请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
对于同一个运算符、函数或对象在不同的场景下,具有不同的作用效果。
第 1 题的答案是:
减少重复的定义函数。
第 2 题的答案是:
可以。
第 3 题的答案是:
不知道。
第 4 题的答案是:
降低了耦合度。
第 5 题的答案是:
不属于。
-------- 动动手 --------
请将第 0 题的代码写在下方:
class Employee:
def __init__(self,name,job,grade,year,uid,salary):
self.name = name
self.job = job
self.grade = grade
self.year = year
self.uid = uid
self.salary = salary
class Salary_calculate:
def calculate(self,job,grade,year):
if job == 'M':
salary = 5000 + 1000 * (int(grade) + int(year))
elif job == 'T':
salary = 4000 + 800 * int(grade) + 100 * int(year)
elif job == 'E':
salary = 3000 + 500 * int(grade) + 50 * int(year)
else:
raise ValueError("输入的数据无效。")
return salary
class Management(Salary_calculate):
def __init__(self):
self.date = {}
self.uid = 10000
def welcome(self):
ins = 0
while ins != '5':
ins = input("1.录入;2.查询;3.升级;4.降级;5.退出:")
if ins == '1':
self.createmployee()
elif ins == '2':
result = self.find()
if result == 1:
print("该工号不存在!")
elif ins == '3':
self.upgrade()
elif ins == '4':
self.downgrade()
elif ins == '5':
print()
def createmployee(self):
name = input("姓名:")
job = input("职位(E.普通员工;T.组长;M.经理):")
year = input("工龄:")
grade = input("级别:")
if job == 'E' and int(grade) > 10:
grade = input("该职位最高级别为10,请重新录入级别:")
elif job == 'T' and int(grade) > 6:
grade = input("该职位最高级别为6,请重新录入级别:")
elif job == 'M' and int(grade) > 3:
grade = input("该职位最高级别为3,请重新录入级别:")
salary = self.calculate(job,grade,year)
employee = Employee(name, job, grade, year, self.uid,salary)
self.date = employee
print(f"录入成功!姓名:{name},工号:{self.uid},薪资:{salary}")
self.uid += 1
def find(self):
s = input("1.员工查询;2.职位查询:")
if s == '1':
uid = int(input("请输入工号:"))
if uid not in self.date.keys():
return 1
print(f"姓名:{self.date.name}")
print(f"职位:{self.date.job}")
print(f"级别:{self.date.grade}")
print(f"工龄:{self.date.year}")
print(f"薪资:{self.date.salary}")
if s == '2':
count1 = 0
date1 = {}
job = input("职位(E.普通员工;T.组长;M.经理):")
if job == 'E':
for i in self.date.values():
if i.job == 'E':
date1 = i.name
count1 += 1
if count1 != 0:
print(f"目前普通员工共有{count1}人:")
for x,y in date1.items():
print(f"{x} - {y}")
else:
print("目前公司没有普通员工!")
if job == 'T':
for i in self.date.values():
if i.job == 'T':
date1 = i.name
count1 += 1
if count1 != 0:
print(f"目前组长共有{count1}人:")
for x, y in date1.items():
print(f"{x} - {y}")
else:
print("目前公司没有组长!")
if job == 'M':
for i in self.date.values():
if i.job == 'M':
date1 = i.name
count1 += 1
if count1 != 0:
print(f"目前经理共有{count1}人:")
for x, y in date1.items():
print(f"{x} - {y}")
else:
print("目前公司没有经理!")
def upgrade(self):
uid = int(input("请输入工号:"))
print(f"{self.date.name},工号:{uid},当前职位:{self.date.job}{self.date.grade},当前薪资:{self.date.salary}")
salary1 = self.date.salary
grade1 = int(input("请输入需要增加的级数:"))
t1 = str(int(self.date.grade) + int(grade1))
if self.date.job == 'E':
if int(self.date.grade) + int(grade1) <= 10:
self.date.grade = t1
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
else:
self.date.grade = '1'
self.date.job = 'T'
self.date.salary = self.calculate(self.date.job,self.date.grade,self.date.year)
print(f"升级成功!\n{self.date.name},工号:{uid},升级后的职位:{self.date.job}{self.date.grade},升级后薪资:{self.date.salary}({self.date.salary - salary1})")
returnNone
if self.date.job == 'T':
if int(self.date.grade) + int(grade1) <= 6:
self.date.grade = t1
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
else:
self.date.grade = '1'
self.date.job = 'M'
self.date.salary = self.calculate(self.date.job,self.date.grade,self.date.year)
print(f"升级成功!\n{self.date.name},工号:{uid},升级后的职位:{self.date.job}{self.date.grade},升级后薪资:{self.date.salary}(+{self.date.salary - salary1})")
return None
if self.date.job == 'M':
if int(self.date.grade) + int(grade1) <= 3:
self.date.grade = t1
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
else:
self.date.grade = '3'
self.date.salary = self.calculate(self.date.job,self.date.grade,self.date.year)
print(f"升级成功!\n{self.date.name},工号:{uid},升级后的职位:{self.date.job}{self.date.grade},升级后薪资:{self.date.salary}({self.date.salary - salary1})")
return None
def downgrade(self):
uid = int(input("请输入工号:"))
print(f"{self.date.name},工号:{uid},当前职位:{self.date.job}{self.date.grade},当前薪资:{self.date.salary}")
salary1 = self.date.salary
grade1 = int(input("请输入需要减少的级数:"))
t2 = str(int(self.date.grade) - int(grade1))
if self.date.job == 'E':
if int(self.date.grade) - int(grade1) > 1:
self.date.grade = t2
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
else:
self.date.grade = '1'
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
print(f"降级成功!\n{self.date.name},工号:{uid},降级后的职位:{self.date.job}{self.date.grade},降级后薪资:{self.date.salary}({self.date.salary - salary1})")
return None
if self.date.job == 'T':
if int(self.date.grade) - int(grade1) > 1:
self.date.grade = t2
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
else:
self.date.grade = '10'
self.date.job = 'E'
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
print(f"降级成功!\n{self.date.name},工号:{uid},降级后的职位:{self.date.job}{self.date.grade},降级后薪资:{self.date.salary}({self.date.salary - salary1})")
return None
if self.date.job == 'M':
if int(self.date.grade) - int(grade1) > 1:
self.date.grade = t2
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
else:
self.date.grade = '6'
self.date.job = 'T'
self.date.salary = self.calculate(self.date.job, self.date.grade, self.date.year)
print(f"降级成功!\n{self.date.name},工号:{uid},降级后的职位:{self.date.job}{self.date.grade},降级后薪资:{self.date.salary}({self.date.salary - salary1})")
return None
def main():
m = Management()
m.welcome()
main()
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
同一个运算符、函数或对象在不同的场景下,具有不同的作用效果
第 1 题的答案是:
多态的好处这样就一目了然了,尽管我们的接口是不变的,但它却可以根据不同的对象执行不同的操作
第 2 题的答案是:
可以
第 3 题的答案是:
When I see a bird that walks like a duck and swims like a duck and quacks like a duck, I call that bird a duck.
第 4 题的答案是:
降低了耦合度
第 5 题的答案是:
不是
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:在类的继承中,子类重写父类,其实就是多态
第 1 题的答案是:简洁,可以去除一些重复的代码
第 2 题的答案是:可以,多态在不同类型中的效果不同,比如"+"符号,在整数运算中就是做加法预算,而在字符串中,就是拼接
第 3 题的答案是:When I see a bird that walks like a duck and swims like a duck and quacks like a duck, I call that bird a duck.
第 4 题的答案是:降低了耦合度
第 5 题的答案是:不是,因为没有出现继承的现象
-------- 动动手 --------
请将第 0 题的代码写在下方:
{:10_249:}
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
不同的类可以使用相同的方法,得到不同的结果
第 1 题的答案是:
灵活
第 2 题的答案是:
是的
第 3 题的答案是:
第 4 题的答案是:
增强了
第 5 题的答案是:
不属于
-------- 动动手 --------
请将第 0 题的代码写在下方:
class user:
def __init__(self,job):
self.job=job
def get_uid(self):
print(f"工号: {self.uid}")
def get_name(self):
print(f"姓名: {self.name}")
def get_job(self):
print(f"职位: {self.job}")
def get_grade(self):
print(f"级别: {self.grade}")
def get_year(self):
print(f"工龄: {self.year}")
def salary(self,jc,gw,gl):
self.mony=jc+gw*int(self.grade)+gl*int(self.year)
"""
if self.job == 'E':
self.mony=3000 + 500*int(self.grade)+50*int(self.year)
elif self.job== 'T':
self.mony=4000 + 800*int(self.grade)+100*int(self.year)
else:
self.mony=5000 + 1000*(int(self.grade)+int(self.year))
"""
class Employee(user):
def __init__(self,name,grade,year,uid):
self.name=name
self.year=year
self.uid=uid
super().__init__("E")
while grade > 10:
grade = input("该职位最高级别为10,请重新录入级别:")
self.grade=grade
def salary(self):
super().salary(3000,500,50)
class Teamleader(user):
def __init__(self,name,grade,year,uid):
self.name=name
self.year=year
self.uid=uid
super().__init__("T")
while grade > 6:
grade = int(input("该职位最高级别为10,请重新录入级别:"))
self.grade=grade
def salary(self):
super().salary(4000,800,100)
class Manager(user):
def __init__(self,name,grade,year,uid):
self.name=name
self.year=year
self.uid=uid
super().__init__("M")
while grade > 3:
grade = input("该职位最高级别为10,请重新录入级别:")
self.grade=grade
def salary(self):
super().salary(5000,1000,1000)
##定义功能函数
def create(count):
name=input("姓名:")
job=input("职位(E.普通员工;T.组长;M.经理):")
year=int(input("工龄:"))
grade=int(input("级别:"))
if job == 'E':
return Employee(name,grade,year,count)
elif job == 'T':
return Teamleader(name,grade,year,count)
else:
return Manager(name,grade,year,count)
def search(uid):
s=input("1.员工查询;2.职位查询:")
if s=='1':
s2=int(input("请输入工号:"))
if s2 not in uid:
print("该工号不存在!")
else:
uid.get_name()
uid.get_job()
uid.get_grade()
uid.get_year()
uid.salary()
print(f"当前薪资: {self.mony}")
else:
s2=input("职位(E.普通员工;T.组长;M.经理):")
num = 0
name=[]
for i in uid:
if uid.job == s2:
num +=1
name.append(.uid,uid.name])
if s2=='E':
print(f"目前普通员工共有{num}人:")
elif s2=="T":
print(f"目前组长共有{num}人:")
else:
print(f"目前经理共有{num}人:")
for i in name:
print(f"{i} - {i}")
def up(uid):
s=int(input("请输入工号:"))
if s not in uid:
print("该工号不存在")
else:
a=uid
print(f"{a.name},工号:{a.uid},当前职位:{a.job}{a.grade}",end=',')
a.salary()
s2=int(input("请输入需增加的级数:"))
moy=a.mony
if uid.job == 'E':
if (uid.grade+s2) > 10:
uid=Teamleader(a.name,0,a.year,a.uid)
else:
a.grade += s2
elif uid.job == 'T':
if (uid.grade+s2) > 6:
uid=Manager(a.name,0,a.year,a.uid)
else:
a.grade += s2
else:
if (uid.grade+s2) > 3:
uid.grade = 3
else:
a.grade += s2
uid.salary()
print(f"升级成功")
print(f"{uid.name},工号:{uid.uid},升级后职位:{uid.job}{uid.grade},升级后薪资:{uid.mony}(+{uid.mony-moy})")
def down(uid):
s=int(input("请输入工号:"))
if s not in uid:
print("该工号不存在")
else:
a=uid
print(f"{a.name},工号:{a.uid},当前职位:{a.job}{a.grade}",end=',')
a.salary()
s2=int(input("请输入需减少的级数:"))
moy=a.mony
if uid.job == 'E':
if (uid.grade-s2) < 0:
uid.grade = 0
else:
a.grade -= s2
elif uid.job == 'T':
if (uid.grade-s2) < 0:
uid=Employee(a.name,10,a.year,a.uid)
else:
a.grade -= s2
else:
if (uid.grade-s2) < 0:
uid=Teamleader(a.name,6,a.year,a.uid)
else:
a.grade -= s2
uid.salary()
print(f"升级成功")
print(f"{uid.name},工号:{uid.uid},升级后职位:{uid.job}{uid.grade},升级后薪资:{uid.mony}({uid.mony-moy})")
def main():
count=10000
uid=dict()
while True:
s=input("1.录入;2.查询;3.升级;4.降级;5.退出:")
if s=='1':
uid=create(count)
uid.salary()
print(f"录入成功! 姓名:{uid.name},工号:{uid.uid},薪资:{uid.mony} ")
count +=1
if s=='2':
search(uid)
if s=='3':
up(uid)
if s=='4':
down(uid)
if s=='5':
break
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
千人千面
第 1 题的答案是:
灵活,影响小
第 2 题的答案是:
可以
第 3 题的答案是:
第 4 题的答案是:
高内聚
第 5 题的答案是:
不
-------- 动动手 --------
请将第 0 题的代码写在下方:
member_list = []
class GetMixin:
def get_uid(self):
uid = input("请输入工号:")
self.uid = uid
def get_name(self):
name = input("姓名;")
self.name = name
def get_job(self):
job = input("职位(E.普通员工;T.组长;M.经理):")
self.job = job
def get_grade(self):
grade = int(input("级别:"))
self.grade = grade
def get_year(self):
year = int(input("工龄:"))
self.year = year
def finding(self,uid,member_list):
for i in member_list:
if i.uid == uid:
return i
print("该工号不存在!")
return None
class Salary(GetMixin):
def __init__(self):
pass
def sala(self,e):
pass
def upper(self,e):
pass
class Employee(Salary):
def __init__(self):
super().__init__()
def sala(self,e):
return 3000 + 500*e.grade + 50*e.year
def upper(self,e):
if e.grade >= 10:
e.job = "T"
e.grade = 1
elif e.grade <= 0:
e.job = "E"
e.grade = 1
class Teamleader(Salary):
def __init__(self):
super().__init__()
def sala(self,e):
return 4000 + 800*e.grade + 100*e.year
def upper(self,e):
if e.grade >= 6:
e.job = "M"
e.grade = 1
elif e.grade <= 0:
e.job = "E"
e.grade = 10
class Manager(Salary):
def __init__(self):
super().__init__()
def sala(self,e):
return 5000 + 1000*(e.grade + e.year)
def upper(self,e):
if e.grade >= 3:
e.job = "M"
e.grade = 3
elif e.grade <= 0:
e.job = "T"
e.grade = 6
JOB_MAP = {"E":Employee(),"T":Teamleader(),"M":Manager()}
class Member(GetMixin):
uid = 10000
def __init__(self):
self.uid = str(Member.uid)
Member.uid += 1
self.name = ''
self.job = ''
self.grade = 1
self.year = 0
self.role = None
class Manage(GetMixin):
def welcome(self):
i = 0
while i != '5':
i = input("\n1.录入;2.查询;3.升级;4.降级;5.退出程序:")
if i == '1':
self.enter()
elif i == '2':
self.find()
elif i == '3':
self.up()
elif i == '4':
self.low()
elif i == '5':
break
def enter(self):
e = Member()
e.get_name()
e.get_job()
e.get_grade()
e.get_year()
e.role = JOB_MAP
member_list.append(e)
sal = e.role.sala(e)
print(f"录入成功!姓名:{e.name},工号:{e.uid},薪资:{sal}")
def find(self):
o_find = input("1.员工查询;2.职位查询:")
if o_find == '1':
uid = input("请输入工号:")
f = self.finding(uid,member_list)
if f:
new_sal = f.role.sala(f)
print(f"姓名:{f.name}\n职位:{f.job}\n级别:{f.grade}\n工龄:{f.year}")
else:
j = input("职位(E.普通员工;T.组长;M.经理):")
p = 0
for n in member_list:
if n.job == j:
p += 1
print(f"目前共有{p}人:\n{n.uid} - {n.name}")
def up(self):
uid = input("请输入工号:")
f = self.finding(uid,member_list)
g = int(input("请输入需要增加的级数:"))
f.grade += g
f.role.upper(f)
f.role = JOB_MAP
new_sal = f.role.sala(f)
print("升级成功!")
print(f"{f.name},工号:{f.uid},升级后职位:{f.job}{f.grade},升级后薪资:{new_sal}")
def low(self):
uid = input("请输入工号:")
f = self.finding(uid,member_list)
l = int(input("请输入需要减少的级数:"))
f.grade -= l
f.role.upper(f)
f.role = JOB_MAP
new_sal = f.role.sala(f)
print("降级成功!")
print(f"{f.name},工号:{f.uid},降级后职位:{f.job}{f.grade},降级后薪资:{new_sal}")
def main():
m = Manage()
m.welcome()
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
针对同一个函数/方法,根据传入的对象不同会有不同的结果
第 1 题的答案是:
可以直接封装好一段代码,然后调用不同类的对象可以实现不同的功能,一段代码就可以实现多种功能
第 2 题的答案是:
可以
第 3 题的答案是:
只要一只鸟走起来像鸭子,叫起来像鸭子,那么它就是鸭子
第 4 题的答案是:
增强了内聚
第 5 题的答案是:
属于,这里相当于传入了不同的对象能够有不同的结果
-------- 动动手 --------
请将第 0 题的代码写在下方:
class Staff:
def __init__(self, uid, name, job, grade, year):
self.uid = uid
self.name = name
self.job = job
self.grade = grade
self.year = year
def get_uid(self):
return self.uid
def get_name(self):
return self.name
def get_job(self):
return self.job
def get_grade(self):
return self.grade
def get_year(self):
return self.year
def get_salary(self):
pass
class Employee(Staff):
def __init__(self, uid, name, job, grade, year):
super().__init__(uid, name, job, grade, year)
def get_salary(self):
return 3000+500*self.grade+50*self.year
class Teamleader(Staff):
def __init__(self, uid, name, job, grade, year):
super().__init__(uid, name, job, grade, year)
def get_salary(self):
return 4000+800*self.grade+100*self.year
class Manager(Staff):
def __init__(self, uid, name, job, grade, year):
super().__init__(uid, name, job, grade, year)
def get_salary(self):
return 5000+1000*(self.grade+self.year)
def enter(staff_lis):
i = 0
while True:
uid = 10000 + i
name = input("姓名:")
job = input("职位(E.普通员工;T.组长;M.经理):")
year = int(input("工龄:"))
grade = int(input("级别:"))
if job == "E":
while grade > 10:
grade = int(input("该职位最高级别为10,请重新录入级别:"))
staff = Employee(uid, name, job, grade, year)
elif job == "T":
while grade > 6:
grade = int(input("该职位最高级别为6,请重新录入级别:"))
staff = Teamleader(uid, name, job, grade, year)
elif job == "M":
while grade > 3:
grade = int(input("该职位最高级别为3,请重新录入级别:"))
staff = Manager(uid, name, job, grade, year)
staff_lis.append(staff)
print(f"录入成功!姓名:{staff.name}, 工号:{staff.uid},薪资:{staff.get_salary()}")
yield staff
i += 1
def query(staff_lis):
cmd = int(input("1.员工查询;2.职位查询:"))
if cmd == 1:
uid_q = int(input("请输入工号:"))
sum = 0
for x in staff_lis:
if uid_q == x.uid:
sum += 1
print("姓名:", x.name)
print("职位:", x.job)
print("级别:", x.grade)
print("工龄:", x.year)
print("薪资:", x.get_salary())
if sum == 0:
print("该工号不存在!")
elif cmd == 2:
job_dic = {"E":"普通员工", "T":"组长", "M":"经理"}
job_q = input("职位(E.普通员工;T.组长;M.经理):")
job_name = job_dic
staff_q_lis =
if len(staff_q_lis) == 0:
print(f"目前公司没有{job_name}")
else:
print(f"目前{job_name}共有{len(staff_q_lis)}人")
for each in staff_q_lis:
print(f"{each.uid} - {each.name}")
def up(staff_lis):
uid = int(input("请输入工号:"))
sum = 0
for x in staff_lis:
if x.uid == uid:
index = staff_lis.index(x)
sum += 1
salary_org = x.get_salary()
print(x.name, "工号:", x.uid, "当前职位:", f"{x.job}", x.grade,
"当前薪资:", salary_org)
up_num = int(input("请输入需要增加的级数:"))
level = x.grade + up_num
if x.job == "E":
if level > 10:
x.grade = 1
x.job = "T"
staff_lis = Teamleader(x.uid, x.name, x.job, x.grade, x.year)
x = staff_lis
else:
x.grade = level
print("升级成功!")
elif x.job == "T":
if level > 6:
x.grade = 1
x.job = "M"
staff_lis = Manager(x.uid, x.name, x.job, x.grade, x.year)
x = staff_lis
else:
x.grade = level
print("升级成功!")
elif x.job == "M":
if x.grade == 3:
print("您已到最高级别,无法再升级")
elif level > 3:
x.grade = 3
print("升级成功!")
else:
x.grade = level
print("升级成功!")
print(x.name, "工号:", x.uid, "升级后职位:", f"{x.job}", x.grade,
"当前薪资:", x.get_salary(), f"(+{x.get_salary() - salary_org})")
if sum == 0:
print("查无此人")
def down(staff_lis):
uid = int(input("请输入工号:"))
sum = 0
for x in staff_lis:
if x.uid == uid:
index = staff_lis.index(x)
sum += 1
salary_org = x.get_salary()
print(x.name, "工号:", x.uid, "当前职位:", f"{x.job}:", x.grade,
"薪资:", salary_org)
down_num = int(input("请输入需要减少的级数:"))
level = x.grade - down_num
if x.job == "E":
if level <= 0:
x.grade = 1
x.job = "E"
else:
x.grade = level
print("降级成功!")
elif x.job == "T":
if level <= 0:
x.grade = 10
x.job = "E"
staff_lis = Employee(x.uid, x.name, x.job, x.grade, x.year)
x = staff_lis
else:
x.grade = level
print("降级成功!")
elif x.job == "M":
if level <= 0:
x.grade = 6
x.job = "T"
staff_lis = Teamleader(x.uid, x.name, x.job, x.grade, x.year)
x = staff_lis
else:
x.grade = level
print("降级成功!")
print(x.name, "工号:", x.uid, "升级后职位:", f"{x.job}:", x.grade,
"薪资:", x.get_salary(), f"({x.get_salary() - salary_org})")
if sum == 0:
print("查无此人")
def main():
staff_lis = []
iterator = enter(staff_lis)
while True:
command = int(input("1.录入;2.查询;3.升级;4.降级;5.退出:"))
if command == 1:
next(iterator)
elif command == 2:
query(staff_lis)
elif command == 3:
up(staff_lis)
elif command == 4:
down(staff_lis)
elif command == 5:
break
if __name__ == '__main__':
main()
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
在子类中对父类的方法再次定义
第 1 题的答案是:
减少重复操作可以依次输入根据情况有不同结果
第 2 题的答案是:
可以
第 3 题的答案是:
看着是鸭子.........
第 4 题的答案是:
鸭子类型和其他模块毫无瓜葛只是被顶层调用降低了耦合
第 5 题的答案是:
不是
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:可以
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:根据不同的对象执行不同的操作
第 1 题的答案是:减少代码重复
第 2 题的答案是:可以
第 3 题的答案是:看起来像鸭子,吃起来也像,那就是鸭子
第 4 题的答案是:增强内聚
第 5 题的答案是:是
-------- 动动手 --------
请将第 0 题的代码写在下方:
class people:
def __init__(self,name,job,grade,year,uid):
self.name = name #姓名
self.job = job #职位
self.grade = grade #级别
self.year = year #工龄
self.uid = uid #工号
def salary(self):
if self.job == 'E':
return 3000 + 500 * int(self.grade) + 50 * int(self.year)
if self.job == 'T':
return 4000 + 800 * int(self.grade) + 100 * int(self.year)
if self.job == 'M':
return 5000 + 1000 * (int(self.grade) + int(self.year))
def main():
word = {}
uid = 10000
while 1:
print()
x = input("1.录入;2.查询;3.升级;4.降级;5.退出:")
if x == '1':
name = input("姓名:")
job = input("职位(E.普通员工;T.组长;M.经理):")
year = input("工龄:")
grade = input("级别:")
while job == "E" and int(grade) > 10:
grade = input("该职位最高级别为10,请重新录入级别:")
while job == "T" and int(grade) > 6:
grade = input("该职位最高级别为6,请重新录入级别:")
while job == "M" and int(grade) > 3:
grade = input("该职位最高级别为3,请重新录入级别:")
word = people(name,job,grade,year,uid)
print(f"录入成功!姓名:{name},工号:{uid},薪资:{word.salary()}")
uid += 1
elif x == '2':
y = input("1.员工查询;2.职位查询:")
if y == "1":
z = input("请输入工号:")
for i in word:
if i == z:
print(f"姓名:{word.name}")
print(f"职位:{word.job}")
print(f"级别:{word.grade}")
print(f"工龄:{word.year}")
print(f"薪资:{word.salary()}")
break
else:
print("该工号不存在!")
elif y == "2":
job = input("职位(E.普通员工;T.组长;M.经理):")
while 1:
num = []
for i in word:
if word.job == job:
num.append(i)
if job == "E":
print(f"目前普通员工共有{len(num)}人:")
for i in num:
print(f"{i} - {word.name}")
break
elif job == "T":
print(f"目前组长共有{len(num)}人:")
for i in num:
print(f"{i} - {word.name}")
break
elif job == "M":
print(f"目前经理共有{len(num)}人:")
for i in num:
print(f"{i} - {word.name}")
break
else:
job = input("职位错误!请重新输入职位:")
elif x == '3':
y = input("请输入工号:")
try:
print(f"{word.name},工号:{word.uid},当前职位:{word.job + word.grade},当前薪资:{word.salary()}")
z = input("请输入需要增加的级数:")
if word.job == 'E' and int(word.grade) + int(z) > 10:
word.job = 'T'
word.grade = '1'
print("升级成功!")
elif word.job == 'T' and int(word.grade) + int(z) > 6:
word.job = 'M'
word.grade = '1'
print("升级成功!")
elif word.job == 'M' and int(word.grade) + int(z) > 3:
print("升级失败!经理最高只能为3级")
else:
word.grade = str(int(word.grade) + int(z))
print(f"{word.name},工号:{word.uid},升级后职位:{word.job + word.grade},升级后薪资:{word.salary()}")
except:
print("错误!该工号不存在!")
elif x == '4':
y = input("请输入工号:")
try:
print(f"{word.name},工号:{word.uid},当前职位:{word.job + word.grade},当前薪资:{word.salary()}")
z = input("请输入需要减少的级数:")
if word.job == 'E' and int(word.grade) - int(z) < 1:
word.pop(y)
print("降级成功!该员工已被辞退!")
elif word.job == 'T' and int(word.grade) - int(z) < 1:
word.job = 'E'
word.grade = '10'
print("降级成功!")
elif word.job == 'M' and int(word.grade) - int(z) < 1:
word.job = 'T'
word.grade = '6'
print("降级成功!")
else:
word.grade = str(int(word.grade) - int(z))
print(f"{word.name},工号:{word.uid},降级后职位:{word.job + word.grade},降级后薪资:{word.salary()}")
except:
print("错误!该工号不存在!")
elif x == '5':
break
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:可以
第 4 题的答案是:降低耦合
第 5 题的答案是:不是
-------- 动动手 --------
请将第 0 题的代码写在下方:
class Employee:
def __init__(self,name,job,grade,year,uid):
self.name = name
self.job = job
self.grade = grade
self.year = year
self.uid = uid
def get_uid(self):
return self.uid
def get_name(self):
return self.name
def get_job(self):
return self.job
def get_year(self):
return self.year
def get_grade(self):
return self.grade
def salary(self):
return 3000 + 500*self.grade +50*self.year
class Teamleader(Employee):
super().__init__()
super().get_uid()
super().get_name()
super().get_job()
super().get_year()
super().get_grade()
def salary(self):
return 4000 + 800*self.grade + 100*self.year
class Manager(Employee):
super().__init__()
super().get_uid()
super().get_name()
super().get_job()
super().get_year()
super().get_grade()
def salary(self):
return 5000 + 1000*(self.grade + self.year)
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
第 1 题的答案是:
第 2 题的答案是:
第 3 题的答案是:
第 4 题的答案是:
第 5 题的答案是:
-------- 动动手 --------
请将第 0 题的代码写在下方:
请回复您的答案^_^
-------- 问答题 --------
第 0 题的答案是:
靠鸭子类型(不强制继承 / 接口)
第 1 题的答案是:
让代码更灵活、更好扩展、更易维护,同时减少重复逻辑。
第 2 题的答案是:
可以
第 3 题的答案是:
如果它走起路来像鸭子,叫起来也像鸭子,那它就是鸭子。
第 4 题的答案是:
鸭子类型优先体现为降低耦合,同时有利于高内聚。
第 5 题的答案是:
不是
-------- 动动手 --------
请将第 0 题的代码写在下方:
不会