乘以倍数？；“什么也不做”；创建除法或类似除法的算法

Question

还在解决啤酒问题。重新开始。我尽量不买半瓶啤酒，所以我从这里开始：

#import math

def buybeers(wallet):
    beers = 0
    for(int in xrange(0,wallet,2)) #xrange not defined
        beers += 1
        wallet -= 2
    return beers, wallet

wallet = int(input("How many wallet do you have?"))

print(buybeers(wallet))

我在想怎么花两美元买一瓶啤酒，然后把零钱还回去。你知道吗

第二次尝试，尝试查找逻辑错误：

def buybeers(wallet):
    beers = 0
    for i in range(1,wallet,3):
        beers += 1
        wallet -= 3
    return beers, wallet

wallet = int(input("How many wallet do you have?"))

print(buybeers(wallet))

当我有14美元的时候，一个问题出现了。最后我得到了-1美元和5瓶啤酒（一瓶我不该喝的啤酒，还欠我1美元）。虽然这是正确的，但一些便利店不接受标签。你知道吗

最终提交（啤酒3美元）

def buybeers(wallet):
    beers = 0
    for i in range(1,wallet,3):
        if wallet >= 3:  #if you got enough for a beer, buy one.
            beers += 1
            wallet -= 3
    #else:
    #    pass #otherwise, "do nothing"/pass (it seems this is automated).
    return beers, wallet

wallet = int(input("How many wallet do you have?"))

print(buybeers(wallet))

Answer 1

你少了几个括号，你的逻辑是错误的。假设你有$5，每瓶啤酒的价格是$2，那么你就可以有2瓶啤酒，剩下$1，但是，你的代码会给你3瓶啤酒，现在你又欠了一美元。你可以用除法和模来简化这个

def buybeers(wallet):
    beers = wallet//2
    wallet = wallet % 2

    return beers, wallet

wallet = int(input("How many wallet do you have?"))

print(buybeers(wallet)) #5 as input

(2, 1)

只是修复语法错误

def buybeers(wallet):
    beers = 0
    for i in xrange(0,wallet,2):
        beers += 1
        wallet -= 2
    return beers, wallet

wallet = int(input("How many wallet do you have?"))

print(buybeers(wallet)) #5 as input

(3, -1)

因为for-loop逻辑已关闭，所以无法提供正确的输出。你知道吗