[已解决]问题求助

代码小白liu

'''
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[i] == "\n":
            continue
        if content[i].islower() and content[i+1:i+4].isupper() and len(content[i+1:i+4]) == 3 \
                and content[i-3:i].isupper() and len(content[i-3:i]) == 3:

            if content[i+4].islower() and content[i-4].islower():
       
                password +=content[i]

    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

最佳答案

月排行榜 / 总排行榜

isdkz

2022-4-13 21:02:10

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[i] == "\n":
            continue
        if content[i].islower() and content[i+1:i+4].isupper() and len(content[i+1:i+4]) == 3 \
                and content[i-3:i].isupper() and len(content[i-3:i]) == 3:

            if content[i+4].islower() and content[i-4].islower():
       
                password +=content[i]

    print(password)

跳转到最佳答案楼层

isdkz

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[i] == "\n":
            continue
        if content[i].islower() and content[i+1:i+4].isupper() and len(content[i+1:i+4]) == 3 \
                and content[i-3:i].isupper() and len(content[i-3:i]) == 3:

            if content[i+4].islower() and content[i-4].islower():
       
                password +=content[i]

    print(password)

代码小白liu

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

isdkz

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[i].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