因此，您将获得一个以10 为基数的正数。您的工作是反转二进制数字并返回该基数10。

例子：

1 => 1 (1 => 1)
2 => 1 (10 => 01)
3 => 3 (11 => 11)
4 => 1 (100 => 001)
5 => 5 (101 => 101)
6 => 3 (110 => 011)
7 => 7 (111 => 111)
8 => 1 (1000 => 0001)
9 => 9 (1001 => 1001)
10 => 5 (1010 => 0101)

这是一个代码挑战，因此使用最少字节数的解决方案将获胜。

这是OEIS中的A030101。

Python，29个字节

lambda n:int(bin(n)[:1:-1],2)

在线尝试！

这是一个匿名的，未命名的函数，它返回结果。

首先，bin(n)将参数转换为二进制字符串。通常，我们将使用切片符号将其反转[::-1]。这将以-1的步长读取字符串，即向后读取。但是，Python中的二进制字符串以开头0b，因此我们将切片的第二个参数设置为1，告诉Python向后读取终止于索引1的内容，因此不读取索引1和0。

现在我们有了向后的二进制字符串，我们将其传递给int(...)第二个参数为2。这会将字符串读取为以2为底的整数，然后由lambda表达式隐式返回。

Python，29个字节

lambda n:int(bin(n)[:1:-1],2)

在线尝试

JavaScript（ES6），30 28字节

@Arnauld节省了2个字节

f=(n,q)=>n?f(n>>1,q*2|n%2):q

这基本上一次计算一次反向：我们从q = 0开始；而Ñ是积极的，我们乘q 2，服务器关闭的最后一位ñ用n>>1，并将其添加到q用|n%2。当n达到0时，该数字已成功反转，我们返回q。

得益于JS内置的长名称，轻松解决此难题需要44个字节：

n=>+[...n.toString(2),'0b'].reverse().join``

使用递归和字符串，您可以得到一个执行相同操作的32字节解决方案：

f=(n,q='0b')=>n?f(n>>1,q+n%2):+q

Java 8，53 47 46 45字节

• -4字节感谢Titus
• -1字节感谢Kevin Cruijssen

这是一个lambda表达式，其原理与ETH的答案相同（尽管递归在Java中本来就太冗长，所以我们改为循环执行）：

x->{int t=0;for(;x>0;x/=2)t+=t+x%2;return t;}

在线尝试！

可以使用进行分配IntFunction<Integer> f = ...，然后使用进行调用f.apply(num)。展开，展开和注释，如下所示：

x -> { 
    int t = 0;           // Initialize result holder   
    while (x > 0) {      // While there are bits left in input:
        t <<= 1;         //   Add a 0 bit at the end of result
        t += x%2;        //   Set it to the last bit of x
        x >>= 1;         //   Hack off the last bit of x
    }              
    return t;            // Return the final result
};

J，6个字节

|.&.#:

|. 相反

&. 下

#: 基数2

果冻，3个字节

BUḄ

在线尝试！

B   # convert to binary
 U  # reverse
  Ḅ # convert to decimal

Mathematica，19个字节

#~IntegerReverse~2&

迷宫，23字节

?_";:_2
  :   %
 @/2_"!

好吧，这很尴尬...这将返回反向二进制数...感谢@Martin Ender指出了我的错误和ID 10T错误。所以这行不通，我必须找到另一个解决方案。

C，48 44 43 42字节

-1字节归功于gurka，-1字节归功于anatolyg：

r;f(n){for(r=n&1;n/=2;r+=r+n%2);return r;}

以前的44字节解决方案：

r;f(n){r=n&1;while(n/=2)r=2*r+n%2;return r;}

以前的48字节解决方案：

r;f(n){r=0;while(n)r=2*(r+n%2),n/=2;return r/2;}

脱胶和用法：

r;
f(n){
 for(
  r=n&1;
  n/=2;
  r+=r+n%2
 );
 return r;}
}


main() {
#define P(x) printf("%d %d\n",x,f(x))
P(1);
P(2);
P(3);
P(4);
P(5);
P(6);
P(7);
P(8);
P(9);
P(10);
}

Ruby，29个 28个字节

->n{("%b"%n).reverse.to_i 2}

“％b”％n将输入n格式化为二进制字符串，取反，然后转换回数字

用法/测试用例：

m=->n{("%b"%n).reverse.to_i 2}
m[1] #=> 1
m[2] #=> 1
m[3] #=> 3
m[4] #=> 1
m[5] #=> 5
m[6] #=> 3
m[7] #=> 7
m[8] #=> 1
m[9] #=> 9
m[10] #=> 5

05AB1E，3个字节

bRC

在线尝试！

Java（OpenJDK），63个字节

a->a.valueOf(new StringBuffer(a.toString(a,2)).reverse()+"",2);

在线尝试！

感谢-12字节的戳和-8字节的Cyoce！

Perl 6，19个字节

{:2(.base(2).flip)}

Haskell，36个字节

0!b=b
a!b=div a 2!(b+b+mod a 2)
(!0)

与ETHproductions的JavaScript答案相同的算法（和长度！）。

Bash / Unix实用程序，24 23字节

dc -e2i`dc -e2o?p|rev`p

在线尝试！

PHP，33字节

<?=bindec(strrev(decbin($argn)));

转换为base2，反向字符串，转换为十进制。保存到文件并使用管道运行-F。

没有内置函数：

迭代的，41个字节

for(;$n=&$argn;$n>>=1)$r+=$r+$n%2;echo$r;

当输入已设置位时，从输入弹出一个位并将其推入输出。与一起作为管道运行-nR。

递归，52字节

function r($n,$r=0){return$n?r($n>>1,$r*2+$n%2):$r;}

MATL，4个字节

BPXB

在线尝试！

说明

B     % Input a number implicitly. Convert to binary array
P     % Reverse array
XB    % Convert from binary array to number. Display implicitly

Pyth，6个字节

i_.BQ2

测试套件在这里可用。

说明

i_.BQ2
    Q     eval(input())
  .B      convert to binary
 _        reverse
i    2    convert from base 2 to base 10

Japt，5个字节

¢w n2

在线尝试！

Scala, 40 bytes

i=>BigInt(BigInt(i)toString 2 reverse,2)

Usage:

val f:(Int=>Any)=i=>BigInt(BigInt(i)toString 2 reverse,2)
f(10) //returns 5

Explanation:

i =>          // create an anonymous function with a parameter i
  BigInt(       //return a BigInt contructed from
    BigInt(i)     //i converted to a BigInt
    toString 2    //converted to a binary string
    reverse       //revered
    ,           
    2             //interpreted as a binary string
  )

Mathematica, 38 bytes

#+##&~Fold~Reverse[#~IntegerDigits~2]&

Groovy, 46 bytes

{0.parseInt(0.toBinaryString(it).reverse(),2)}

CJam, 8 bytes

ri2bW%2b

Try it online!

Explanation

ri          e# Read integer
  2b        e# Convert to binary array
    W%      e# Reverse array
      2b    e# Convert from binary array to number. Implicitly display

Batch, 62 bytes

@set/an=%1/2,r=%2+%1%%2
@if %n% gtr 0 %0 %n% %r%*2
@echo %r%

Explanation: On the first pass, %1 contains the input parameter while %2 is empty. We therefore evaluate n as half of %1 and r as +%1 modulo 2 (the % operator has to be doubled to quote it). If n is not zero, we then call ourselves tail recursively passing in n and an expression that gets evaluated on the next pass effectively doubling r each time.

C#, 98 bytes

using System.Linq;using b=System.Convert;a=>b.ToInt64(string.Concat(b.ToString(a,2).Reverse()),2);

R, 55 bytes

sum(2^((length(y<-rev(miscFuncs::bin(scan()))):1)-1)*y)

Reads input from stdin and consequently uses the bin function from the miscFuncs package to convert from decimal to a binary vector.

Pushy, 19 bytes

No builtin base conversion!

$&2%v2/;FL:vK2*;OS#

Try it online!

Pushy has two stacks, and this answer makes use of this extensively.

There are two parts two this program. First, $&2%v2/;F, converts the number to its reverse binary representation:

            \ Implicit: Input is an integer on main stack.
$      ;    \ While i != 0:
 &2%v       \   Put i % 2 on auxiliary stack
     2/     \   i = i // 2 (integer division)
        F   \ Swap stacks (so result is on main stack)

Given the example 10, the stacks would appear as following on each iteration:

1: [10]
2: []

1: [5]
2: [0]

1: [2]
2: [0, 1]

1: [1]
2: [0, 1, 0]

1: [0]
2: [0, 1, 0, 1]

We can see that after the final iteration, 0, 1, 0, 1 has been created on the second stack - the reverse binary digits of 10, 0b1010.

The second part of the code, L:vK2*;OS#, is taken from my previous answer which converts binary to decimal. Using the method decsribed and explained in that answer, it converts the binary digits on the stack into a base 10 integer, and prints the result.

k, 18 bytes

{2/:|X@&|\X:0b\:x}

Example:

k){2/:|X@&|\X:0b\:x}6
3

C#, 167 bytes

 for(int i = 1; i <= 10; i++)
 {
 var bytes= Convert.ToString(i, 2);
 var value= Convert.ToInt32(byteValue.Reverse()); 
 console.WriteLine(value);
}

Explanation:

Here I will iterate n values and each time iterated integer value is convert to byte value then reverse that byte value and that byte value is converted to integer value.

c/c++ 136 bytes

uint8_t f(uint8_t n){int s=8*sizeof(n)-ceil(log2(n));n=(n&240)>>4|(n&15)<<4;n=(n&204)>>2|(n&51)<<2;n=(n&172)>>1|(n&85)<<1;return(n>>s);}

It's not going to win, but I wanted to take a different approach in c/c++ 120 bytes in the function

#include <math.h>
#include <stdio.h>
#include <stdint.h>

uint8_t f(uint8_t n){
    int s=8*sizeof(n)-ceil(log2(n));
    n=(n&240)>>4|(n&15)<<4;
    n=(n&204)>>2|(n&51)<<2;
    n=(n&172)>>1|(n&85)<<1;
    return (n>>s);
}

int main(){
    printf("%u\n",f(6));
    return 0;
}

To elaborate on what I am doing, I used the log function to determine the number of bits utilized by the input. Than a series of three bit shifts left/right, inside/outside, even/odd which flips the entire integer. Finally a bit shift to shift the number back to the right. Using decimals for bit shifts instead of hex is a pain but it saved a few bytes.