题解 | #字符串合并处理#

while True:
    try:
        str1, str2 = input().split()
        combine = str1 + str2
        
        # 排序
        combinels = []
        even = []
        odd = []
        nums = len(combine)
        
        # 偶数排序
        for i in range(0, nums, 2):
            even.append(combine[i])
        even = sorted(even)
        
        # 奇数排序
        for i in range(1, nums, 2):
            odd.append(combine[i])
        odd = sorted(odd)
        
        # 得到排序后的列表
        for i in range(0, nums):
            if i%2 == 0:
                combinels.append(even[i//2])
            else:
                combinels.append(odd[i//2])
            
        temp = combinels
        chars = ""
        # 进制转换(只对A-F和0-9)
        pattern = list("ABCDEFabcdef0123456789")
        for i in temp:
            if i in pattern:
                char_16to10 = int(i,16)

                char_10to2 = list(str(bin(char_16to10)[2:]))# 在这里,可能有的数字只有一位(比如1),我们需要添加前面的项000
                char_10to2.reverse() # 翻转过后就是在后面添加000
                if len(char_10to2)<4:
                    char_10to2.extend(["0"]*(4-len(char_10to2)))

                char_10to2 = "".join(char_10to2)
                char_2to16 = hex(int(char_10to2,2))[2:]

                chars += char_2to16.upper() # A-F需要转换为大写
                
            else:
                chars += i
        print("".join(chars))

    except:
        break