# 1 使用遞歸打印斐波那契數列(前兩個數的和得到第三個數，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 一個嵌套很多層的列表，如l=［1,2,[3,[4,5,6,[7,8,[9,10,[11,12,13,[14,15]]]]]]］，用遞歸取出所有的值
# 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 編寫用戶登錄裝飾器，在登錄成功後無需重新登錄，同一賬號重復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(‘已登陸‘)
#                                 return
#                             print(‘登陸成功‘)
#                             res=func(*args,**kwargs)
#                             return res
#                         else:
#                             print(‘密碼錯誤‘)
#                             count+=1
#     return wrapper
#
# @outter
# def index():
#     print(‘.......‘)
#     time.sleep(4)
# index()

# 4、求文件a.txt中總共包含的字符個數？思考為何在第一次之後的n次sum求和得到的結果為0？（需要使用sum函數）
# with open(‘db.txt‘,mode=‘r‘,encoding=‘utf-8‘) as f:
#     print(sum(len(line) for line in f))

# 5、文件shopping.txt內容如下
# 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)


    # 打印出所有商品的信息，格式為[{‘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)

8.2每日作業系列之函數的遞歸調用