动动手:
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
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第一次看题,觉得简单,于是写出以下代码:
- l = len(a)
- for i in range(3, l - 3):
- if a[i].islower():
- for j in range(1, 4):
- if a[i + j].isupper() + a[i - j].isupper() == 2:
- if a[i + 4].isupper() + a[i - 4].isupper() == 0:
- print(a[i], end = '')
结果如下:
- asaazzllqqqpbbiidllmmbbismrrppqqrzpztijjcntjauuiiiooyaasookttqqfaauconeitwwhhoddduaayyvzmoumfaallldubbkkkljlqzqyyarrswvapxmcoqzftzwwhhggfrzzoootvvoyfasgnebxxllvvrrghwsvvvzojbooiwwffyhhptqqzqseeenzmmyxlwwthhodcvyyvjwdccbmtnfcifenpzhhqvwclfffoozzwwowccbstuqivqmtenwwfnqkkvkknaljlegeffccnmeebiisnuuflwkmazookbvtyjpeeddavxxttpprwwiiimmddzzgbooddbbguyiikljdzzvvharxgukyyojtqqcchdlsubnfniuuzzqhljiwdtccfyuunldaagscjsssxtnoysfkhoocxgfuymmcxooqiblufovmddgvjxalhwwdddemtccyyuhqqppddirfyeouuaamznaabnlfphhwzgxuzzaycfewyvqqppvuuwwnnwwyfwwddssjhuppvvmmvvfjdczgkkhuutrjffyvvggvvzfhhhzzaddcczqjxifekxxhpggracllvvivccllppxbthjulrqqggxjuthamgttvjwwgenhvqggkkqceyyhhdfzlyyieruxcbbqqxxjoouuovcrrjnnesjnddazznmhhecffajppzebbppktfdtkkdendnsqqbbcvccchhrivocnfzzvvooeehqxntbbodfiiklritidapgvnnlbffpuvvfifnaddnyuuuaaed
这哪是什么密码,简直就是乱码,找了半天没找到原因,死马当活医,于是将代码修改如下:
- l = len(a)
- for i in range(3, l - 3):
- if a[i].islower():
- if a[i + 1].isupper() + a[i - 1].isupper() == 2:
- if a[i + 2].isupper() + a[i - 2].isupper() == 2:
- if a[i + 3].isupper() + a[i - 3].isupper() == 2:
- if a[i + 4].isupper() + a[i - 4].isupper() == 0:
- print(a[i], end = '')
结果如下:
- ilovefishc
经过仔细研究,终于找到原因了,这段代码的问题在于条件只满足了两个条件:
1.小写字母前后的第4个为小写字母
2.小写字母前后的第j个为大写字母
应该改为小写字母前后的3个都为大写字母,直接上代码:
1.小写字母前后的第4个为小写字母
2.小写字母前后的第j个为大写字母
应该改为小写字母前后的3个都为大写字母,直接上代码:
- l = len(a)
- x = 0
- for i in range(3, l - 3):
- if a[i].islower():
- if a[i + 4].isupper() + a[i - 4].isupper() == 0:
- for j in range(1, 4):
- if a[i + j].isupper() + a[i - j].isupper() == 2:
- x = 1
- else:
- x = 0
- break
- if x == 1:
- print(a[i], end = '')
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