wmy1212 发表于 2021-6-28 22:43:40

关于20讲动动手01找出密码的问题

# 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

str2 = r'''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
countA = 0
countB = 0
countC = 0
length = len(str2)
for i in range(length):
    if str2 == '\n':
      continue
    if str2.isupper():#如果是个大写，判断是A部分还是C部分
      if countB:#说明是C部分
            countC += 1
      else:
            countA += 1
            countC = 0
    if str2.islower == 1:#如果是个小写，要看A部分是不是3，B是不是0
      if countA != 3:
            countA = 0
            countB = 0
            countC = 0
      else:
            if countB:
                countA = 0
                countB = 0
                countC = 0
            else:#满足条件（A=3，B=0）
                countB += 1
                countC = 0
                target = i
    if countA == 3 and countC == 3:
      if i+1 != length and str2.isupper():
            countB = 0
            countC = 0
      else:
            print(str2,end='')
            countA = 3
            countB = 0
            countC = 0
我的问题：
1.我照着答案写的，为什么跑不出来呀？不知道哪里错了。。
2.倒数第8行这里， if i+1 != length and str2.isupper():这时候为什么不用让countA也=0呢？

Twilight6 发表于 2021-6-28 22:43:41


第 122 行这里你忘记给函数加上 () 括号表示调用了

if str2.islower() == 1


wmy1212 发表于 2021-6-30 14:49:47

Twilight6 发表于 2021-6-29 12:10
第 122 行这里你忘记给函数加上 () 括号表示调用了

谢谢！！那倒数第8行这里， if i+1 != length and str2.isupper():这时候为什么不用让countA也=0呢？

jackz007 发表于 2021-7-2 14:14:35

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qKWGc'''
s = 'a' + '' . join(str1 . split('\n')) + 'a'
p = ''
for k in range(4 , len(s) - 4):
    if s . islower() and s . isupper() and s . islower() and s . isupper() and s . islower():
      p += s
print(p)
页: [1]
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