受C指令启发#define。

挑战

给定一个带有一些别名的短语，以及每个别名文本一个数组。输出初始短语，将每个别名替换为其相应的文本。

别名由一个尖锐字符定义，#后跟其在数组中的索引（索引可以从零或一开始）定义。别名可以在其文本内包含另一个别名，您必须解析所有别名（可能是递归的）。您可以假设别名永远不会陷入无限循环。别名不会有前导零（#02不是index 2处的别名，而是index 0后面是文本的别名2）。

您可以假设数组长度不超过20个项目。

您可以编写程序，函数或什至#define-都不错:)

您还可以使用另一种更适合您的语言的输入法。

例

phrase: "#0 & #3"
array: [
    "Programming #1",
    "Puzzles",
    "Code",
    "#2 Golf"
]
output: "Programming Puzzles & Code Golf"

一步步：

0> "#0 & #3"
1> "Programming #1 & #2 Golf"
2> "Programming Puzzles & Code Golf"

由于这是 代码高尔夫球，最短答案以字节为单位！

另一个样品

phrase: "#0!"
array: [
    "We are #1",
    "#2",
    "#3",
    "#4 !",
    "graduating"
]
output: "We are graduating !!"

phrase: "##0#1#0#21#3#4"
array: [
    "a",
    "m",
    "z",
    "n",
    "g"
]
output: "#amaz1ng"

phrase: "##1#23"
array: [
    "WEIRD",
    "0 C",
    "AS"
]
output: "WEIRD CAS3"

phrase: "#1#7#6y#4#7#10s#7b#11#0#0#11r#7#0h#6#5#2#5#9#4."
array: [
    "t",
    "#12#3",
    "#11ga#3",
    "#0#10v#11",
    "#0h#10#8g",
    "#7#8",
    "a#8",
    " ",
    "n",
    "o",
    "i",
    "e",
    "P#9s#10"
]
output: "Positive anything is better than negative nothing."

上面的示例使用索引从零开始的Array。

JavaScript（ES6）58

递归函数

f=(s,h,r=s.replace(/#(1?\d)/g,(_,x)=>h[x]))=>r==s?r:f(r,h)

测试

f=(s,h,r=s.replace(/#(1?\d)/g,(_,x)=>h[x]))=>r==s?r:f(r,h)

// Version without default parameters, same length but using a global
// f=(s,h)=>(r=s.replace(/#(1?\d)/g,(_,x)=>h[x]))==s?r:f(r,h)

console.log=x=>O.textContent+=x+'\n'

;[
  ['##1#23',['WEIRD','0 C','AS']],
  ["#0!",["We are #1","#2","#3","#4 !","graduating"]],
  ["##0#1#0#21#3#4",["a","m","z","n","g"]],
  ["##0#1#0#21#13#4",["a","m","z","","g","","","","","","","","","n"]],
  ["#1#7#6y#4#7#10s#7b#11#0#0#11r#7#0h#6#5#2#5#9#4.",
    ["t","#12#3","#11ga#3","#0#10v#11","#0h#10#8g","#7#8","a#8"," ","n","o","i","e","P#9s#10"]
  ],
  ["#0 & #3",["Programming #1","Puzzles","Code","#2 Golf"]]
].forEach(t=>{
  var a=t[0],b=t[1]
  console.log(a+' ['+b+']\n -> '+f(a,b))
})

<pre id=O></pre>

Mathematica，74个字节

FixedPoint[#~StringReplace~a&,a=0;a=Reverse["#"<>ToString@a++->#&/@#2];#]&

不太复杂。其中大多数只是用于创建索引。

朱莉娅112 107 66字节

f(s,x)=(r=replace(s,r"#1?\d",m->x[1+parse(m[2:end])]))==s?s:f(r,x)

这是一个递归函数，它接受一个字符串和一个数组并返回一个字符串。它使用基于0的索引。

我们首先构造一个字符串r作为输入字符串s，用正则表达式的所有匹配项#1?\d替换为x的元素，该元素对应于1 +从匹配项中解析出的整数。如果等于s，则返回s，否则递归，将r作为字符串传递。

C，269 232

#define f(p,a,l)char*n,o[999],i=0,c,x;for(strcpy(o,p);o[i];)o[i]==35&isdigit(o[i+1])?c=o[i+1]-48,c=isdigit(o[i+2])&c?10*c+o[i+2]-48:c,n=a[c<l?c:c/10],x=strlen(n),memmove(o+i+x,o+i+2+c/10,strlen(o+i)),i=!memmove(o+i,n,x):++i;puts(o);

根据要求，单一#define解决问题！C宏不能递归，因此必须迭代解决该问题。该宏带有3个参数；短语p，数组a和数组的长度l。 我只从非解决方案中删除了空格；我知道我可以保存更多的字符，但是我认为这不会使我的字符数降至200以下。这绝对不是一个有竞争力的解决方案。 解决方案已全面解决。函数形式的非解决方案：

f(char*p,char**a,int l){
  char o[999]={0},i=0;
  strcpy(o,p);
  while(o[i]){
    if(o[i]=='#'&&isdigit(o[i+1])){
      int c = o[i+1]-'0';
      if(isdigit(o[i+2])&&c)
        c=10*c+o[i+2]-'0';
      if(c>=l)
        c/=10;
      char *n=a[c];
      memmove(o+i+strlen(n),o+i+2+c/10,strlen(o+i));
      memmove(o+i,n,strlen(n));
      i=0;
    }else{
      i++;
    }
  }
  puts(o);
}

和测试代码：

void test(char *phrase, char **array, int length) {
  f(phrase, array, length);
}

main() {
  const char *t1[] = {
    "Programming #1","Puzzles","Code","#2 Golf"
  };
  test("#0 & #3", t1, 4);

  const char *t2[] = {
    "We are #1","#2","#3","#4 !","graduating"
  };
  test("#0!", t2, 5);

  const char *t3[] = {
    "a","m","z", "n","g"
  };
  test("##0#1#0#21#3#4", t3, 5);

  const char *t4[] = {
    "WEIRD","0 C","AS"
  };
  test("##1#23", t4, 3);

  const char *t5[] = {
    "t","#12#3","#11ga#3","#0#10v#11","#0h#10#8g","#7#8","a#8"," ","n","o","i","e","P#9s#10"
  };
  test("#1#7#6y#4#7#10s#7b#11#0#0#11r#7#0h#6#5#2#5#9#4.", t5, 13);
}

编辑：工作一些高尔夫魔术。正如我认为的那样，它是如此简短和难以理解。