题解 | #字符串加解密#

需要往后移动的都将字母转为assi码再加1，然后再转回对应字母 如果是a和z就单独判断。是否需要大小写转化，对应字母assi码大于97就转大写，否则就转小写

def jiami(string_):

d = ''
for i in string_:
    if i.isalpha():
        if i == 'z':
            c = 'A'
        elif i == 'Z':
            c = 'a'
        else:
            c = ord(i) + 1
            if c > 97:  # a 的assi是97
                c = chr(c).upper()
            else:
                c = chr(c).lower()
        d += c
    elif i.isdigit():
        if i == "9":
            c = "0"
        else:
            c = int(i) + 1
        d += str(c)
return d

def jiemi(_string):

d = ''
for i in _string:
    if i.isalpha():
        if i == 'a':
            c = 'Z'
        elif i == 'A':
            c = 'z'
        else:
            c = ord(i) - 1
            if c >= 97:
                c = chr(c).upper()
            else:
                c = chr(c).lower()
        d += c
    elif i.isdigit():
        if i == "0":
            c = "9"
        else:
            c = int(i) - 1
        d += str(c)
return d

if name == 'main':

while True:
    try:
        a = input()
        b = input()
        print(jiami(a))
        print(jiemi(b))
    except:
        break