函数课后作业
#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
length2 = len(longstr)
password = ''
for each in (4,length2 - 3):
if longstr.islower() == True: #判断是否为小写字母
if longstr.islower == True and longstr.islower() == True:#判断右/左侧第四个是小写
count = 0 #初始化count
for nums in range(1,4):
if longstr.isupper == True and longstr.isupper == True: #判断右/左侧是否为大写字母
count += 1
if count == 3:
password += longstr
print(password)
每次运行都是''
服气了哦
思路如下：
1，获取字符串长度，依次迭代出每一个元素
2，判断是否为小写，并从这开始对前后的元素是否符合要求进行判断（毕竟密码全是小写么）
3，判断左右第四个是否为小写，如果不是那就pass
4，初始化count，并迭代1~3三个数，进行对前后三个元素是否符合进行判断
5，每迭代一次如果符合全是大写，count+1
6，最后判断count是否为三，如果是就说明前后三个全是大写
实在想不到哪里错了{:10_266:} {:10_334:} python 的方法都是带括号的
楼主很多处的 islower() isupper() 等缺失了 () Daniel_Zhang 发表于 2021-2-25 19:08
python 的方法都是带括号的
楼主很多处的 islower() isupper() 等缺失了 ()
额......
改正之后还是''{:10_243:} 本帖最后由 Daniel_Zhang 于 2021-2-25 22:16 编辑
跟我一起敲代码 发表于 2021-2-25 21:54
额......
改正之后还是''
那你把你改了之后的在下面放出来呀，直接在我这个回复下面点击回复按钮，这样我就可以看到了{:10_297:}
顺便说一下这是小甲鱼第几节的作业，我去看看 Daniel_Zhang 发表于 2021-2-25 22:03
那你把你改了之后的在下面放出来呀，直接在我这个回复下面点击回复按钮，这样我就可以看到了 ...
#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
length2 = len(longstr)
password = ''
for each in (4,length2 - 3):
if longstr.islower() == True: #判断是否为小写字母
if longstr.islower() == True and longstr.islower() == True:#判断右/左侧第四个是小写
count = 0 #初始化count
for nums in range(1,4):
if longstr.isupper() == True and longstr.isupper() == True: #判断右/左侧是否为大写字母
count += 1
if count == 3:
password += longstr
print(password)
这个应该是旧版python第二十讲：内嵌函数和闭包
小甲鱼的做法和我的不一样（他的太高级，还在研究中...{:10_266:}）
跟我一起敲代码 发表于 2021-2-25 22:33
这个应该是旧版python第二十讲：内嵌函数和闭包
小甲鱼的做法和我的不一样（他的太高级，还在研究中 ...
那就正好，给你个简单的
他的代码我还没研究，感觉有点复杂就是
# 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
# 源代码并未定义 longstr 变量
longstr = 'ABSaDKSbRIHcRHGcdDIF'
length2 = len(longstr)
password = ''
# 此处源代码少了一个 range, 没有 range 表明在 3 和 length - 2 这两个数里面运行 for，而非组合两个数的范围之间运行
for each in range(3, length2 - 3):# 此处应该为 3 而不是 4，因为 下标从 0 开始的
# 为了简略，可以直接写 if longstr.islower():
# 不需要if longstr.islower() == True:
# 因为只有当返回值不为 0 才会通过 if 语句，返回值为 0 则不通过
# True 在 python 中可以认为是 1， False 可以认为是 0
# 如果需要的是非小写字母通过 if 语句，则使用 if !longstr.islower():
# !True 表示的就是 0
if longstr.islower():# 判断是否为小写字母
# 避免越界，即不超过右边界。如果超过右边界，则退出循环，即 each + 3 超过了 longstr 的长度，不需要继续检查了
if each + 3 < length2:
# 如果当前位置左边的三个均为大写字母则通过 if 循环
if longstr.isupper() and longstr.isupper() and longstr.isupper():
# 如果当前位置右边的三个均为大写字母则通过 if 循环, 并且将当前字符插入到 password 中，即找到了需要的密码
if longstr.isupper() and longstr.isupper() and longstr.isupper():
password += longstr
else:
break
print(password)
你别看我这么多行，其实注释占了一大半{:10_263:} Daniel_Zhang 发表于 2021-2-25 22:44
那就正好，给你个简单的
他的代码我还没研究，感觉有点复杂就是
非常感谢！懂了{:10_275:}
（那个for 循环没加range是真的自闭了，居然没看出来） Daniel_Zhang 发表于 2021-2-25 22:44
那就正好，给你个简单的
他的代码我还没研究，感觉有点复杂就是
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qKWGc'''
>>> length2 = len(longstr)
>>> password = ''
>>> for each in range(3, length2 - 3):# 此处应该为 3 而不是 4，因为 下标从 0 开始的
# 为了简略，可以直接写 if longstr.islower():
# 不需要if longstr.islower() == True:
# 因为只有当返回值不为 0 才会通过 if 语句，返回值为 0 则不通过
# True 在 python 中可以认为是 1， False 可以认为是 0
# 如果需要的是非小写字母通过 if 语句，则使用 if !longstr.islower():
# !True 表示的就是 0
if longstr.islower():# 判断是否为小写字母
# 避免越界，即不超过右边界。如果超过右边界，则退出循环，即 each + 3 超过了 longstr 的长度，不需要继续检查了
if each + 3 < length2:
# 如果当前位置左边的三个均为大写字母则通过 if 循环
if longstr.isupper() and longstr.isupper() and longstr.isupper():
# 如果当前位置右边的三个均为大写字母则通过 if 循环, 并且将当前字符插入到 password 中，即找到了需要的密码
if longstr.isupper() and longstr.isupper() and longstr.isupper():
password += longstr
else:
break
>>> print(password)
llpuiallouovvhezfbilbjiirmsqqxthtgcnkdljoasncexw
好像不太对啊{:10_319:} Daniel_Zhang 发表于 2021-2-25 22:44
那就正好，给你个简单的
他的代码我还没研究，感觉有点复杂就是
但是代码运行出来的结果是llpuiallouovvhezfbilbjiirmsqqxthtgcnkdljoasncexw
不应该是ilovefishc这个吗{:10_255:} 跟我一起敲代码 发表于 2021-2-25 23:20
但是代码运行出来的结果是
不应该是这个吗
?
你是说 llpuiallouovvhezfbilbjiirmsqqxthtgcnkdljoasncexw 放进去运行吗
问题是，大写字母呢？
# 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 Daniel_Zhang 发表于 2021-2-25 23:39
?
你是说 llpuiallouovvhezfbilbjiirmsqqxthtgcnkdljoasncexw 放进去运行吗
不，你理解错了，给你整个代码吧longstr = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
length2 = len(longstr)
password = ''
# 此处源代码少了一个 range, 没有 range 表明在 3 和 length - 2 这两个数里面运行 for，而非组合两个数的范围之间运行
for each in range(3, length2 - 3):# 此处应该为 3 而不是 4，因为 下标从 0 开始的
# 为了简略，可以直接写 if longstr.islower():
# 不需要if longstr.islower() == True:
# 因为只有当返回值不为 0 才会通过 if 语句，返回值为 0 则不通过
# True 在 python 中可以认为是 1， False 可以认为是 0
# 如果需要的是非小写字母通过 if 语句，则使用 if !longstr.islower():
# !True 表示的就是 0
if longstr.islower():# 判断是否为小写字母
# 避免越界，即不超过右边界。如果超过右边界，则退出循环，即 each + 3 超过了 longstr 的长度，不需要继续检查了
if each + 3 < length2:
# 如果当前位置左边的三个均为大写字母则通过 if 循环
if longstr.isupper() and longstr.isupper() and longstr.isupper():
# 如果当前位置右边的三个均为大写字母则通过 if 循环, 并且将当前字符插入到 password 中，即找到了需要的密码
if longstr.isupper() and longstr.isupper() and longstr.isupper():
password += longstr
else:
break
print(password)
#结果如下
llpuiallouovvhezfbilbjiirmsqqxthtgcnkdljoasncexw 不好意思哈，前两天没看到，我的上一个回答确实有点问题，没考虑到密码是左右有且只有三个大写字母
没判断第四个是否是字母
自行添加一下 longstr，longstr 太长，没给过审
longstr= ''''''
length2 = len(longstr)
password = ''
# 此处源代码少了一个 range, 没有 range 表明在 3 和 length - 2 这两个数里面运行 for，而非组合两个数的范围之间运行
for each in range(4, length2 - 4):
# 为了简略，可以直接写 if longstr.islower():
# 不需要if longstr.islower() == True:
# 因为只有当返回值不为 0 才会通过 if 语句，返回值为 0 则不通过
# True 在 python 中可以认为是 1， False 可以认为是 0
# 如果需要的是非小写字母通过 if 语句，则使用 if !longstr.islower():
# !True 表示的就是 0
if longstr.islower():# 判断是否为小写字母
# 避免越界，即不超过右边界。如果超过右边界，则退出循环，即 each + 3 超过了 longstr 的长度，不需要继续检查了
if each + 4 < length2:
# 如果当前位置左边的三个均为大写字母则通过 if 循环
# 此处需要判断第四个是否为小写字母
# 如果第四个是小写字母，则满足密码的条件
# 因为密码左右两边有且只有三个大写字母
if longstr.isupper() and longstr.isupper() and longstr.isupper() and (longstr.isupper() == False):
# 如果当前位置右边的三个均为大写字母则通过 if 循环, 并且将当前字符插入到 password 中，即找到了需要的密码
if longstr.isupper() and longstr.isupper() and longstr.isupper() and (longstr.isupper() == False):
password += longstr
else:
break
print(password)
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