Question

我在进行编码挑战时遇到了我从未遇到过的问题。问题要求以最有效的方式退还变更。用户输入价格，并将支付的金额作为数组（有点像向自动售货机捐款）

由于某些原因，当剩余更改为0.01时，代码不会通过最后一个if块

def ChangeMaker(price, payment):
    # Write your code here
    totalPayment=0
    for i in payment:
        totalPayment+=i
    change = totalPayment-price
    coinChange=change-int(change)

    def helper(remain, result):
        holder= result
        newRemain = remain
        if remain>=.25:
            holder[3]+=1
            newRemain-=.25
            return helper(newRemain, holder)
        elif remain>=.10:
            holder[2]+=1
            newRemain -=.10
            return helper(newRemain, holder)
        elif remain>=.5:
            holder[1]+=1
            newRemain -=.05
            return helper(newRemain, holder)
        elif remain>.01 or remain==.01:
            holder[0]+=1
            newRemain -=.01
            return helper(newRemain, holder)
        else:
            return holder;
    return helper(coinChange, [0,0,0,0])

print(ChangeMaker(1.87, [5, .25,.25,.25,.25, 1]))

Answer 1

正如@gmds所说，这纯粹是一个浮点精度问题。这是一种计算货币（仅保留小数点后两位）的快速解决方案，方法是将每个数字计时100，然后将其转换为整数。

def ChangeMaker(price, payment):
    # Write your code here
    totalPayment=int(sum(payment) * 100)
    price = int(price * 100)
    change = totalPayment-price
    coinChange=change % 100

    def helper(remain, result):
        holder= result
        newRemain = remain
        if remain>=25:
            holder[3]+=1
            newRemain-=25
            return helper(newRemain, holder)
        elif remain>=10:
            holder[2]+=1
            newRemain -= 10
            return helper(newRemain, holder)
        elif remain>=5:
            holder[1]+=1
            newRemain -= 5
            return helper(newRemain, holder)
        elif remain >= 1:
            holder[0]+=1
            newRemain -= 1
            return helper(newRemain, holder)
        else:
            return holder
    return helper(coinChange, [0,0,0,0])

print(ChangeMaker(1.87, [5, .25,.25,.25,.25, 1]))

为什么阻止不通过？

1 个答案: