to computer science:第7课练习2

Question

以下是我如何回答这个问题。我怎样才能更好地解决这个问题？

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定义一个过程，stamps，它以正整数为输入，以便士为单位，并返回组成该值所需的5p、2p和1p邮票(p为便士)的数目。返回值应该是三个数字的元组(也就是说，您的返回语句后面应该是5p、2p和1p邮票的编号)。你的答案应该使用尽可能少的总邮票，首先使用尽可能多的5便士邮票，然后是2便士邮票，最后是1便士邮票，以构成总数。(对于USians来说，仅仅说使用“永久”邮票并完成它是不公平的！)

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这里是我的解决方案

def stamps(i):
    # Your code here
    five = 0
    two = 0
    one = 0
    while i > 0:
        if i == 0:
            break
        if i >= 5:
            five = five + 1
            i = i - 5
        if i == 0:
            break
        if i < 5 or i == 2:
            two = two + 1
            i = i - 2
        if i == 0:
            break
        if i < 2 or i == 1:
            one = one + 1
            i = i - 1
    return five,two,one

下面是练习中的测试

print stamps(8)
#>>> (1, 1, 1)  # one 5p stamp, one 2p stamp and one 1p stamp
print stamps(5)
#>>> (1, 0, 0)  # one 5p stamp, no 2p stamps and no 1p stamps
print stamps(29)
#>>> (5, 2, 0)  # five 5p stamps, two 2p stamps and no 1p stamps
print stamps(0)
#>>> (0, 0, 0) # no 5p stamps, no 2p stamps and no 1p stamps

Answer 1

我将使用模块操作和剩余操作：

def modulo_and_remainder(a, b):
    return a//b, a %b

def stamps(a):
    five, rem = modulo_and_remainder(a, 5)
    two, one = modulo_and_remainder(rem, 2)
    return five, two, one

或者(甚至不知道这一点)您可以使用内置的divmod：

def stamps(a):
    five, rem = divmod(a, 5)
    two, one = divmod(rem, 2)
    return five, two, one

Answer 2

为了使它更加通用，一个函数需要一个邮票类型的元组：

def stamps(postage_cost,stamps):
        stamps_required = []
        for stamp in stamps:
            (num_stamps,remainder) = divmod(postage_cost,stamp)
            stamps_required.append(num_stamps)
            postage_cost = remainder
        return tuple(stamps_required)

stamp_types = (5,2,1)
required = stamps(8,stamp_types)
print(required)

Answer 3

def stamps(x): return (x / 5, (x - x / 5 * 5) / 2, x - (x / 5 * 5 + (x - x / 5 * 5) / 2 * 2))