Python---面向对象---龟鱼游戏

一、定义一个门票系统

门票的原价是100元

当周末的时候门票涨价20%

小孩子半票

计算2个成人和1个小孩的平日票价

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class Ticket():
    def __init__(self, weekend=False, child=False):
        self.exp = 100
        if weekend:
            self.inc = 1.2
        else:
            self.inc = 1

        if child:
            self.discount = 0.5
        else:
            self.discount = 1

    def cal_price(self, num):
        return self.exp * self.inc * self.discount * num

adult = Ticket()
child = Ticket(child=True)

print("两个成年人和一个小孩子平日的价格是{}".format(adult.cal_price(2)+ child.cal_price(1)))
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游戏编程：按一下要求定义一个乌龟类和鱼类并尝试编程
一、假设游戏场景为范围（x,y）为0<=x<=10,0<=y<=10
二、游戏生成1只乌龟和10条鱼
三、他们的移动方向均随机
四、乌龟的最大移动能力是2（乌龟可以随机选择移动是1还是2），鱼的最大移动能力是1
五、当移动到场景边缘，自动向反方向移动
六、乌龟没移动一次，体力消耗1
七、当乌龟和鱼重叠，乌龟吃掉鱼，乌龟体力增加20
八、鱼不计算体力
九、当乌龟体力值为0或者鱼的数量为0时，游戏结束
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import random as r

class Turtle(object):
    def __init__(self):
        self.power = 100

        #初始化乌龟的位置
        self.x = r.randint(0, 10)
        self.y = r.randint(0, 10)

    def move(self):
        new_x = r.choice([1, 2, -1, -2]) + self.x
        new_y = r.choice([1, 2, -1, -2]) + self.y

        # 判断   乌龟的移动是否超出了边界

        if new_x < 0:
            self.x = 0 - (new_x - 0)
        elif new_x > 10:
            self.x = 10 - (new_x - 10)
        else:
            self.x = new_x

        if new_y < 0:
            self.y = 0 - (new_y - 0)
        elif new_y > 10:
            self.y = 10 - (new_y - 10)
        else:
            self.y = new_y

        self.power -= 1
        return (self.x, self.y)

    def eat(self):
        self.power += 20
        if self.power >= 100:
            self.power = 100

class Fish(object):

    def __init__(self):
        self.x = r.randint(0, 10)
        self.y = r.randint(0, 10)

    def move(self):
        new_x = self.x + r.choice([1, -1])
        new_y = self.y + r.choice([1, -1])

        if new_x < 0:
            self.x = 0 - (new_x - 0)
        elif new_x > 10:
            self.x = 10 - (new_x - 10)
        else:
            self.x = new_x

        if new_y < 0:
            self.y = 0 - (new_y - 0)
        elif new_y > 10:
            self.y = 10 - (new_y - 10)
        else:
            self.y = new_y

        return (self.x, self.y)

turtle = Turtle()
fish = []
for i in range(10):
    new_fish = Fish()
    fish.append(new_fish)

while True:
    if not len(fish):
        print("鱼被吃完了，游戏结束")
        break
    if not turtle.power:
        print("乌龟体力被耗尽了，游戏结束了")
        break

    pos = turtle.move()

    # 在迭代中做列表的删除元素是非常危险的，经常会出现一些意想不到的问题，因为迭代器是直接引用列表元素的数据做的操作
    # 所以 我们这里把列表拷贝一份传给迭代器，然后再对原列表做操作
    for each_fish in fish[:]:
        if each_fish.move() == pos:
            turtle.eat()
            fish.remove(each_fish)
            print("有一条鱼被吃掉了")
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定义一个点（point）和直线（Line）类，使用getLen方法获取两点构成直线的长度
-----------------------------------------

import math

class Point(object):
    def __init__(self, x=0, y=0):
        self.x = x
        self.y = y

    def get_x(self):
        return self.x

    def get_y(self):
        return self.y

class Line(object):
    def __init__(self, p1, p2):
        self.x = p1.get_x() - p2.get_x()
        self.y = p1.get_y() - p2.get_y()

        self.len = math.sqrt(self.x*self.x + self.y*self.y)

    def get_len(self):
        return self.len

p1 = Point(2,3)
p2 = Point(5,7)
line = Line(p1, p2)
line.get_len()