Divide

传入一组商数与一组除数，返回字典，键为每个商数，值为表示每个商数能整除的除数的列表

def divide(quotients, divisors):
    """Return a dictonary in which each quotient q is a key for the list of
    divisors that it divides evenly.

    >>> divide([3, 4, 5], [8, 9, 10, 11, 12])
    {3: [9, 12], 4: [8, 12], 5: [10]}
    >>> divide(range(1, 5), range(20, 25))
    {1: [20, 21, 22, 23, 24], 2: [20, 22, 24], 3: [21, 24], 4: [20, 24]}
    """
    return {i: [x for x in divisors if x%i==0] for i in quotients}

注意用表达式建立列表与字典的方法

Buying Fruit

实现buy函数，通过给定的水果与价格，用恰好为传入的total_amount的价格购买指定的水果（每种指定水果至少买一次）用display函数输出所有结果

def buy(required_fruits, prices, total_amount):
    """Print ways to buy some of each fruit so that the sum of prices is amount.

    >>> prices = {'oranges': 4, 'apples': 3, 'bananas': 2, 'kiwis': 9}
    >>> buy(['apples', 'oranges', 'bananas'], prices, 12)
    [2 apples][1 orange][1 banana]
    >>> buy(['apples', 'oranges', 'bananas'], prices, 16)
    [2 apples][1 orange][3 bananas]
    [2 apples][2 oranges][1 banana]
    >>> buy(['apples', 'kiwis'], prices, 36)
    [3 apples][3 kiwis]
    [6 apples][2 kiwis]
    [9 apples][1 kiwi]
    """
    def add(fruits, amount, cart):
        if fruits == [] and amount == 0:
            print(cart)
        elif fruits and amount > 0:
            fruit = fruits[0]
            price = prices[fruit]
            for k in range(1,amount//price+1):
                add(fruits[1:], amount-k*price, cart+display(fruit,k))
    add(required_fruits, total_amount, '')

def display(fruit, count):
    """Display a count of a fruit in square brackets.

    >>> display('apples', 3)
    '[3 apples]'
    >>> display('apples', 1)
    '[1 apple]'
    """
    assert count >= 1 and fruit[-1] == 's'
    if count == 1:
        fruit = fruit[:-1]  # get rid of the plural s
    return '[' + str(count) + ' ' + fruit + ']'

返回条件是fruits为空或钱被花光，且递归时每次都取fruits的首个元素可以推断递归时对fruits数组进行了切片，每次往后切一个元素 for循环遍历选择每个水果的个数，从至少选一个到最多能选的个数递归时传入总价格减去已经使用的价格数的不同情况，并使用display函数进行字符串拼接来显示

Cities ADT

以下问题基于建立的该ADT 一个城市由以下参数描述：名称、经度、维度 包括一个构造函数 - make_city(name, lat, lon)：建立一个城市对象，存储其名称、经度、维度以下选择器 - get_name(city)：获取城市名称 - get_lat(city)：获取城市经度 - get_lon(city)：获取城市维度该抽象数据类型已经在文件中实现，你不需要知道是怎么实现的

Distance

计算并返回两城市的距离

from math import sqrt
def distance(city_a, city_b):
    """
    >>> city_a = make_city('city_a', 0, 1)
    >>> city_b = make_city('city_b', 0, 2)
    >>> distance(city_a, city_b)
    1.0
    >>> city_c = make_city('city_c', 6.5, 12)
    >>> city_d = make_city('city_d', 2.5, 15)
    >>> distance(city_c, city_d)
    5.0
    """
    return sqrt(abs(get_lat(city_a)-get_lat(city_b))**2+abs(get_lon(city_a)-get_lon(city_b))**2)

Closer City

比较两个城市离给定经纬度的远近，返回更近的那个城市

def closer_city(lat, lon, city_a, city_b):
    """
    Returns the name of either city_a or city_b, whichever is closest to
    coordinate (lat, lon). If the two cities are the same distance away
    from the coordinate, consider city_b to be the closer city.

    >>> berkeley = make_city('Berkeley', 37.87, 112.26)
    >>> stanford = make_city('Stanford', 34.05, 118.25)
    >>> closer_city(38.33, 121.44, berkeley, stanford)
    'Stanford'
    >>> bucharest = make_city('Bucharest', 44.43, 26.10)
    >>> vienna = make_city('Vienna', 48.20, 16.37)
    >>> closer_city(41.29, 174.78, bucharest, vienna)
    'Bucharest'
    """
    fake_city = make_city('fake', lat, lon)
    if distance(fake_city,city_a)<distance(fake_city,city_b):
        return get_name(city_a)
    else:
        return get_name(city_b)

可以将指定的经纬度看作第三个城市并构造这么一个对象通过上面已经写完的distance函数计算两城市与第三个城市的距离并比较，输出更近的若相同输出city_b

本作品采用知识共享署名-相同方式共享 4.0 国际许可协议进行许可

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【CS61A】CS61A——Lab4

Divide

Buying Fruit

Cities ADT

Distance

Closer City