[已解决]第20讲 课后题

小时候可帅了

我看了答案 感觉这道题我的方法能解出来，但是遇到了点问题，总是报错
str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''

str2 = 'abcdefghijklmnokprstuvwxyz'
str3 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

list1= []
x = 4
for each1 in str1:
    for each2 in str2:
        for each3 in str3:
            a = len(each1)
            if str2 not in list1:
                if each1 in str2 :
                    if each1[a-1-x-1] in each3 and each1[a-1-x-2] in each3 and each1[a-1-x-3]in each3 and each1[a-1-x+1]in each3 and each1[a-1-x+2]in each3 and each[a-1-x+3]in each3:
                        if each in str2:
                            print(each[a])
                        else:
                            x += 1
            list1.append(each1)
                                

最佳答案

月排行榜 / 总排行榜

jackz007

2020-3-5 20:10:31

         请关注这个条件：
        b) 每位密码的左右两边均有且只有三个大写字母
         这个条件告诉我们，密码字符的两边如果有第四个字符，那就一定是小写字母。
         试试下面的代码吧：

str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''

s = 'a' + '' . join(str1 . split('\n')) + 'a'
p = ''
for k in range(4 , len(s) - 4):
    if s[k - 4] . islower() and s[k - 3 : k] . isupper() and s[k] . islower() and s[k + 1 : k + 4] . isupper() and s[k + 4] . islower():
        p += s[k]
print(p)

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小时候可帅了

题目是这样的：
1. 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Powered by bbs.fishc.com
     b) 每位密码的左右两边均有且只有三个大写字母qH{]*ZR
Suj|vEA'h<^1eF]7i*(tbZU+$H yx

jackz007

         请关注这个条件：
        b) 每位密码的左右两边均有且只有三个大写字母
         这个条件告诉我们，密码字符的两边如果有第四个字符，那就一定是小写字母。
         试试下面的代码吧：

str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''

s = 'a' + '' . join(str1 . split('\n')) + 'a'
p = ''
for k in range(4 , len(s) - 4):
    if s[k - 4] . islower() and s[k - 3 : k] . isupper() and s[k] . islower() and s[k + 1 : k + 4] . isupper() and s[k + 4] . islower():
        p += s[k]
print(p)

小时候可帅了

jackz007 发表于 2020-3-5 20:10
请关注这个条件：
        b) 每位密码的左右两边均有且只有三个大写字母
         这个条件告 ...

大哥 我看了您的代码 不是很懂，能否讲解一下呢

jackz007

         应该是这一句看不懂吧

s = 'a' + '' . join(str1 . split('\n')) + 'a'

         因为 str1 是三引号跨越多行的字符串，在每一行的末尾，都是以一个换行符 '\n' 来结束的，这个字符肯定不是我们希望存在的，必须加以去掉。

'' . join(str1 . split('\n'))

         这一句就是根据换行字符 '\n'，对 str1 进行分段，得到由每行字符串为元素构成的列表，然后，通过字符串方法 '' . join() 把这些字符串元素拼接在一起，成为一个统一的字符串。

s = 'a' + '' . join(str1 . split('\n')) + 'a'

        在密码字符两侧有且仅有 3 个大写字符的意思，是在字符串的首、尾只需要判断密码字符两边的 3 个字符，而在字符串内部，是需要判断 4 个字符的。
        完成新字符串拼接后，在字符串的首尾各添加一个小写的字母 'a' ，这样，就可以统一按密码字符两边各 4 个字符进行判断了，字符串的首尾就不再需要特殊对待了。从而，实质性简化了代码。

小时候可帅了

jackz007 发表于 2020-3-5 23:21
应该是这一句看不懂吧

         因为 str1 是三引号跨越多行的字符串，在每一行的末尾，都是以 ...

这个方法好，感觉自己还要努力学