密码验证合格程序_

Python 3

开挂了的牛牛很讲文明

while True:
    try:
        S = input()
        d, x, n, q = 0, 0, 0, 0
        if len(S) < 8:
            print('NG')
        else:
            for s in S:
                if s.isupper() == True and d != 1:
                    d = 1
                elif s.islower() == True and x != 1:
                    x = 1
                elif s.isdigit() == True and n != 1:
                    n = 1
                elif s.isalnum() == True and s != ' ' and s != '\n' and q != 1:
                    q = 1
            if d + x + n + q < 3:
                print('NG')
            else:
                i  = 3
                pd = False
                while i < len(S) / 2:
                    for j in range(len(S) - i):
                        Sq = S[j:j + i]
                        Sy = S[0:j] + S[j + i:]
                        if Sq in Sy:
                            pd = True
                            break
                    if pd == True:
                        print('NG')
                        break
                    else:
                        i += 1
                if pd == False:
                    print('OK')
    except:
        break

编辑于 2024-01-31 14:45:39 回复(0)

Python 3

TUB-Lis

Q~7T&(4^$OXz725(3!Xl(3+s*

这组数据有子串‘(3’重复了，为什么答案是OK

发表于 2023-10-23 23:51:13 回复(0)

Python 3

牛客897231835号

import sys

for line in sys.stdin:

a = line.rstrip('\n')

a1, a2, a3, a4, a5 = 0, 0, 0, 0, 0

if len(a) >= 8:

if a.count(' ') == 0 or a.count('\n') == 0:

for i in range(0, len(a)):

if a[i] >= '0' and a[i] <= '9':

a1 = 1

elif a[i] >= 'a' and a[i] <= 'z':

a2 = 1

elif a[i] >= 'A' and a[i] <= 'Z':

a3 = 1

else:

a4 = 1

if a1 + a2 + a3 + a4 >= 3:

for i in range(0, len(a) - 3):

for j in range(i + 3, len(a) - 3):

if a[i:i + 3] == a[j:j + 3]:

a5 = 1

else:

continue

if a5 == 1:

print('NG')

else:

print('OK')

else:

print('NG')

else:

print('NG')

else:

print('NG')

发表于 2023-07-21 21:24:26 回复(0)

Python 3

Shiningyoo

import re
import sys
for s in sys.stdin:
    flag=0
    sign=set()
    if len(s)>8:
        flag+=1
    for i in s:
        if i.isdigit():
            sign.add("D")
        elif i.isupper():
            sign.add("U")
        elif i.islower():
            sign.add("L")
        elif not i.isalnum() and not i.isspace() and i!="\n":
            sign.add("S")
    if len(sign)>=3:
        flag+=1
    if not re.findall(r"(.{3,}).*\1",s):
        flag+=1
    print("OK" if flag==3 else "NG")

发表于 2023-06-06 14:34:57 回复(0)

Python 3

ui=f(data)

import sys


def pwd_is_ok():
    for line in sys.stdin:
        a = line.split()
        s = a[0]
        if is_len_8(s) and is_3_kinds_str(s) and is_repeat_str(s):
            print('OK')
        else:
            print('NG')
    
def is_len_8(s):
    if len(s) < 8: return False
    return True

def is_3_kinds_str(s):
    kind_count = 0
    for x in s:
        if ord(x) in range(33, 47)&nbs***bsp;ord(x) in range(58, 64)&nbs***bsp;ord(x) in range(91, 95)&nbs***bsp;ord(x) in range(123, 126):
            kind_count += 1
        if ord(x) in range(48, 57):
            kind_count += 1
        if ord(x) in range(65, 90):
            kind_count += 1
        if ord(x) in range(97, 122):
            kind_count += 1
    if kind_count > 2: return True
    return False


def is_repeat_str(s):
    tmp_list = list()
    for i in range(len(s)):
        if i < len(s) - 1:
            tmp_list.append(s[i:i+3])
    # print(tmp_list, len(tmp_list))
    # print(set(tmp_list), len(set(tmp_list)))
    if len(tmp_list) > len(set(tmp_list)):
        return False
    return True


def main():
    pwd_is_ok()


if __name__ == '__main__':
    main()

发表于 2023-02-16 17:31:16 回复(0)

Python 3

wawaha

n = input()

z = []

q = []

if 1<=len(n)<=100:

if len(n)>8:

x = set(n)

for y in x:

if n.count(y)>=3:

t = n.index(y)

if n[t] == n[t+1] == n[t+2]:

z.append("NG")

break

else:

q.append("OK")

else:

z.append("NG")

else:

z.append("NG")

qq = 0

ww = 0

ee = 0

rr = 0

for p in n:

if p.islower()==True:

qq+=1

elif p.isupper()==True:

ww+=1

elif p.isdigit()==True:

ee+=1

else:

rr+=1

yy = [qq,ww,ee,rr]

if yy.count(0)>=2 or len(z)!=0:

print("NG")

elif yy.count(0)<2 and len(z) == 0:

print("OK")

发表于 2023-02-10 10:51:54 回复(0)

Python 3

朱亚琼

while True:
    try:
        import re

        a = input()
        m, n, p, q = 0, 0, 0, 0
        if len(a) > 8:
            for t in a:
                b1 = re.search("[a-z]", t)
                b2 = re.search("[A-Z]", t)
                b3 = re.search("[0-9]", t)
                b4 = re.search("[^a-zA-Z0-9]", t)
                if b1:
                    m = 1
                elif b2:
                    n = 1
                elif b3:
                    q = 1
                else:
                    p = 1
            if m + n + p + q >= 3:
                i = 0
                k = 0
                for i in range(len(a)):
                    for j in range(i, len(a)):
                        if a[i : j + 3] not in a[j + 3 : len(a)]:
                            i += 1
                        else:
                            k += 1
                if k == 0:
                    print("OK")
                else:
                    print("NG")
            else:
                print("NG")
        else:
            print("NG")
    except:
        break

发表于 2023-01-17 13:59:17 回复(0)

Python 3

牛客246867805号

import sys

passwords =  list(sys.stdin.readlines())

def isValid(s:str) -> bool:
    if len(s) <= 8:
        return False
    # 检查第二个条件
    a,b,c,d = 0,0,0,0
    for ch in s:
        if ord("a") <= ord(ch) <= ord("z"):
            a = 1
        elif ord("A") <= ord(ch) <= ord("Z"):
            b = 1
        elif ord("0") <= ord(ch) <= ord("9"):
            c = 1
        else:
            if ch not in [" ","\n"]:
                d = 1
    if a+b+c+d <3:
        return False
    # 检查第三个条件
    for i in range(len(s)-3+1):
        if len(s.split(s[i:i+3])) >=3:
            return False
    return True

for password in passwords:
    if isValid(password):
        print("OK")
    else:
        print("NG")

发表于 2022-10-09 18:17:12 回复(0)

Python 3

FTY001

import re

while True:
    try:
        raw_strs = input()
        raw_set = "".join(set(raw_strs)).replace(' ', '').replace('\n', '')
        lenght = len(raw_strs)
        mid_lst = []
        for i in range(lenght-2):
            mid_lst.append(raw_strs[i:i+3])
        if len(set(mid_lst)) != len(mid_lst):
            print("NG")
        elif len(raw_strs) <8:
            print("NG")
        else:
            mid_dict = {
                'dig':0
                , "alpha_low":0
                , "other":0
                , "alpha_upper":0
            }
            pattern_low = re.compile(r'([a-z]+)')
            pattern_upper = re.compile(r'([A-Z]+)')
            pattern_num = re.compile(r'(\d+)')
            if re.findall(pattern_low, raw_set):
                mid_dict["alpha_low"] += 1
                raw_set = re.sub(pattern_low, "", raw_set)
            if re.findall(pattern_upper, raw_set):
                mid_dict["alpha_upper"] += 1
                raw_set = re.sub(pattern_upper, "", raw_set)
            if re.findall(pattern_num, raw_set):
                mid_dict["dig"] += 1
                raw_set = re.sub(pattern_num, "", raw_set)
            mid_len = len(raw_set)
            if len(raw_set)>0:
                mid_dict["other"] = mid_len
            if sum(i for i in mid_dict.values())>=3:
                print("OK")
            else:
                print("NG")
                
        
        
        
        
    except:
        break

代码长了点，实在是正则学得太烂了😑

发表于 2022-08-14 19:02:47 回复(0)

Python 3

心静-静心

# 模仿写的，应该是好理解的思路

def code(a):

# str = "~!@#$%^&*()_+-*/<>,.[]\/"
if len(a) <= 8: # 密码必须超过8位
return False

checks = [0, 0, 0, 0] # 必须有大写字母、小写字母、数字、符号四种中的三种
for i in a:
if i.isupper():
checks[0] = 1
elif i.islower():
checks[1] = 1
elif i.isdigit():
checks[2] = 1
else:
checks[3] = 1
if sum(checks) < 3:
return False

for i in range(len(a) - 2): # 不能有长度大于2的包含公共元素的子串重复
if a[i:i+3] in a[i+3:]: # 最少3位开始
return False
return True

while True:
try:
b = input()
if code(b):
print('OK')
else:
print('NG')
except:
break

发表于 2022-08-08 09:44:57 回复(0)

Python 3

牛客179825537号

# 新手的笨办法

while True:

try:
str = input()
# 判断是否大于8位，否则直接返回NG
if len(str) <= 8:
print("NG")
else:
# 判断是否满足大小写字母、数字、特殊符号中的三种，不符合返回NG，符合了再判断是否有子串重复
a = {"xx":0,"dx":0,"sz":0,"fh":0} # 记录大小写字母、数字、符号对应的个数
for i in str:
if i.islower():
a["xx"] += 1
elif i.isupper():
a["dx"] += 1
elif i.isdigit():
a["sz"] += 1
else:
a["fh"] += 1
count = 0 # 记录字典a中，值为0的个数，如果超过两个元素的值为0，则没有满足大小写字母、数字、特殊符号必须满足三种的条件，返回NG
for i in a:
if a[i] == 0:
count += 1
if count >= 2:
print("NG")
else:
# 定义一个列表，将str每三个字符截取一次，保存到列表，再来判断是否有重复的，有则返回NG，没有返回OK
ls = []
for i in range(len(str)-2):
ls.append(str[i:i+3])
# 通过判断列表和转换为集合的长度(集合会去重)，来判断是否有重复
if len(ls) == len(set(ls)):
print("OK")
else:
print("NG")
except:
break

发表于 2022-08-07 23:09:36 回复(0)

Python 3

牛客245788614号

def check_password(password:str):
    #判断字符串长度
    if len(password)<= 8:
        return False
    check_re = [0,0,0,0]
    #使用四个判断来看符不符合“包括大小写字母.数字.其它符号,以上四种至少三种”条件     for letter in password: #一个一个字符遍历
        if letter.isupper(): #判断是否有大写
            check_re[0] = 1
        elif letter.islower():#判断是否有小写
            check_re[1] = 1
        elif letter.isdigit(): #判断是否为数字
            check_re[2] = 1
        else: #判断是否有其他字符
            check_re[3] = 1
        
    if sum(check_re) < 3:
        return False
    
        #使用逐位增加，每次取三位，然后保存在字典dic里，然后每次取出来都进行判断字典dic里面you'w     dic = {}
    str_check = True
    for i in range(len(password) - 2):
        cut_str = password[i:i+3]
        if cut_str in dic.keys():
            #dic[cut_str] += 1
            return False 
        else:
            dic[cut_str] = 1
            
    return True

while True: 
    try:
        password = input()
        if check_password(password):
            print("OK")
        else:
            print("NG")
    except:
        break

发表于 2022-07-28 15:42:44 回复(0)

Python 3

来个offer叭！

import sys
# 判断密码是否合法
def is_legal(str1):
    lst1 = list(str1)
    # pop掉最后一个换行符
    lst1.pop()
    # 长度判断
    if len(lst1) <= 8:
        return 'NG'
    # 公共子串判断
    for i in range(len(lst1)-2):
        # 长度为3的子串, 重复出现则返回NG
        if str1[i:i+3] in str1[i+1:]:
            return 'NG'
    # 四种情况判断
    kind_dict = {'dig': 0, 'upper': 0, 'lower': 0, 'other': 0}
    for i in lst1:
        # 数字
        if i.isdigit():
            kind_dict['dig'] = 1
        # 大写字母
        elif ord('Z') >= ord(i) and ord('A') <= ord(i):
            kind_dict['upper'] = 1
        # 小写字母
        elif ord('z') >= ord(i) and ord('a') <= ord(i):
            kind_dict['lower'] = 1
        # 其他
        else:
            kind_dict['other'] = 1
    # 满足三种以上
    if sum(list(kind_dict.values())) >= 3:
        return 'OK'
    else:
        return 'NG'
# 读取
lines = list(sys.stdin.readlines())
for i in lines:
    print(is_legal(i))

发表于 2022-07-25 20:33:55 回复(0)

Python 3

大海里捞offer

while True:
    try:
        s = input()
        if len(s) <= 8:
            print('NG')
        else:
            a1 = 0
            a2 = 0
            a3 = 0
            a4 = 0
            active = False
            for i in range(len(s)):
                if s[i].isalpha():
                    if s[i].isupper():
                        a1 = 1
                    else:
                        a2 = 1
                elif s[i].isnumeric():
                    a3 = 1
                elif s[i] != ' ' and s[i] != '\n':
                    a4 = 1
                if a1 + a2 + a3 + a4 >= 3:
                    active = True
                    break
            if not active:
                print('NG')
            else:
                se = set()
                for j in range(len(s) - 2):
                    s1 = s[j:j+3]
                    se.add(s1)
#                 print(se)
                if len(se) <= len(s)-3:
                    print('NG')
                else:
                    print('OK')
    except:
        break

发表于 2022-07-25 09:59:13 回复(0)

Python 3

哇哈哈201904101142231

class Test:
    m = "!@#$%^&*()_-+=.?/|"
    l1 = []
    while True:
        # try:
        try:
            s = input()
        except:
            break
        n = "OK"
        l = [False, False, False, False]

        if len(s) < 8:
            n = "NG"
        if n != "NG":
            for i in s:
                if i.islower():
                    l[0] = True
                if i.isupper():
                    l[1] = True
                if i.isnumeric():
                    l[2] = True
                if i in m:
                    l[3] = True
            if l.count(True) < 3:
                n = "NG"
        if n != "NG":
            for i in range(0, len(s) - 2):
                for j in range(i + 2, int(len(s) / 2) + 1):
                    if s.count(s[i:j+1]) > 1:
                        # print(s[i:j+1])
                        n = "NG"
                       
                        break

        l1.append(n)

    for i in l1:
            print(i)

发表于 2022-07-19 23:55:24 回复(0)

Python 3

AlexLyn_00

python3 三条规则

while True:
    try:
        input_string = input()
        rule_1, rule_2, rule_3 = 0,0,0
        # rule_1
        length_string = len(input_string)
        if length_string > 8:
            rule_1 = 1
        # rule_2
        sign_1, sign_2, sign_3, sign_4 = 0,0,0,0
        for i in range(length_string):
            if "a"<=input_string[i]<="z":
                sign_1 = 1
            elif "A"<=input_string[i]<="Z":
                sign_2 = 1
            elif input_string[i].isdigit():
                sign_3 = 1
            else:
                sign_4 = 1
        if sign_1+sign_2+sign_3+sign_4 >= 3:
            rule_2 = 1
        # rule_3
        sign_5 = 0
        for i in range(0, length_string-2):
            temp_string = input_string[i:i+3]
            for j in range(i+3, length_string-2):
                if temp_string == input_string[j:j+3]:
                    sign_5 = 1
                    break
        if sign_5 == 0:
            rule_3 = 1
        else:
            rule_3 = 0
        
        sign = rule_1 + rule_2 + rule_3
        if sign == 3:
            print("OK")
        else:
            print("NG")
        
    except:
        break

发表于 2022-07-13 16:38:10 回复(0)

密码验证合格程序

输入描述:

输出描述:

输入

输出

问题信息

热门推荐

通过挑战的用户

相关试题