问题求助
'''请用已学过的知识编写程序,找出小甲鱼藏在下边这个长字符串中的密码,密码的埋藏点符合以下规律:Y5d|MAp
Y5d|MAp
a) 每位密码为单个小写字母
b) 每位密码的左右两边均有且只有三个大写字母
'''
password = ""
with open("string2.txt","r",encoding="utf-8") as f:
content = f.read()
length = len(content)
for i in range(length):
if content == "\n":
continue
if content.islower() and content.isupper() and len(content) == 3 \
and content.isupper() and len(content) == 3:
if content.islower() and content.islower():
password +=content
print(password)
content 如下
ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
lqNsYCyaQXBzrFUbkAUAWAKrDgDtAlGMBqWQhpEwquZqWZJpslUfMllCwWptqINjrOBTLuPzwvXNbLCx
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qKWGc
和答案差了一点,不知道是哪里没考虑全 因为content中有换行,得先把换行去掉:
'''
请用已学过的知识编写程序,找出小甲鱼藏在下边这个长字符串中的密码,密码的埋藏点符合以下规律:Y5d|MAp
Y5d|MAp
a) 每位密码为单个小写字母
b) 每位密码的左右两边均有且只有三个大写字母
'''
password = ""
with open("string2.txt","r",encoding="utf-8") as f:
content = f.read().replace('\n', '') # 注意这里
length = len(content)
for i in range(length):
if content == "\n":
continue
if content.islower() and content.isupper() and len(content) == 3 \
and content.isupper() and len(content) == 3:
if content.islower() and content.islower():
password +=content
print(password)
isdkz 发表于 2022-4-13 21:02
因为content中有换行,得先把换行去掉:
原本代码的12行不是将已经考虑了吗 代码小白liu 发表于 2022-4-13 21:07
原本代码的12行不是将已经考虑了吗
这样子考虑不太合理,因为换行影响的不是对当前字符的判断,
所以你有没有跳过没有区别,因为换行也不满足后面的 content.islower(),
但是它会影响对当前字符的前 3 个字符和后 3 个字符的判断,换行并不是字母,但是却可以忽略,
你不去掉换行的话,当前字符的前三个字符或后三个字符包括换行显然是不合理的
页:
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