除非您有修改站点设计的用户脚本（即使这样做），否则您应该已经注意到我们拥有站点设计！

_{^{（现在是十一月）}}

因此，为了庆祝这一点，让我们制作一个追溯应用此设计的程序（当然，这过于简化了！）！

因此，最重要的更改是：

• 我们现在的新名称是Code Golf和Coding挑战赛
• 我们的网站不再是默认的蓝色，而是深绿色

因此，给定一个字符串，请更改：

• PPCG 至 CGCC
• Programming Puzzles {non-whitespace} Code Golf到Code Golf {same-thing} Coding Challenges（例如，Programming Puzzles + Code Golf，and Code Golf，和& Code Golf将所有被改变为Code Golf + Coding Challenges，and Coding Challenges，和& Coding Challenges。
• #abc到#acb（交换蓝色到绿色，反之亦然-只是忽视的事实是绿色的应该不是逻辑变成蓝色，但我不想色移复杂）
• #abcdef到#abefcd（与上面相同）

请注意，对于颜色交换，您应该接受从0-9到的任何十六进制数字a-f。您可以选择所需的十六进制数字大小写，但输入与输出之间必须保持一致且相同。

您的替换内容区分大小写，如果不区分大小写，请指定输出的工作方式。

仅当字符串被单词边界包围时（包括的开始），才应进行替换#。换句话说，只有当匹配项位于字符串的边缘或以非字母数字字符为边界（在两侧）时，才应进行每个指定的替换。

有标准漏洞。这是一个代码挑战，所以最短的代码胜出！

例子

PPCG -> CGCC
Programming Puzzles or Code Golf -> Code Golf or Coding Challenges
PPCG stands for Programming Puzzles and Code Golf and its site color is #abf -> CGCC stands for Code Golf and Coding Challenges and its site color is #afb
The color #00f is much more intense than #5568ed -> The color #0f0 is much more intense than #55ed68
Programming Puzzles and No Substitution -> Programming Puzzles and No Substitution
No Substitution and Code Golf -> No Substitution and Code Golf
Programming Puzzles and no substitution Code Golf -> Programming Puzzles and no substitution Code Golf
Code Golf and Programming Puzzles -> Code Golf and Programming Puzzles
Programming Puzzles and Programming Puzzles and Code Golf -> Programming Puzzles and Code Golf and Coding Challenges

（对于最后一个测试用例，重要的是要注意，转换后的字符串可以再次转换，但是转换必须恰好应用一次）

重要更新

（感谢@tsh）

该Programming Puzzles ... Code Golf置换规则可以包括它里面的其他规则（Programming Puzzles PPCG Code Golf有效）。在这种情况下，您可以选择是否应用规则，但是必须确定。我并不需要您之间保持一致PPCG，#...因为答案似乎会在我的列表顺序中实施规则，从而导致不一致。这只是一个澄清；我相信所有当前答案仍然有效。

红宝石 -p，165个 164 159字节

最终它与sed答案非常相似，但是它滥用Ruby的字符串插值法([\da-f]{1,2})将第三次正则表达式中的十六进制组匹配项重复了三遍，而无需再次重复整个过程。

• @ randomdude999的-1个字节。
• 利用@Xcali的Perl解决方案获得-5个字节

gsub /\bPPCG\b/,"CGCC"
gsub /\bProgramming Puzzles( \S+ )(Code Golf)\b/,'\2\1Coding Challenges'
[1,2].map{|i|gsub /(^|\s)#\K#{'([\da-f]{%d})'%i*3}\b/,'\2\4\3'}

在线尝试！

C ++（GCC） ，270个 285 283字节

感谢Neil指出错误。

-2个字节，感谢ceilingcat。

#import<regex>
#import<string>
auto f=[](auto s){typeof(s)R[][2]{"bPPCG","CGCC","bProgramming Puzzles( \\S+ )(Code Golf)","$2$1Coding Challenges","B#(?=([\\da-f]{3}){1,2}\\b)(.+?)(..??)(..??)","#$2$4$3"};for(auto r:R)s=std::regex_replace(s,std::regex('\\'+*r+"\\b"),r[1]);return s;};

在线尝试！

视网膜0.8.2，153个 130字节

\bPPCG\b
CGCC
\bProgramming Puzzles( \S+ )(Code Golf)\b
$2$1Coding Challenges
\B(#(?=([\da-f]{3}){1,2}\b).+?)(..??)(..??)\b
$1$4$3

在线尝试！链接包括测试用例。所有替换均区分大小写。假定正常的正则表达式单词字符是可接受的，因此\B#仅匹配#不跟随单词字符的s。编辑：由于@tsh，节省了22个字节。

GNU sed -E，198个字符

s/\bPPCG\b/CGCC/g
s/\bProgramming Puzzles( \S* Cod)e Golf\b/Code Golf\1ing Challenges/g
s/((^|\W)#[0-9a-f])([0-9a-f])([0-9a-f])\b/\1\4\3/g
s/((^|\W)#[0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})\b/\1\4\3/g

可使用例如sed -E 'the above';; 换行符可以按原样包含，也可以根据需要替换为;。两者都可以。

@HyperNeutrino认为单词边界规则很愚蠢。看看我在这种#情况下要做的。

是的，我什至没有尝试。：P

@Value墨水+9

Stax，85 个字节

ì▀¼FΣ¼≤C╛╓ÄydîNû►┘Δ▲Bd♫|α╒oñFτ▒t!↑▌╩┘♦pMc6Hèé▄·│╝∙↔¥^4b5╠·9█&╨^╨♂═î£ε■屫\┴J₧å♠Å≡z╜û♀

运行并调试

05AB1E，123个 109 105 110 114 字节

žKISå_Å¡JεÐć'#QsžhA6£«sSåP*i3äćsRJ«ë"PPCG"Qi"CGCC"]J”–±ÇÀ”DU¡ćsε”ƒËŠˆ”©¡DVćDÁ2ôεðå}ćs_P*YyÊP*i”Âïªï”«s®ý«®ìëyXì]J«

+5字节修复测试用例，如Programming Puzzles and no substitution Code Golf和Programming Puzzles and Programming Puzzles and Code Golf。
+4字节固定测试用例，例如color-#00f（颜色周围带有空格/换行符的颜色）。感谢@Grimy引起我的注意。

区分大小写。十六进制值带有小写abcdef; Programming Puzzles ... Code Golf在标题中；PPCG大写。

在线尝试。

绝对不适合工作的正确的语言..模仿单词边界和更换Programming Puzzles \S+ Code Golf，但不Code Golf \S+ Programming Puzzles还是Programming Puzzles \S+ \S+ Code Golf没有任何的正则表达式是相当困难的（做短）。>>

说明：

žK                # Push "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
                  # (NOTE: if '_' also count as word boundary, use `žj` instead)
  IS              # Push the input as a list of characters
    å_            # Check for each character if they are NOT in the string
      Å¡          # Split the (implicit) input on truthy values
        J         # Join each inner character list to a string again
ε                 # Map each word to:
 Ð                #    Triplicate the word
  ć               #    Extract head; push remainder and head separately to the stack
   '#Q           '#    Check if the head equals "#"
    žh            #    Push string "0123456789"
      A6£«        #    Append the first 6 letters of the alphabet: "0123456789abcdef"
   s      sSåP    #    Check if the characters of the remainder are all in this string
   *i             #    If both are truthy:
     3ä           #     Split the 'word' into three parts
                  #      i.e. "#ab12cd" → ["#ab","12","cd"]
       ćs         #     Extract head, and swap
         R        #     Reverse the remainder list
          J«      #     Join them together, and merge them to the head again
    ë"PPCG"Qi     #    Else-if the word is "PPCG":
     "CGCC"       #     Push "CGCC"
                  #    (Implicit else:
                  #      Use the word that's still there from the initial triplicate)
]                 # Close all if statements and the nested map
 J                # Join the mapped words together again
”–±ÇÀ”            # Push dictionary string "Programming Puzzles"
      DU          # Save a copy in variable `X`
        ¡         # Split the string by this
         ćs       # Extract head & swap; pushing head and remainder to the stack
ε                 # Map each substring `y` in the remainder to:
 ”ƒËŠˆ”           #  Push dictionary string "Code Golf"
       ©          #  Save it in variable `®` (without popping)
        ¡         #  Split the current substring we're mapping by it
         DV       #  Save a copy of this list in variable `Y`
           ćD     #  Extract the head and duplicate
                  #  (so the stack is: remainder, head, head)
 Á                #  Rotate the characters in the head-string once towards the right
  2ô              #  Split it into parts of size 2
    εðå}          #  Check in each part if it contains a space
        ćs        #  Extract head and swap again
          _       #  Check if all values are 0
           P      #  And check if this is truthy for all
          *       #  And check if this is truthy, as well as the head
                  #  (this means the current string has a leading and trailing space,
                  #   and no other spaces)
 YyÊP             #   Check that none of the susbtrings in variable `Y`
                  #   are equal to the current substring `y`
 *i               #   If both checks above are truthy:
   ”Âïªï”«        #    Prepend "Coding Challenges" to the duplicated head
          s®ý     #    Join the remainder by variable `®` ("Code Golf")
             «    #    Append it
              ®ì  #    And prepend an additional variable `®` ("Code Golf")
  ë               #   Else:
   y              #    Simply keep the substring `y` as is
    Xì            #    And prepend variable `X` ("Programming Puzzles") 
                  #    since we've split by it
]                 # Close all if-else statements and the map
 J                # Join the mapped substrings together to a single string
  «               # And append it to the initially extracted head
                  # (then output the resulting string implicitly as result)

见矿（部此次05AB1E提示如何使用字典？）理解为什么”–±ÇÀ”是"Programming Puzzles"; ”ƒËŠˆ”是"Code Golf"; 并且”Âïªï”是"Coding Challenges"。

Python 2，240个字节

import re
lambda x,s=re.sub,b='(?<!\w)',e='(?!\w)',h='([\da-f]',t=r'#\1\3\2':s(b+'#%s{2})'%h+h+'{2})%s{2})'%h+e,t,s(b+'#%s)'%h+h+')%s)'%h+e,t,s(b+'Programming Puzzles( \S+ Cod)e Golf'+e,r'Code Golf\1ing Challenges',s(b+'PPCG'+e,'CGCC',x))))

在线尝试！

JavaScript（Node.js），174个字节

s=>s[R='replace'](/\bPPCG\b/g,'CGCC')[R](/\bProgramming Puzzles( \S+ )(Code Golf)\b/g,'$2$1Coding Challenges')[R](/\B#(?=([\da-f]{3}){1,2}\b)(.+?)(..??)(..??)\b/ig,'#$2$4$3')

在线尝试！

Pyth，177个 173 162 142字节

J::jb.z"\\bPPCG\\b""CGCC"." z¶NZI°Pÿúd(MÜ_BöIkxnqä'u)"." s6#~ÍN³=<nñu/GÎg"VS2~:J%"(^|\W)#%s\\b"*3%"([\da-f]{%d})"N$r"\1#\2\4\3"$)J

这是一个没有Pyth的字符串压缩机制的版本（aka复制粘贴是安全的）：

J::jb.z"\\bPPCG\\b""CGCC""\\bProgramming Puzzles( \S+ )(Code Golf)\\b""\\2\\1Coding Challenges"VS2~:J%"(^|\W)#%s\\b"*3%"([\da-f]{%d})"N$r"\1#\2\4\3"$)J

在线尝试！

由于我试图对正则表达式尽可能学究，所以最终的结果确实很长。我尝试压缩所有可能的字符串，但是其中大多数都没有变小或无法正确粘贴到TIO中。

说明：

J::                      # definition of J to the following 2 regex replacements
   jb.z                  # input to first regex replacement: all input lines joined together
   "\\bPPCG\\b"          # first regex
   "CGCC"                # first replacement
   ."<compressed>"       # second regex: "\\bProgramming Puzzles( \S+ )(Code Golf)\\b"
   ."<compressed>"       # second replacement: "\\2\\1Coding Challenges"
VS2                      # loop twice, N = 1 or 2
  ~:J                    # some weird assignment: J = regex replace in J (would be J := (regex, replacement) if : was python's regex replace operator)
    %                    # third regex: string format
     "(^|\W)#%s\\b"      # format string
     *3                  # repeat 3 times:
       %"([\da-f]{%d})"N # string format, replace %d with N (the loop counter)
    $r"\1#\2\4\3"$       # third replacement: uses python raw literals because it's shorter than escaping the backslashes
    )                    # end for loop
J                        # print J

• -11感谢Value Ink的Ruby答案提供更好的正则表达式
• -20感谢在Ruby和Perl答案的启发下对两个十六进制替换使用循环

Perl 5中 -p，152个 145字节

@ValueInk保存7个字节

s/\bPPCG\b/CGCC/g;s/\bProgramming Puzzles( \S+ )(Code Golf)\b/$2$1Coding Challenges/g;for$a(1,2){$p="([a-f0-9]{$a})"x3;s/(^|\s)#\K$p\b/$2$4$3/gi}

在线尝试！

Java 8，192字节

s->s.replaceAll("\\bPPCG\\b","CGCC").replaceAll("\\bProgramming Puzzles( \\S+ )(Code Golf)\\b","$2$1Coding Challenges").replaceAll("\\B(#(?=([0-9a-f]{3}){1,2}\\b).+?)(..??)(..??)\\b","$1$4$3")

@Neil的Retina答案端口，请确保对他进行投票！

在线尝试。