[已解决]第二十讲课后作业最后一题

Enril

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：
    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

str1 = """ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKElqNsYCyaQXBzrFUbkAUAWAKrDgDtAlGMBqWQhpEwquZqWZJpslUfMllCwWptqINjrOBTLuPzwvXNbLCx
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我写了两个第一种把str换成list了，结果第一种是ilovefishc，第二种是uilovefisdhc，哪里不一样呢。。。
如果是没算空格的话输出结果应该也一样啊。。

第一种：

num = 3
list1 = list(str1)
for each in list1:
    if str(list1[num]).islower() and\
   str(list1[(num - 3):(num)]).isupper() and\
    str(list1[(num + 1):(num + 4)]).isupper() and\
    str(list1[num-4]).islower() and\
    str(list1[num+4]).islower():
        print(str(list1[num]),end = '')
    num += 1
    if num ==len(list1)-3:
        break



第二种：

num = 3
for each in str1:
    if str1[num].islower() and\
   str1[(num - 3):(num)].isupper() and\
    str1[(num + 1):(num + 4)].isupper() and\
    str1[num - 4].islower() and\
    str1[num + 4].islower():
        print(str1[num],end = '')
    num += 1
    if num ==len(str1)-3:
        break

最佳答案

月排行榜 / 总排行榜

flamezyy

2020-3-13 12:33:26

1. iD S K J K L
   
2. lT J N H Z Y
   
3. oS K L I O S
   
4. vT U P O C P
   
5. eI E K G G G
   
6. fH I R D B Y
   
7. iB G T A B T
   
8. sI O K D P O
   
9. hJ F F C S V
   
10. cN T L A E I

复制代码

这是第一个找出密码的前三位和后三位

1. uI Y
   
2. A B N
   
3. iD S K J K L
   
4. lT J N H Z Y
   
5. oS K L I O S
   
6. vT U P O C P
   
7. eI E K G G G
   
8. fH I R D B Y
   
9. iB G T A B T
   
10. sI O K D P O
    
11. d
    
12. W F Z N R
    
13. hJ F F C S V
    
14. cN T L A E I

复制代码

这是第二个，可以发现比第一个多出的'u'和'd'算入了'\n'，我们来看个例子：

1. >>> "['B', 'C', '\\n']".isupper()
   
2. False

复制代码

'\\n'并不会被当做大写，在第一个程序str（list（））语句中，原有字符串中的'\n'会被转化为'\\n'：

1. >>> list1 = ['A','B','C','\n']
   
2. >>> str(list1[1:4])
   
3. "['B', 'C', '\\n']"

复制代码

而这会被isupper方法判定为假。
再来看第二个：

1. >>> str1 = 'ABC\n'
   
2. >>> str1[1:4].isupper()
   
3. True
   
4. >>> str1[1:4]
   
5. 'BC\n'
   
6. >>>

复制代码

字符串中即使带有'\n'，因为被转义了，所以会被忽略，只要其他字符是大写，isupper就会判定为真
但要注意，单个的'\n'不会被当做大写：

1. >>> '\n'.isupper()
   
2. False

复制代码

跳转到最佳答案楼层

jackz007

       第一种：'ilovefishc' 是正确的
       试试这个代码，str1 用你的，由于太长，这里就不拷贝了

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

复制代码

Hoiste

第二种，在num = 3前面加上一句

1. str1 = str1.replace('\n','').replace('\r','')

复制代码

，这里的问题其实是代码中字符处理将回车也作为一个大写字符判断为符合条件了，所以需要首先把换行符‘\n’和回车符号'\r'替换成空字符‘’再进行处理，而对于列表则不会接受到这种问题

flamezyy

1. iD S K J K L
   
2. lT J N H Z Y
   
3. oS K L I O S
   
4. vT U P O C P
   
5. eI E K G G G
   
6. fH I R D B Y
   
7. iB G T A B T
   
8. sI O K D P O
   
9. hJ F F C S V
   
10. cN T L A E I

复制代码

这是第一个找出密码的前三位和后三位

1. uI Y
   
2. A B N
   
3. iD S K J K L
   
4. lT J N H Z Y
   
5. oS K L I O S
   
6. vT U P O C P
   
7. eI E K G G G
   
8. fH I R D B Y
   
9. iB G T A B T
   
10. sI O K D P O
    
11. d
    
12. W F Z N R
    
13. hJ F F C S V
    
14. cN T L A E I

复制代码

这是第二个，可以发现比第一个多出的'u'和'd'算入了'\n'，我们来看个例子：

1. >>> "['B', 'C', '\\n']".isupper()
   
2. False

复制代码

'\\n'并不会被当做大写，在第一个程序str（list（））语句中，原有字符串中的'\n'会被转化为'\\n'：

1. >>> list1 = ['A','B','C','\n']
   
2. >>> str(list1[1:4])
   
3. "['B', 'C', '\\n']"

复制代码

而这会被isupper方法判定为假。
再来看第二个：

1. >>> str1 = 'ABC\n'
   
2. >>> str1[1:4].isupper()
   
3. True
   
4. >>> str1[1:4]
   
5. 'BC\n'
   
6. >>>

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字符串中即使带有'\n'，因为被转义了，所以会被忽略，只要其他字符是大写，isupper就会判定为真
但要注意，单个的'\n'不会被当做大写：

1. >>> '\n'.isupper()
   
2. False

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mozhihui

Hoiste 发表于 2020-3-13 12:30
第二种，在num = 3前面加上一句，这里的问题其实是代码中字符处理将回车也作为一个大写字符判断为符合条件 ...

谢谢！意思是这里必须要把回车符转换成空字符，这样切片时才不会把回车符当成判断条件？
答案写的太复杂了..觉得还是切片来寻找比较好理解。