Dictionaries 部分跳过
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
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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
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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\(\frac{amount}{price}+1\) 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
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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
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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