Newbie_XZW 发表于 2020-11-27 17:03:23

大佬们,帮忙修改下代码

代码在字符串下面,麻烦向下拖一下

str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''

list1 = list(str1)
ler1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
answer = []
def subgroup(sample,list):
    x = []
    for i in range(sample, len(list)):
      x.append(list)
    return x

soup = subgroup(7,list1)

for each in soup:
    if each in ler1:
      if each in ler1:
            if each in ler1:
                if each in ler1:
                  if each in ler1:
                        if each in ler1:
                            if each not in ler1:
                              if each != '\n':
                                    print(each,end='')

大佬们,帮忙看看代码哪有问题

正确答案:ilovefishc

代码答案:llpuiallouovvhezfbilbjiirmsqqxthtgcnkdljoasncexw

Newbie_XZW 发表于 2020-11-28 17:30:57

帅哥美女们,快来帮忙看一下

suchocolate 发表于 2020-11-28 18:18:18

题目是啥?

Newbie_XZW 发表于 2020-11-28 20:50:24

suchocolate 发表于 2020-11-28 18:18
题目是啥?

第020讲:函数:内嵌函数和闭包 | 课后测试题及答案

动动手 最后一题

题目: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:

                  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

suchocolate 发表于 2020-11-28 21:05:02

本帖最后由 suchocolate 于 2020-11-28 22:04 编辑

import re
result = re.findall(r'{3}(){3}', str1)
print(result)

昨非 发表于 2020-11-29 00:44:43

suchocolate 发表于 2020-11-28 21:05


啊这,人家问你拧螺丝,你这直接造火箭啊{:10_256:}

太阳总会升起 发表于 2020-11-29 11:47:53

你可以说一下你的思路吗,for循环里面看不太懂
enmmmm

太阳总会升起 发表于 2020-11-29 12:28:32

错误的地方:
判断不全

题目要求小写字母前三位和后三位有且仅有三个大写字母,也就是说小写字母的前面第四位和后面第四位是小写,所以要每9位分开组成大列表来判断前后第四位是否为小写

你的源代码应该是的字符串每7位分开组成一个大列表,然后判断第1, 2, 3, 5, 6, 7位是大写,第4位不是大写也不是换行就打印第三位

下面是你的源代码修改的
list1 = list(str1)
ler1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

def subgroup(sample, list):
    x = []
    for i in range(sample, len(list)):
      x.append(list)
    return x


soup = subgroup(9, list1)

for each in soup:
    if len(each) == 9:
      if each not in ler1:
            if each in ler1:
                if each in ler1:
                  if each in ler1:
                        if each in ler1:
                            if each in ler1:
                              if each in ler1:
                                    if each not in ler1:
                                        if each not in ler1:
                                          if each != '\n':
                                                print(each, end='')

因为我没看懂你的for循环那里,所以下面是我按照你的思路写的函数和调用
def choice(list):
    x = []
    length = len(list1)
    for i in range(length):
      x.append(list)
    return x

soup = choice(list1)

lijiachen 发表于 2020-11-29 16:43:44

昨非 发表于 2020-11-29 00:44
啊这,人家问你拧螺丝,你这直接造火箭啊

正则有点秀{:10_256:}

python_zbl 发表于 2020-11-29 22:11:57

{:10_256:}
页: [1]
查看完整版本: 大佬们,帮忙修改下代码