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[已解决]第20讲 动动手1 后边看不太懂

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发表于 2020-11-16 00:10:43 | 显示全部楼层 |阅读模式

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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:i=>|3KYD
i=>|3KYD
    a) 每位密码为单个小写字母Powered by bbs.fishc.com
    b) 每位密码的左右两边均有且只有三个大写字母j~:y
3u=9m7_0 Pzy"pCV#RdwA,21^[t
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                                        ##这里看不太明白 为什么如果后边有四个大写字母的话 countA为什么不清零 没有意义了啊
    |-- 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='')                                                                ##这里也是 为什么countA 不清零 清零不是更好吗
            countA = 3
            countB = 0
            countC = 0
最佳答案
2020-11-16 01:14:36
如果在后面第4个大写字母清零countA,那么如果有第5,6,7个大写字母,就会被当成3个大写字母。     
而有下面这几行:
if str1.islower():
        if countA != 3:
            countA = 0
            countB = 0
            countC = 0
这几行保证了,不清零countA也没关系,因为!=3的情况也包含大于3,所以在这里(再次遇到小写时)会重新清零的。
想知道小甲鱼最近在做啥?请访问 -> ilovefishc.com
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 楼主| 发表于 2020-11-16 00:11:19 | 显示全部楼层
今天问题巨多 感谢各位大佬
想知道小甲鱼最近在做啥?请访问 -> ilovefishc.com
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发表于 2020-11-16 00:37:12 | 显示全部楼层
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qKWGc'''

str = ''
for x in str1:
    str += x . strip()
str = 'a' + str + 'a'
passwd = ''
for k in range(4 , len(str) - 4):
    if str[k - 4] . islower() and str[k - 3 : k] . isupper() and str[k] . islower() and str[k + 1 : k + 4] . isupper() and str[k + 4] . islower():
        passwd += str[k]
print(passwd)
想知道小甲鱼最近在做啥?请访问 -> ilovefishc.com
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发表于 2020-11-16 01:14:36 | 显示全部楼层    本楼为最佳答案   
如果在后面第4个大写字母清零countA,那么如果有第5,6,7个大写字母,就会被当成3个大写字母。     
而有下面这几行:
if str1.islower():
        if countA != 3:
            countA = 0
            countB = 0
            countC = 0
这几行保证了,不清零countA也没关系,因为!=3的情况也包含大于3,所以在这里(再次遇到小写时)会重新清零的。
想知道小甲鱼最近在做啥?请访问 -> ilovefishc.com
回复 支持 反对

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