題目描述
按照指定規則對輸入的字串進行處理。

詳細描述：

將輸入的兩個字串合併。

對合並後的字串進行排序，要求為：下標為奇數的字元和下標為偶數的字元分別從小到大排序。這裡的下標意思是字元在字串中的位置。

對排序後的字串進行操作，如果字元為‘0’——‘9’或者‘A’——‘F’或者‘a’——‘f’，則對他們所代表的16進位制的數進行BIT倒序的操作，並轉換為相應的大寫字元。如字元為‘4’，為0100b，則翻轉後為0010b，也就是2。轉換後的字元為‘2’； 如字元為‘7’，為0111b，則翻轉後為1110b，也就是e。轉換後的字元為大寫‘E’。

舉例：輸入str1為”dec”，str2為”fab”，合併為“decfab”，分別對“dca”和“efb”進行排序，排序後為“abcedf”，轉換後為“5D37BF”

輸入描述:

輸入兩個字串

輸出描述:

輸出轉化後的結果

示例1
輸入

dec fab

輸出

5D37BF

C++：

#include<iostream>
#include<string>
#include"math.h"
#include<algorithm>
using namespace std;

string presort(string);
char process(char a);
int main()
{
    string s, s1, s2;
    while
 
 (cin >> s1)
    {
        cin.get();
        cin >> s2;
        s = presort(s1+s2);
        string output = "";
        for (int i = 0; i < s.size(); i++)
        {
            if (s[i] >= '0' && s[i] <= '9' || s[i] >= 'A' && s[i] <= 'F' || s[i] >= 'a' && s[i] <= 'f'
 
)
            {
                s[i] = process(s[i]);
            }
            output += s[i];
        }
        cout << output << endl;
    }
    return 0;
}
string presort(string s)
{
    string s1, s2;
    s1 = ""; s2 = "";
    string result = "";
    for (int i = 0; i < s.size(); i++)
    {
        if (i % 2 == 0)
            s1 += s[i];
        else
            s2 += s[i];
    }
    sort(s1.begin(), s1.end());
    sort(s2.begin(), s2.end());
    int n = s.size();
    if (n % 2 == 0)
    {
        for (int i = 0; i < s1.size(); i++)
        {
            result += s1[i];
            result += s2[i];
        }

    }
    else
    {
        for (int i = 0; i < s2.size(); i++)
        {
            result += s1[i];
            result += s2[i];
        }
        result += s1[s2.size()];
    }
    return result;
}
int ch2int(char a)
{
    int b = 0;
    if (a >= '0' && a <= '9')
        b = a - '0';
    else if (a >= 'a'&&a <= 'f')
        b = a - 'a' + 10;
    else if (a >= 'A'&&a <= 'F')
        b = a - 'A' + 10;
    return b;
}
int reverse_ten2two(int a)
{
    int ch[4] = { 0 };
    int sum = 0;
    for (int i = 0; i < 4; i++)
    {
        ch[i] = a % 2;
        a /= 2;
        sum += ch[i] * pow(2, 3 - i);
    }
    return sum;
}
char int2ch(int a)
{
    char b;
    if (a >= 0 && a <= 9)
        b = a + '0';
    else if (a >= 10 &&a <= 15)
        b = a - 10 + 'A';
    return b;
}
char process(char a)
{
    int i = ch2int(a);
    int ri = reverse_ten2two(i);
    char b = int2ch(ri);
    return b;
}

Python：

while True:
    try:
        def preinput(s):
            s = s.replace(' ','')
            a = ''; b = '';
            for i,v in enumerate(s): #這邊可以直接a = s[::2],b=a[1::2]
                if i % 2 == 0:
                    a = a + v
                else:
                    b = b + v
            a = ''.join(sorted(a))
            b = ''.join(sorted(b))
            result = ''
            for i in range(len(b)):
                result = result + a[i]
                result = result + b[i]
            if(len(a) - len(b)):
                result += a[len(b)]
            return result

        def transform(a):
            if '0' <= a <= '9':
                a = int(a)
            elif 'a' <= a <= 'f':  #這裡可以直接對a.upper()進行判斷，減少一次條件判斷
                a = ord(a) - ord('a') + 10
            elif 'A' <= a <= 'F':
                a = ord(a) - ord('A') + 10
            a = bin(a).lstrip('0b')
            if(len(a) < 4):  #這邊可以直接s.rjust(4,'0')
                a = (4 - len(a)) * '0' + a
            a = a[::-1]
            sum = 0
            for i,v in enumerate(a):
                sum = sum + int(v) * pow(2, 3 - i)
            a = sum
            if a > 9:  
                a = chr(a - 10 + ord('A'))
            else:
                a = '%s'%a
            return a
        s = input()
        a = preinput(s)
        s = ""
        for i,v in enumerate(a):
            if '0' <= v <= '9' or 'a' <= v <= 'f' or 'A' <= v <= 'F':
                s = s + transform(v)
            else:
                s = s + v
        print(s)
    except:
        pass