import re def kind_is_ok(strs): # 判断种类是否大于等于3种 kind = [0, 0, 0, 0] for s in strs: m = kind.count(0) if m <= 1: # 0的数量小于等于1,说明已经有三种了 return True else: if s.isdecimal(): kind[0] += 1 continue if re.match("[A-Z]", s): kind[1] += 1 continue if re.match("[a-z]", s): kind[2] += 1 co...