高尔夫代码:七个细分 [英] Code Golf: Seven Segments

问题描述

按字符计数的最短代码，用于生成给定十六进制数字的七段显示表示.

The shortest code by character count to generate seven segment display representation of a given hex number.

仅由数字[0-9]和十六进制字符(大写和小写[a-fA-F])组成.无需处理特殊情况.

Input is made out of digits [0-9] and hex characters in both lower and upper case [a-fA-F] only. There is no need to handle special cases.

使用那些 ASCII 面，输出将是输入的七段表示:

Output will be the seven segment representation of the input, using those ASCII faces:

  _       _   _       _   _   _   _   _   _       _       _   _  
 | |   |  _|  _| |_| |_  |_    | |_| |_| |_| |_  |    _| |_  |_  
 |_|   | |_   _|   |  _| |_|   | |_|  _| | | |_| |_  |_| |_  | 

限制

禁止使用以下内容:eval，exec，system，figlet，厕所和外部库.

Restrictions

The use of the following is forbidden: eval, exec, system, figlet, toilet and external libraries.

Input:
    deadbeef

Output:
        _  _        _  _  _ 
     _||_ |_| _||_ |_ |_ |_ 
    |_||_ | ||_||_||_ |_ |  


Input:
    4F790D59

Output:
        _  _  _  _     _  _ 
    |_||_   ||_|| | _||_ |_|
      ||    | _||_||_| _| _|

代码计数包括输入/​​输出(即完整程序).

Code count includes input/output (i.e full program).

推荐答案

Perl，134个字符

所有换行符都可以删除.

Perl, 134 characters

All linebreaks may be removed.

@s=unpack"C*",~"P\xdbLI\xc3a`[\@AB\xe0t\xc8df";
$_=<>;for$s(6,3,0){print map$s[hex$&]&1<<$_+$s?$_%2?"_":"|":" ",
0..2while/./g;print$/}

说明

@s 存储每个段的位.条目的顺序是从0到F(感谢 hex())，并且位按以下顺序映射到段:

Explanation

@s stores the bits for each segment. The entries are in order from 0 to F (thanks to hex()) and the bits map to segments in this order:

6 7 x
3 4 5
0 1 2

，其中 0 是LSB.(未使用位6).值存储在位反转的字符串中，因此有更多可打印的字符.〜运算符将这些位翻转并解压缩为我提供数字(涉及字符串时，perl的按位运算符更加笨拙).

with 0 being the LSB. (Bit 6 is unused). The values are stored packed in the string bit-inverted so there are a lot more printable characters; the ~ operator flips the bits and unpack gives me numbers (perl's bitwise operators are much clumsier when it comes to strings).

掌握数据后，我读取了输入，并进行了三次遍历.三个循环之间的唯一区别是所需的位掩码.对于输入的每个字符，将打印输出的三个字符.要打印的字符是

With the data in hand I read the input and proceed to loop over it three times; the only difference between the three loops is the bitmask required. For each character of input three characters of output are printed. The character to be printed is

$s[ hex $& ] & (1 << ($_ + $s) )
? ($_ % 2 ? "_" : "|" )
: " "

其中 @s 是查找表， $ s 是根据行而产生的转移，而 $ _ 是是否在行中打印第一个，第二个或第三个字符.如果查找表条目中的右位为false，则会打印一个空格；否则，它将显示一个空格.否则，它会打印一个"|"在侧面或中间是"_".

where @s is the lookup table, $s is the shift in effect depending on the row, and $_ is whether we're printing the 1st, 2nd, or 3rd character in the row. If the right bit in the lookup table entry is false it prints a space; otherwise it prints a "|" on the sides or a "_" in the middle.

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