8.2每日做業系列之函數的遞歸調用

# 1 使用遞歸打印斐波那契數列(前兩個數的和獲得第三個數，如：0 1 1 2 3 4 7...)# def func(x,y):#     res=x+y#     print(x)#     if res<100:#         return func(y,res)#     else:#         return y# print(func(0,1))# 2 一個嵌套不少層的列表，如l=［1,2,[3,[4,5,6,[7,8,[9,10,[11,12,13,[14,15]]]]]]］，用遞歸取出全部的值# l=[1,2,[3,[4,5,6,[7,8,[9,10,[11,12,13,[14,15]]]]]]]# def func(l):#     for i in l:#         if type(i) is not list:#             print(i)#         else:#             func(i)# func(l)# 3 編寫用戶登陸裝飾器，在登陸成功後無需從新登陸，同一帳號重複輸錯三次密碼則鎖定5分鐘# import time# def outter(func):#     def wrapper(*args,**kwargs):#         count = 0#         while True:#             if count == 3:#                 time.sleep(10)#                 wrapper()#             name = input('用戶名:').strip()#             pwd = input('密碼:').strip()#             with open('db.txt',mode='r',encoding='utf-8') as f:#                 for line in f:#                     if line.startswith(name):#                         line = line.strip('\n').split(',')#                         if line[1]==pwd:#                             if line[3]:#                                 print('已登錄')#                                 return#                             print('登錄成功')#                             res=func(*args,**kwargs)#                             return res#                         else:#                             print('密碼錯誤')#                             count+=1#     return wrapper## @outter# def index():#     print('.......')#     time.sleep(4)# index()# 四、求文件a.txt中總共包含的字符個數？思考爲什麼在第一次以後的n次sum求和獲得的結果爲0？（須要使用sum函數）# with open('db.txt',mode='r',encoding='utf-8') as f:#     print(sum(len(line) for line in f))# 五、文件shopping.txt內容以下# mac,20000,3# lenovo,3000,10# tesla,1000000,10# chicken,200,1# 求總共花了多少錢？with open('info.txt',mode='r',encoding='utf-8') as f:#     # info=[line.strip('\n').split(',') for line in f]#     # cost=sum(float(price)*int(count) for _,price,count in info)#     # print(cost)#    l=[]    for line in f:        info=line.strip('\n').split(',')        name,price,count=info        res=float(price)*int(count)        l.append(res)    cost=sum(l)    print(cost)#     for line in f:#         info=line.strip('\n').split(',')#         for _, price, count in info:#             print(_, price, count)#         for i in info:#             print(i)    # 打印出全部商品的信息，格式爲[{'name':'xxx','price':333,'count':3},...]# with open('info.txt',mode='r',encoding='utf-8') as f:#     info=[{#         'name': line.strip('\n').split(',')[0],#         'price': float(line.strip('\n').split(',')[1]),#         'count': int(line.strip('\n').split(',')[2]),#     } for line in f]#     print(info)# 求單價大於10000的商品信息,格式同上# with open('info.txt',mode='r',encoding='utf-8') as f:#     info=[{#         'name': line.strip('\n').split(',')[0],#         'price': float(line.strip('\n').split(',')[1]),#         'count': int(line.strip('\n').split(',')[2]),#     } for line in f if float(line.strip('\n').split(',')[1]) > 10000]#     print(info)# 明日默寫：# 二分查找# l=[1,2,10,30,33,99,101,200,301,311,402,403,500,900,1000] #從小到大排列的數字列表# def search(n,l):#     print(l)#     if len(l) == 0:#         print('not exists')#         return#     mid_index=len(l) // 2#     if n > l[mid_index]:#         #in the right#         l=l[mid_index+1:]#         search(n,l)#     elif n < l[mid_index]:#         #in the left#         l=l[:mid_index]#         search(n,l)#     else:#         print('find it')# search(3,l)