题解 | #密码验证合格程序#

# s.isupper(), s.islower(),s.isdigit(), isalnum()
# try, except break
# 难点在于如何判断是否有子串重复

def has_repeat_substring(s):
# 判断方法：双指针，i指向第一个指针，j指向第二个指针。每到一个新的i，j就在i+3往后遍历一遍，把子串记录下来
# 如果当前子串曾经出现过，则return true，都循环完则返回false
    sub_string = []
    for i in range(len(s)):
        for j in range(i+3,len(s)+1): #字串长度大于2
            a = s[i:j]
            if a in sub_string:
                return True
            else:
                sub_string.append(a)
    return False

while True:
    try:
        s = input()
        flag = [False]*5
        if len(s) > 8:
            flag[0] = any(char.isupper() for char in s)
            flag[1] = any(char.islower() for char in s)
            flag[2] = any(char.isdigit() for char in s)
            flag[3] = any(not char.isalnum() for char in s)
            flag[4] = not has_repeat_substring(s)
            if sum(flag[0:4])>= 3 and flag[4]:
                print('OK')
            else:
                print('NG')
        else:
            print('NG')
    except:
        break