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本帖最后由 liyidong3000 于 2021-9-1 11:03 编辑
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
这是小甲鱼的答案,算法比较复杂。
- str1 = '''ABSaDKSbRIHcRHGcdDIF'''
- countA = 0 # 统计前边的大写字母
- countB = 0 # 统计小写字母
- countC = 0 # 统计后边的大写字母
- length = len(str1)
- for i in range(length):
- if str1[i] == '\n':
- continue
- """
- |如果str1[i]是大写字母:
- |-- 如果已经出现小写字母:
- |-- -- 统计后边的大写字母
- |-- 如果未出现小写字母:
- |-- -- 清空后边大写字母的统计
- |-- -- 统计前边的大写字母
- """
- if str1[i].isupper():
- if countB:
- countC += 1
- else:
- countC = 0
- countA += 1
- """
- |如果str1[i]是小写字母:
- |-- 如果小写字母前边不是三个大写字母(不符合条件):
- |-- -- 清空所有记录,重新统计
- |-- 如果小写字母前边是三个大写字母(符合条件):
- |-- -- 如果已经存在小写字母:
- |-- -- -- 清空所有记录,重新统计(出现两个小写字母)
- |-- -- 如果该小写字母是唯一的:
- |-- -- -- countB记录出现小写字母,准备开始统计countC
- """
- if str1[i].islower():
- if countA != 3:
- countA = 0
- countB = 0
- countC = 0
- else:
- if countB:
- countA = 0
- countB = 0
- countC = 0
- else:
- countB = 1
- countC = 0
- target = i
- """
- |如果前边和后边都是三个大写字母:
- |-- 如果后边第四个字母也是大写字母(不符合条件):
- |-- -- 清空记录B和C,重新统计
- |-- 如果后边仅有三个大写字母(符合所有条件):
- |-- -- 打印结果,并清空所有记录,进入下一轮统计
- """
- if countA == 3 and countC == 3:
- if i+1 != length and str1[i+1].isupper():
- countB = 0
- countC = 0
- else:
- print(str1[target], end='')
- countA = 3
- countB = 0
- countC = 0
复制代码
我自己写了一个
- string='''ABSaDKSbRIHcRHGcdDIF'''
- length=len(string)
- list1=[]
- for i in range(length):
- if string[i] == '\n':
- continue
- if string[i].islower() and string[i-3:i].isupper() and string[i+1:i+4].isupper() and string[i-4:i].isupper() == 0 and string[i+1:i+5].isupper() == 0:
- list1.append(string[i])
- print(list1)
复制代码
感觉逻辑没问题,但是运行结果跟小甲鱼的答案不同。
字符编码这个简短的是没问题的。题目原始编码为
ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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uABNuEDKFWiUaqQiBHGsgEmDSKiJKLkkzFWlobXIyUlDzGcwDnEOwwWxgFpebsslmKItazIYctPROSJp
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qKWGc
你这个程序的问题出在字符串里的回车上。字符串里有大量的'\n',当'\n'前面的字符为2个大写字母时,str[i-3:i].isupper()会为True,就是把'\n'也算为大写字母了
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