小甲魚課後練習題
各位魚友悶~這題可以用切片解出來嗎? 我自己寫的code在最下面!!!!!!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
str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
-------------------------------------------------------------------------------------------------------------------------
str2 = 'a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z'
str3 = 'A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z'
list1 = []
for each in str1:
if each not in list1:
if each in str2:
idx = str1.index(each)
a,b,c = str1
a, b, c in str3
if each in str2:
idx = str1.index(each)
d,e,f = str1
d, e, f in str3
list1.append(each)
print(each) 本帖最后由 _2_ 于 2020-5-9 13:54 编辑
import re
str1="…" #这里自行补充
re.findall(r"(A-Z)\1\1(a-z)\1\1\1", str1) 你這個繁體中文看的我很難受的 说几个显性错误：
str1.index(each)获得索引值不靠谱，因为无论each出现多少次，只会返回each再str1中第一次出现的位置的值；
a, b, c in str3、a, b, c in str3，这两句没有任何意义，我猜你是想要：if a in str3 and b in str3 and c in str3:
str2、str3中的字母用逗号隔开没有必要，相反如果str1中有逗号反而容易误判。 txxcat 发表于 2020-5-9 11:03
说几个显性错误：
str1.index(each)获得索引值不靠谱，因为无论each出现多少次，只会返回each再str1中第一 ...
這題用切片解的出來嗎 TsaiYen 发表于 2020-5-9 12:01
這題用切片解的出來嗎
切片是解这题的必要操作，算法就各有不同了。 txxcat 发表于 2020-5-9 12:06
切片是解这题的必要操作，算法就各有不同了。
只跑出 l不知道哪裡出問題!!!!!
str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
str2 = 'a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z'
str3 = 'A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z'
list1 = [] #創建空列表
for each in str1: #把字串中的字符給each
if each not in list1: #判斷each是不是沒有在列表中(沒有的話)
if each in str2: #each 判斷是不是小寫字母
idx = str1.index(each) #找出他在str1裡面的位移植丟給變數
a,b,c = str1
d,e,f = str1
if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
list1.append(each)
print(list1) #修改過後只跑出一個 l
str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
str2 = 'abcdefghijklmnopqrstuvwxyz'
str3 = 'ABCDEFGHJIKLMNOPQRSTUVWXYZ'
list1 = [] #創建空列表
for each in str1: #把字串中的字符給each
if each not in list1: #判斷each是不是沒有在列表中(沒有的話)
if each in str2: #each 判斷是不是小寫字母
idx = str1.index(each) #找出他在str1裡面的位移植丟給變數
a,b,c = str1
d,e,f = str1
if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
list1.append(each)
print(list1)
你更新的代码还是没有解决index只能得到第一个字母位置的问题，然后if each not in list1完全没必要，没有说密码不能有重复的字母，最后是 你判断前三后三是大写，却没有判断前三之前还后三之后那个是不是大写，这也会导致问题。根据你的思路改写了一下代码，可以跑出来，不过有一点不严谨的地方，看看你能找出来不。
还有个建议，判断大小写用isupper()和islow()更好。
list1 = []
lenth=len(str1) #获取字符串长度
for idx in range(3,lenth-3): #从第四个开始算到倒数第三个字符
if str1 in str2 and str1 not in str3 and str1 not in str3: #除了判断是小写，加入判断前4后4的字符是否是大写的判断
a,b,c = str1
d,e,f = str1
if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
list1.append(str1)
print(''.join(list1)) #优化一下输出，其实不用列表，用字符相加更方便 txxcat 发表于 2020-5-9 18:20
你更新的代码还是没有解决index只能得到第一个字母位置的问题，然后if each not in list1完全没必要，没有 ...
想請問下不用這種思路還能用甚麼方法嗎?
for循環不是把str1一個個丟給each為甚麼還是只能得到第一個字母位置
萬分的感謝您!!!!!!!!!!!!!!!!!!!!! TsaiYen 发表于 2020-5-10 16:49
想請問下不用這種思路還能用甚麼方法嗎?
for循環不是把str1一個個丟給each為甚麼還是只能得到第一個字母 ...
遍历字符串可以获得每一个字符，但你要获得字符串的位置，使用了str1.index(each)，比如'a'这个字符，第一个a在第31个，获得的值是31，这是对的，但是后面的'a'呢，str1.index(each)还是会返回第一个'a'的位置，不会返回你希望的值。如果你希望遍历时获得索引值，还可以这么写：
idx=0
for each in str1:
pass
idx+=1
这样可以获得正确的索引值，str1等于each，但是因为有索引值-3、+3、-4、+4的操作，要正确处理索引值溢出的问题，最后需要的代码会比我的方法要多不少。
这里有几种算法，其实大同小异：
lenth = len(str1)
for i in range(3,lenth-3):
allupper=True #定义一个标量标志全部是大写
if str1.islower():
for each in str1+str1: #遍历前3个+后3个字符
if not each.isupper(): #有1个不是大写就跳出判断，标志改为False
allupper=False
break
if allupper and (str1.islower()) and (str1.islower()):
print(str1,end="")
暴力简单去除\n法：
str1=str1.replace('\n','')
length = len(str1)
for i in range(3, length-3):
if str1.islower() andstr1.isupper() and str1.isupper() andnot str1.isupper() and not str1.isupper():
print(str1,end='') txxcat 发表于 2020-5-10 20:51
遍历字符串可以获得每一个字符，但你要获得字符串的位置，使用了str1.index(each)，比如'a'这个字符，第 ...
OK感恩!!! 還在就學嗎?!好厲害 TsaiYen 发表于 2020-5-11 02:49
OK感恩!!! 還在就學嗎?!好厲害
No，作为孩子她爹，对年轻人们充满了羡慕！{:10_262:} 是軟體工程師嗎哈哈！！ MIke_python小小 发表于 2020-5-9 10:26
你這個繁體中文看的我很難受的
哈哈,台灣人? 對！
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