题解 | #简单密码#

想法大概都差不多，就是在循环里面加一个break来节约一些资源

chars = ['abc','def','ghi','jkl','mno','pqrs','tuv','wxyz']
string = input()
newpswd = ''
for i in string:
    if ord(i) >=ord('0') and ord(i)<=ord('9'):
        newpswd += i
    elif ord(i)>=ord('a') and ord(i)<=ord('z'):
        for w in chars:
            if i in w:
                newpswd += str(chars.index(w)+2)
                break
    elif ord(i)>=ord('A') and ord(i)<=ord('Y'):
        newpswd += chr(ord(i)+ord('a')-ord('A')+1)
    elif i =='Z':newpswd += 'a'
print(newpswd)