这是一个很好的简单挑战：

给定一个代表未知基数的字符串，请确定该数字可能位于的最低基数。该字符串仅包含0-9, a-z。如果您愿意，可以选择使用大写字母而不是小写字母，但是请指定。您必须以十进制输出这个最低的底数。

这是一个更具体的例子。如果输入字符串为“ 01234”，则该数字不可能为二进制，因为2、3和4都未定义为二进制。同样，此数字不能以3或4为底。因此，此数字必须以5或更高的底数为基础，因此应输出“ 5”。

您的代码必须适用于以1为底（一元，全为0）和以36为底（“ 0-9”和“ a-z”）的任何底。

您可以接受输入并以任何合理的格式提供输出。允许进行基本转换内置函数。像往常一样，存在标准漏洞，而最短的答案是以字节为单位！

测试IO：

#Input          #Output
00000       --> 1
123456      --> 7
ff          --> 16
4815162342  --> 9
42          --> 5
codegolf    --> 25
0123456789abcdefghijklmnopqrstuvwxyz    --> 36

果冻，4 字节

ṀØBi

需要大写。在线尝试！或验证所有测试用例。

怎么运行的

ṀØBi  Main link. Arguments: s (string)

Ṁ     Yield the maximum of s.
 ØB   Yield "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".
   i  Find the 1-based index of the maximum in that string.

Python，27 22字节

lambda s:(max(s)-8)%39

这要求输入是字节串（Python 3）或字节数组（Python 2和3）。

感谢@AleksiTorhamo打高尔夫球5个字节！

在Ideone上进行测试。

怎么运行的

我们从获取字符串的最大值开始。这是字母的代码点高于数字的代码点，此最大字符也是最大的基数36位。

的代码点“0” - “9”是48 - 57，所以我们必须减去48从它们的代码点来计算对应的数字，或47来计算最低的可能基。同样，字母的代码点“A” - “Z”是97 - 122。由于“ a”代表数值为10的数字，我们必须从其代码点中减去87来计算相应的数字，或者减去86来计算可能的最低基数。实现此目的的一种方法如下。

之间的差值97和58（“：”后的文字，“9”）是39，因此服用代码点模39所能达到的减法。由于48％39 = 9，并为字符所需的结果“0”是1，我们首先减去8取模的结果之前39。首先必须减去，因为否则'u'％39 = 117％39 = 0。

c    n    n-8    (n-8)%39
0    48    40     1
1    49    41     2
2    50    42     3
3    51    43     4
4    52    44     5
5    53    45     6
6    54    46     7
7    55    47     8
8    56    48     9
9    57    49    10
a    97    89    11
b    98    90    12
c    99    91    13
d   100    92    14
e   101    93    15
f   102    94    16
g   103    95    17
h   104    96    18
i   105    97    19
j   106    98    20
k   107    99    21
l   108   100    22
m   109   101    23
n   110   102    24
o   111   103    25
p   112   104    26
q   113   105    27
r   114   106    28
s   115   107    29
t   116   108    30
u   117   109    31
v   118   110    32
w   119   111    33
x   120   112    34
y   121   113    35
z   122   114    36

Python，25个字节

lambda x:int(max(x),36)+1

定义一个接受字符串的lambda x。查找字符串中最大的数字（按python的默认设置，在数字上方用字母排序），然后转换为以36为底。加1，因为8不在8底。

Haskell，34个字节

f s=length['\t'..maximum s]`mod`39

使用mod(ord(c)-8,39)丹尼斯的想法。

41个字节

g '0'=1
g 'W'=1
g x=1+g(pred x)
g.maximum

45个字节：

(`elemIndex`(['/'..'9']++['a'..'z'])).maximum

输出如Just 3。

切达（Cheddar），34 29 21字节

由于丹尼斯节省了8个字节！！！

s->(s.bytes.max-8)%39

使用小写字母

在线尝试

说明

s -> (      // Input is `s`
  s.bytes    // Returns array of char codes
   .max      // Get maximum item in array
) % 39      // Modulus 39

05AB1E，6个字节

{¤36ö>

大写字母。

说明

{       # sort
 ¤      # take last
  36ö   # convert from base 36 to base 10
     >  # increment

在线尝试

实际上，6个字节

M6²@¿u

在线尝试！

Julia, 22 bytes

!s=(maximum(s)-'')%39

There's a BS character (0x08) between the quotes. Try it online!

JavaScript (ES6), 41 37 bytes

s=>parseInt([...s].sort().pop(),36)+1

Edit: Saved 4 bytes thanks to @edc65.

Haskell, 55 40 bytes

f=(\y->mod(y-8)39).Data.Char.ord.maximum

Thanks @Dennis for his approach. (take that, @xnor ;))

Perl 6: 18 bytes

{:36(.comb.max)+1}

Defines a lambda that takes a single string argument, and returns an integer. It splits the string into characters, finds the "highest" one, converts it to base 36, adds 1.

{(.ords.max-8)%39}

This one uses the modulo approach from Dennis. Same length.

Retina, 28 bytes

O`.
.\B

{2`
$`
}T01`dl`_o
.

Try it online! (The first line enables a linefeed-separated test suite.)

Explanation

O`.

This sorts the characters of the input.

.\B

This removes all characters except the last, so the first two stages find the maximum character.

{2`
$`
}T01`dl`_o

These are two stages which form a loop. The first one duplicates the first character and the second one "decrements" it (replacing e.g. x with w, a with 9 and 1 with 0). The latter stage encounters a zero as the first character, it removes it instead. This is a standard technique for generating a range of characters, given the upper end. Hence, this generates all "digits" from 0 to the maximum digit.

.

Finally, we simply count the number of digits, which gives us the base.

R, 99 89 85 bytes

Look ! Less than 100 bytes !
Look ! 10 bytes off !
Look ! 4 bytes off !

ifelse((m=max(strsplit(scan(,''),"")[[1]]))%in%(l=letters),match(m,l)+10,strtoi(m)+1)

Ungolfed :

l=letters                  #R's built-in vector of lowercase letters

n=scan(what=characters())  #Takes an input from STDIN and convert it to characters

m=max(strsplit(n,"")[[1]]) #Splits the input and takes to max. 
                           #`letters` are considered > to numbers (i.e. a>1)


ifelse(m%in%l,match(m,l)+10,strtoi(m)+1) #If the max is in `letters`,
                                             #outputs the matching position of `m`in `letters` + 10 (because of [0-9]). 
                                             #Else, outputs `m` (as a number) + 1.

As often, this answer makes use of the ifelse function : ifelse(Condition, WhatToDoIfTrue, WhatToDoElse)

PHP, 51 38 bytes

(From Dennis) ^^

<?=(ord(max(str_split($argv[1])))-8)%39;

Other proposal without Dennis' trick

<?=($a=max(str_split($argv[1])))<a?$a+1:ord($a)-86;

• Takes input as argument $argv[1];
• Take max character (using ASCII) values
• If it is a number (inferior to < 'a' ascii value) then output number+1
• Else output ascii value -86 (97 for 'a' in ascii, -11 for 'a' is 11th base-digit)

Octave, 20 bytes

@(a)mod(max(a)-8,39)

Pyke, 6 bytes

Seb36h

Try it here!

Se     -   sorted(input)[-1]
  b36  -  base(^, 36)
     h - ^ + 1

Java 7, 67 61 bytes

int c(char[]i){int m=0;for(int c:i)m=m>c?m:c;return(m-8)%39;}

(m-8)%39 is thanks to @Dennis' amazing answer.

Ungolfed & test code:

Try it here.

class Main{
  static int c(char[] i){
    int m = 0;
    for(int c : i){
      m = m > c
           ? m
           : c;
    }
    return (m-8) % 39;
  }

  public static void main(String[] a){
    System.out.println(c("00000".toCharArray()));
    System.out.println(c("123456".toCharArray()));
    System.out.println(c("ff".toCharArray()));
    System.out.println(c("4815162342".toCharArray()));
    System.out.println(c("42".toCharArray()));
    System.out.println(c("codegolf".toCharArray()));
    System.out.println(c("0123456789abcdefghijklmnopqrstuvwxyz".toCharArray()));
  }
}

Output:

1
7
16
9
5
25
36

C89, 55 53 52 50 bytes

f(s,b)char*s;{return*s?f(s+1,*s>b?*s:b):(b-8)%39;}

-8%39 shamelessly stolen from Dennis

Test

test(const char* input)
{
    printf("%36s -> %u\n", input, f((char*)input,0));
}

main()
{
    test("00000");
    test("123456");
    test("ff");
    test("4815162342");
    test("42");
    test("codegolf");
    test("0123456789abcdefghijklmnopqrstuvwxyz");
}

Output

                               00000 -> 1
                              123456 -> 7
                                  ff -> 16
                          4815162342 -> 9
                                  42 -> 5
                            codegolf -> 25
0123456789abcdefghijklmnopqrstuvwxyz -> 36

Saved 2 bytes thanks to Toby Speight

Saved 2 bytes thanks to Kevin Cruijssen

C, 55 bytes

This answer assumes that the input is in ASCII (or identical in the numbers and letters, e.g. ISO-8859 or UTF-8):

m;f(char*s){for(m=0;*s;++s)m=m>*s?m:*s;return(m-8)%39;}

We simply iterate along the string, remembering the largest value seen, then use the well-known modulo-39 conversion from base-{11..36}.

Test program

int printf(char*,...);
int main(int c,char **v){while(*++v)printf("%s -> ",*v),printf("%d\n",f(*v));}

Test results

00000 -> 1
123456 -> 7
ff -> 16
4815162342 -> 9
42 -> 5
codegolf -> 25
0123456789abcdefghijklmnopqrstuvwxyz -> 36

Mathematica, 34 32 bytes

2 bytes saved thanks to Martin Ender

Max@Mod[ToCharacterCode@#-8,39]&

I decided the different method deserved a new answer.

_{method stolen inspired by Dennis' solution}

Mathematica, 34 32 bytes

saved 2 bytes thanks to Martin Ender

Max@BaseForm[Characters@#,36]+1&

Defines a pure function that takes a string as input.

Splits the input into characters, converts them to base 36 numbers, and returns the maximum +1.

C# REPL, 17 bytes

x=>(x.Max()-8)%39

Just ported @Dennis's answer to C#.

CJam, 10 bytes

Thanks to Martin Ender for saving me a few bytes!

Uses Dennis's formula

q:e>8-i39%

Try it online

CJam, 18 16 btyes

Alternative solution:

A,s'{,97>+q:e>#)

Try it online

A,s'{,97>+       e# Push the string "0123456789abcdefghijklmnopqrstuvwxyz"
          q      e# Get the input
           :e>   e# Find the highest character in the input
              #  e# Find the index of that character in the string
               ) e# Increment

Scala, 25 bytes

print((args(0).max-8)%39)

Run it like:

$ scala whatbase.scala 0123456789abcdefghijklmnopqrstuvwxyz

R, 62 54 bytes

max(match(strsplit(scan(,''),"")[[1]],c(0:9,letters)))

Ungolfed:

max(
  match( # 2: Finds the respective positions of these characters
    strsplit(scan(,''),"")[[1]], # 1: Breaks the input into characters
                                c(0:9,letters)) # 3: In the vector "0123...yz"
                                                )

Update: shaved off 8 bytes due to the redundancy of na.rm=T under the assumption of input validity.

A 39% improvement in size compared to Frédéric's answer. Besides that, it runs a wee bit faster: 0.86 seconds for 100000 replications versus 1.09 seconds for the competing answer. So the one of mine is both smaller and more efficient.

Dyalog APL, 10 bytes

Prompts for uppercase input.

⌈/⍞⍳⍨⎕D,⎕A

⌈/ maximum

⍞ characters of input

⍳⍨ 1-indexed into

⎕D, all digits followed by

⎕A all characters

TryAPL online!

BASH 70

grep -o .|sort -r|head -c1|od -An -tuC|sed s/$/-86/|bc|sed s/-/39-/|bc

Input letters are lowercase.

JavaScript, 57 50 48 bytes

7 bytes saved thnks to @kamaroso97 2 bytes saved thanks to @Neil

n=>Math.max(...[...n].map(a=>parseInt(a,36))+1)

Original answer:

n=>n.split``.map(a=>parseInt(a,36)).sort((a,b)=>b-a)[0]+1

Perl, 30 27 bytes

Includes +1 for -p

Run with the input on STDIN, e.g.

base.pl <<< codegolf

base.pl:

#!/usr/bin/perl -p
\@F[unpack"W*"];$_=@F%39-9

LiveScript, 32 bytes

->1+parseInt (it/'')sort!pop!,36

A port of this answer in my favorite language that compile to JavaScript. If base~number operator worked with variables I could write ->1+36~(it/'')sort!pop! (23 bytes), but it conflicts with the function bind operator :/