Python入门——练习

# _*_coding:utf-8_*_
# 1.简述编译型与解释型语言的区别，且分别列出你知道的哪些语言属于编译型，哪些属于解释型
# 编译型：需要编译器，执行速度快，调试麻烦，跨平台差
# e.g. C\C++\

# 解释型：不需要编译器，执行一行翻译一行，执行速度慢，调试方便
#        跨平台性好，依赖解释器运行
# e.g. Python\Java

# 2.执行 Python 脚本的两种方式是什么
# 交互式环境：输入代码立即执行，得到结果
# 命令行：python3 test.txt 以文件的方式将代码永久保存

# 3.Pyhton 单行注释和多行注释分别用什么?
# 单行注释用#
# 多行注释用 ''' …………
#              …………
#                   '''

# 4.布尔值分别有什么?
# True False bool值在判断中用

# 5.声明变量注意事项有那些?
# 1.变量的命名应该能够反应出值的状态
# 2.变量名只能是字母、数字、下划线
# 3.变量名的第一个字符不能是数字
# 4.不能是PYTHON的关键字

# 6.如何查看变量在内存中的地址?
# print(id(a))

# 7.写代码
#     1.实现用户输入用户名和密码,当用户名为 seven 且 密码为 123 时,显示登陆成功,否则登陆失败!
# username = 'seven'
# pwd = '123'
# name = input('Please enter your username:')
# password = input('Please enter your password:')
# if name == username and password == pwd:
#     print("login successful!")
# else:
#     print("login failed!")

#     2.实现用户输入用户名和密码,当用户名为 seven 且 密码为 123 时,显示登陆成功,否则登陆失败,失败时允许重复输入三次
#         username = 'seven'
#         pwd = '123'
#
#         count = 1
#         while count < 4:
#             name = input('Please enter your username:')
#             password = input('Please enter your password')
#             if name == username and pwd == password:
#                 print('login successful')
#                 break
#             else:
#                 print('login failed')
#                 count += 1

#     3.实现用户输入用户名和密码,当用户名为 seven 或 alex 且 密码为 123 时,显示登陆成功,否则登陆失败,失败时允许重复输入三次
# username = 'seven'
# username1 = 'alex'
# pwd = '123'
#
# count = 1
# while count < 4:
#     name = input('Please enter your username:')
#     password = input('Please enter your password')
#     if (name == username) or (name == username1) and pwd == password:
#         print('login successful')
#         break
#     else:
#         print('login failed')
#         count += 1
# 8.写代码
#     a. 使用while循环实现输出2-3+4-5+6...+100 的和
# tag = 1
# count = 2
# sum = 0
# while count<=100:
#
#     sum = sum + tag * count
#     tag *= -1
#     count += 1
#
# print(sum)

#     b. 使用 while 循环实现输出 1,2,3,4,5,7,8,9,11,12
# count = 0
# while count<=11:
#     count +=1
#     if count == 6 or count == 10:
#         continue
#     print(count)
#     c.使用 while 循环实现输出 1-100 内的所有奇数
#     count = 0
#     while count <=99:
#         count +=1
#         if count%2 ==1:
#             print(count)

#     d. 使用 while 循环实现输出 1-100 内的所有偶数
# count = 0
# while count <= 99:
#     count += 1
#     if count % 2 == 0:
#         print(count)

# 9.现有如下两个变量,请简述 n1 和 n2 是什么关系?
# n1 = 123456
# n2 = n1
# print(id(n1),id(n2))
# id地址相同 计算值绑定两个变量名

# 作业：编写登陆接口
# 基础需求：
# 让用户输入用户名密码
# 认证成功后显示欢迎信息
# 输错三次后退出程序

# user_info = {
#     'oOC':{'password':'123','count':0},
#     'Egon':{'password':'123','count':0},
#     'David':{'password':'123','count':0},
# }
# while True:
#     name = input('username>>: ')
#
#     if not name in user_info:
#         print('用户不存在')
#         continue
#     if user_info[name]['count']>2:
#         print('尝试次数过多，锁定')
#         continue
#
#     password = input('password>>: ')
#
#     if password == user_info[name]['password']:
#         print('login successful')
#         break
#     else:
#         print('密码输入错误')
#         user_info[name]['count']+=1


# 升级需求

# dic = {
#     'egon1': {'password': '123', 'count': 0},
#     'egon2': {'password': '123', 'count': 0},
#     'egon3': {'password': '123', 'count': 0},
# }
#
# count = 0
# while True:
#     name = input('u>>: ')
#     if name not in dic:
#         print('User does not exist: ')
#         continue
#
#     with open('db.txt', 'r') as f:
#         lock_users = f.read().split('|')
#         if name in lock_users:
#             print('user %s locked' %name)
#             break
#
#     if dic[name]['count'] > 2:
#         print('try too much times,locked')
#         with open('db.txt', 'a') as f:
#             f.write('%s|' % name)
#
#         break
#
#     password = input('p>> ')
#
#     if password == dic[name]:
#         print('login successful')
#         break
#     else:
#         print('login failed')
#         dic[name]['count'] += 1

转载于:https://www.cnblogs.com/OutOfControl/p/9656466.html