给定一个正整数n，输出最小的底数b >= 2，其中n底数b中没有前导零的表示不包含0。您可能会假设b <= 256所有输入都是这样。

测试用例

1 -> 2 (1)
2 -> 3 (2)
3 -> 2 (11)
4 -> 3 (11)
5 -> 3 (12)
6 -> 4 (12)
7 -> 2 (111)
10 -> 4 (22)
17 -> 3 (122)
20 -> 6 (32)
50 -> 3 (1212)
100 -> 6 (244)
777 -> 6 (3333)
999 -> 4 (33213)
1000 -> 6 (4344)
1179360 -> 23 ([12, 9, 21, 4, 4])
232792560 -> 23 ([15, 12, 2, 20, 3, 13, 1])
2329089562800 -> 31 ([20, 3, 18, 2, 24, 9, 20, 22, 2])
69720375229712477164533808935312303556800 -> 101 ([37, 17, 10, 60, 39, 32, 21, 87, 80, 71, 82, 14, 68, 99, 95, 4, 53, 44, 10, 72, 5])
8337245403447921335829504375888192675135162254454825924977726845769444687965016467695833282339504042669808000 -> 256 ([128, 153, 236, 224, 97, 21, 177, 119, 159, 45, 133, 161, 113, 172, 138, 130, 229, 183, 58, 35, 99, 184, 186, 197, 207, 20, 183, 191, 181, 250, 130, 153, 230, 61, 136, 142, 35, 54, 199, 213, 170, 214, 139, 202, 140, 3])

Pyth，6个字节

f*FjQT

验证所有测试用例。

怎么运行的

f * FjQT〜完整程序。

f〜条件为真的第一个正整数。
   jQT〜将输入转换为当前元素的基数。
 * F〜产品。如果列表包含0，则为0，否则严格为正。
          0->虚假；> 0-> Truthy。
        〜隐式输出结果。

尽管Pyth的f运算1, 2, 3, 4, ...从1开始，但Pyth会将以1为底的数字（一元）视为一堆零，因此忽略以1为底的数字。

C， 52  50字节

i,k;f(n){for(i=2,k=n;k;)k=k%i++?k/--i:n;return i;}

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C（gcc）， 47个  45字节

i,k;f(n){for(i=2,k=n;k;)k=k%i++?k/--i:n;n=i;}

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感谢@Nevay对@Kevin Cruijssen的回答，节省了两个字节！

Haskell，56 52 48字节

b#n=n<1||mod n b>0&&b#div n b
f n=until(#n)(+1)2

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很基本，但是想不出什么好办法来缩短它

编辑：感谢Laikoni为我节省了4个字节！不知道为什么我从未想过!!0。我可能应该尝试删除这些括号，但是当您尝试一起使用||和时，我对某些奇怪的错误有模糊的记忆&&。也许我将其与相等运算符混淆了。

编辑2：感谢@Lynn剃了另外4个字节！不知道我以前从未知道until过。

Wolfram语言（Mathematica），33个字节

2//.i_/;DigitCount[#,i,0]>0:>i+1&

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外壳，7个字节

→V▼MBtN

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说明

→V▼MBtN
     tN    list of natural numbers starting from 2
   MB      convert the (implicit) input to each of those bases
 V▼        find the (1-based) index of the first result where the minimum digit is truthy
→          add 1 to this index

Python 2，57字节

n=x=input()
b=2
while x:z=x%b<1;b+=z;x=[x/b,n][z]
print b

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这比递归函数短一个字节：

f=lambda n,b=1,x=1:b*(x<1)or f(n,b+(x%b<1),[x/b,n][x%b<1])

果冻，7个字节

2b@Ạ¥1#

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05AB1E，6个字节

-4个字节感谢Adnan

1µNвPĀ

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外壳，9个字节

←foΠ`B⁰tN

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说明

            -- input N
        tN  -- tail of [1..] == [2..]
←f(    )    -- filter with the following:
    `B⁰     --   convert N to that base
   Π        --   product (0 if it contains 0)
←           -- only keep first element

Java 8，61 56 54字节

n->{int b=2,t=n;for(;t>0;)t=t%b++<1?n:t/--b;return b;}

在这里尝试。

说明：

n->{            // Method with integer as both parameter and return-type
  int b=2,      //  Base-integer, starting at 2
      t=n;      //  Temp-integer, copy of the input
  for(;t>0;)    //  Loop as long as `t` is not 0
    t=t%b++<1?  //   If `t` is divisible by the base `b`
                //   (and increase the base `b` by 1 afterwards with `b++`)
       n        //    Set `t` to the input `n`
      :         //   Else:
       t/--b;   //    Divide `t` by the `b-1`
                //    (by decreasing the base `b` by 1 first with `--b`)
                //  End of loop (implicit / single-line body)
  return b;     //  Return the resulting base
}               // End of method

我觉得可以使用算术方法进行练习。它的确可以，用的端口@Steadybox” C的答案，然后由于2个字节golfed @Nevay。

旧（61字节）答案：

n->{int b=1;for(;n.toString(n,++b).contains("0"););return b;}

在这里尝试。

说明：

n->{         // Method with Integer as both parameter and return-type
  int b=1;   //  Base-integer, starting at 1
  for(;n.toString(n,++b).contains("0"););
             //  Loop as long as the input in base-`b` does contain a 0,
             //  after we've first increased `b` by 1 before every iteration with `++b`
  return b;  //  Return the resulting base
}            // End of method

Japt，8字节

@ìX e}a2

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说明

@    }a2

返回第一个数字（X）以通过该函数，从2

ìX

将输入数字转换为基数数组X。

e

检查所有数字是否真实。

JavaScript（ES6），43 41 37字节

n=>(g=x=>x?g(x%b++?x/--b|0:n):b)(b=1)

测试用例

let f =

n=>(g=x=>x?g(x%b++?x/--b|0:n):b)(b=1)

console.log(f(1   )) // 2 (1)
console.log(f(2   )) // 3 (2)
console.log(f(3   )) // 2 (11)
console.log(f(4   )) // 3 (11)
console.log(f(5   )) // 3 (12)
console.log(f(6   )) // 4 (12)
console.log(f(7   )) // 2 (111)
console.log(f(10  )) // 4 (22)
console.log(f(17  )) // 3 (122)
console.log(f(20  )) // 6 (32)
console.log(f(50  )) // 3 (1212)
console.log(f(100 )) // 6 (244)
console.log(f(777 )) // 6 (3333)
console.log(f(999 )) // 4 (33213)
console.log(f(1000)) // 6 (4344)

Brachylog，11个字节

;.≜ḃ₎¬∋0∧1¬

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说明

;.≜ḃ₎           The Input represented in base Output…
     ¬∋0        …contains no 0
        ∧1¬     And Output ≠ 1

Python 2，57字节

n=m=input()
b=2
while m:c=m%b<1;b+=c;m=(m/b,n)[c]
print b

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-1感谢Felipe Nardi Batista。
-2感谢Lynn（现在这是她的解决方案的伪装：D）

APL（Dyalog），20 19 字节

1+⍣{~0∊⍺⊥⍣¯1⊢n}≢n←⎕

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和往常一样，感谢@Adám在聊天中提供帮助，并使代码在TIO中起作用。另外，保存1个字节。

这tradfn（TRAD itional ˚F unctio Ñ）体。要使用它，您需要为其分配一个名称（在TIO的标题字段中），将其括在∇s中（名称前一个，在TIO的页脚字段中一个），然后使用其名称进行调用。由于它使用四边形（⎕）来接受用户的输入，因此称为，f \n input而不是通常的f input

怎么样？

1+⍣{~0∊⍺⊥⍣¯1⊢n}≢n←⎕  ⍝ Main function.
                  n←⎕ ⍝ Assigns the input to the variable n
1+⍣{           }≢     ⍝ Starting with 1, add 1 until the expression in braces is truthy
    ~0∊               ⍝ returns falsy if 0 "is in"
        ⊥             ⍝ convert
            ⊢n        ⍝ the input
         ⍣¯1          ⍝ to base
       ⍺              ⍝ left argument (which starts at 1 and increments by 1)

然后该函数返回结果基数。

Proton, 40 bytes

x=>filter(y=>all(digits(y,x)),2..x+3)[0]

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R, 79 71 66 63 65 bytes

function(n){while(!{T=T+1;all(n%/%T^(0:floor(log(n,T)))%%T)})T
T}

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This answer is based on Giuseppe's re-arrangement in one single loop.

Saved 8 bytes thanks to JDL, and 6 bytes thanks to Giuseppe.

MATL, 13 12 bytes

`G@Q_YAA~}@Q

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-1 byte thanks to Luis Mendo. This program does not handle testcases bigger than 2^53 (flintmax, the maximum consecutive integer representable by a floating point type), as the default datatype is double in MATL. However, it should be able to find any arbitrary zeroless base below that number.

`            % Do while
 G           %  Push input
  @ _        %  Outputs the iteration number, negate.
     YA      %  Convert input to base given by the iteration number, the negative number is to instruct MATL we want an arbitrary high base with a integer vector rather than the default character vector we know from hexadecimal
       A~    %  If they're not all ones, repeat
         }   % But if they are equal, we finally
          @  %  Push the last base
   Q       Q %  As base 1 makes no sense, to prevent MATL from errors we always increase the iteration number by one.

PHP, 59+1 bytes

using builtins, max base 36:

for($b=1;strpos(_.base_convert($argn,10,++$b),48););echo$b;

no builtins, 6360+1 bytes, any base:

for($n=$b=1;$n&&++$b;)for($n=$argn;$n%$b;$n=$n/$b|0);echo$b;

Run as pipe with -nR or try them online.

Actually,  12  11 bytes

╗1⌠╜¡m'0<⌡╓

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Uses this consensus. Thanks to Mego for byte-saving help in chat.

J, 26 bytes

]>:@]^:(0 e.]#.inv[)^:_ 2:

Would love to know if this can be improved.

The main verb is a the dyadic phrase:

>:@]^:(0 e.]#.inv[)^:_

which is given the input on the left and the constant 2 on the right. That main verb phrase then uses J's Do..While construct, incrementing the right y argument as long as 0 is an element of e. the original argument in base y.

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Lua, 77 76 bytes

b=2n=...N=n
while~~N>0 do
if N%b<1then
b=b+1N=n
else
N=N//b
end
end
print(b)

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Milky Way, 38 bytes

^^'%{255£2+>:>R&{~^?{_>:<;m_+¡}}^^^}

usage: ./mw base.mwg -i 3

Explanation

code                                 explanation                    stack layout

^^                                   clear the preinitialized stack []
  '                                  push the input                 [input]
   %{                              } for loop
     255£                             push next value from 0..254   [input, base-2]
         2+                           add 2 to the get the base     [input, base]
           >                          rotate stack right            [base, input]
            :                         duplicate ToS                 [base, input, input]
             >                        rotate stack right            [input, base, input]
              R                       push 1                        [input, base, input, 1]
               &{~             }      while ToS (=remainder) is true ...
                  ^                    pop ToS                      [input, base, number]
                   ?{         }        if ToS (=quotient) ...
                     _>:<;              modify stack                [input, base, number, base]
                           m            divmod                      [input, base, quotient, remainder]
                           _+¡         else: output ToS (0) + SoS and exit
                                ^^^   pop everything but the input.

I'm sure this can be shortened using a while-loop instead of a for loop, but I couldn't get it to work.

Stacked, 23 bytes

{!2[1+][n\tb all]until}

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This increments ([1+]) J starting from two (2) while the base_J representation of the input has no zeroes (all and until).

Perl 5, 52 + 2 (-pa) = 54 bytes

$\=1;while($_&&=$F[0]){$\++;$_=int$_/$\while$_%$\}}{

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