡Šdesž…“#“‚•„í†ìˆÈŒšï¿Ÿ¯¥Š—¿—ÉŸÄ‹ŒÁà“#âí" of "ýsè

在网上试试！

解释

“»€Å‹¡Šdesž…“#                                          # push list of suits
              “‚•„í†ìˆÈŒšï¿Ÿ¯¥Š—¿—ÉŸÄ‹ŒÁà“#             # push list of ranks
                                           â            # cartesian product
                                            í           # reverse each
                                             " of "ý    # join on " of "
                                                    sè  # index into cardlist with input

Answer 2

Emojicode，202个字节

itwo.three.four.five.six.seven.eight.nine.ten.jack.queen.king.ace.i 13 of hearts.diamonds.spades.clubs.➗i 13

0索引。在网上试试！

解释

		start of the closure block
  i		 closure takes an integer argument i
  		 print:
    		  concatenate these strings:
      ....i 13  [a]
       of 
      ....➗i 13  [b]
    


[a]:
		tell Emojicode to dereference without checking
		 get the nth element of the following array
  		  create an array using the following string and separator
    ...
    .
   i 13	n, i mod 13

[b]
....➗i 13
same but with ⌊i÷13⌋

Answer 3

R，154个字节

paste(el(strsplit("Two,Three,Four,Five,Six,Seven,Eight,Nine,Ten,Jack,Queen,King,Ace",",")),"of",rep(c("Hearts","Diamonds","Spades","Clubs"),e=13))[scan()]

在网上试试！

从STDIN获取输入(1-索引)并使用source(...,echo=T)将结果打印到控制台。

它不是很漂亮，但它比我使用outer的最佳解决方案短了2个字节(如下所示)，所以让我们来提醒您检查另一种方法吧！

paste(                          # concatenate together, separating by spaces,
                                # and recycling each arg to match the length of the longest
el(strsplit("Two,...",",")),    # split on commas and take the first element
"of",                           # 
 rep(c("Hearts",...),           # replicate the suits (shorter as a vector than using strsplit
               e=13)            # each 13 times
                    )[scan()]   # and take the input'th index.

R，156个字节

outer(el(strsplit("Two,Three,Four,Five,Six,Seven,Eight,Nine,Ten,Jack,Queen,King,Ace",",")),c("Hearts","Diamonds","Spades","Clubs"),paste,sep=" of ")[scan()]

在网上试试！

本质上和上面一样；然而，outer会正确地进行回收，但是必须为paste设置sep=" of "使这个过程变得更长。