题解 | #密码强度等级#
密码强度等级
http://www.nowcoder.com/practice/52d382c2a7164767bca2064c1c9d5361
#小白做题 写这个题的时候感觉奖励逻辑部分有问题，参考答案也有点问题，求大神指教
while True:
try:
mima = input()
count = 0
if len(mima) <= 4:
count += 5
elif 4 < len(mima) < 8:
count += 10
else:
count += 25
#字母判断
xiaoxie = []
daxie = []
for i in mima:
if i.islower():
xiaoxie.append(i)
elif i.isupper():
daxie.append(i)
if len(xiaoxie) + len(daxie) == 0:
count += 0
elif (len(xiaoxie) > 0 and len(daxie) == 0) or (len(daxie) > 0 and len(xiaoxie) == 0):
count += 10
elif len(xiaoxie) > 0 and len(daxie) > 0:
count += 20
#数字判断
shuzi = []
for i in mima:
if i.isdigit():
shuzi.append(i)
if len(shuzi) == 0:
count += 0
elif len(shuzi) == 1:
count += 10
elif len(shuzi) > 1:
count += 20
#符号判断
fuhao = []
fuhao1 = '!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'
for i in mima:
if i in fuhao1:
fuhao.append(i)
if len(fuhao) == 0:
count += 0
elif len(fuhao) == 1:
count += 10
elif len(fuhao) > 1:
count += 25
#奖励判断 该部分有问题 根据公布的答案也不行，python答案中最多浏览那个，他是把+5那个选项放到了前面 所以把28示例测试过去了，
#但是把他的顺序调整后，在28也是不能通过。
if (len(xiaoxie) + len(daxie)) > 0 and len(shuzi) > 0 and len(fuhao) == 0:
count += 2
elif (len(xiaoxie) + len(daxie)) > 0 and len(shuzi) > 0 and len(fuhao) > 0:
count += 3
elif len(xiaoxie) > 0 and len(daxie) > 0 and len(shuzi) > 0 and len(fuhao) > 0:
count += 5
#这一部分是我最早写的，在29示例出现问题，但是29示例我在jupyter通过了，第二个符号是1就通过，这个是示例里面的，
#但是把第二个符号改为小写l就出问题，牛客这个29示例应该是1而不是小写字母l。
# if len(xiaoxie) + len(shuzi) == len(mima) or len(daxie) + len(shuzi) == len(mima) or len(daxie) + len(xiaoxie) + len(shuzi) == len(mima):
# count += 2
# elif len(xiaoxie) + len(shuzi) + len(fuhao) == len(mima) or len(daxie) + len(shuzi) + len(fuhao) == len(mima):
# count += 3
# elif len(xiaoxie) + len(daxie) + len(shuzi) + len(fuhao) == len(mima):
# count += 5
if count >= 90:
print('VERY_SECURE')
elif 80 <= count < 90:
print('SECURE')
elif 70 <= count < 80:
print('VERY_STRONG')
elif 60 <= count < 70:
print('STRONG')
elif 50 <= count < 60:
print('AVERAGE')
elif 25 <= count < 50:
print('WEAK')
elif 0 <= count < 25:
print('VERY_WEAK')
except:
break
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