Question

我正在为我的CS课程制作一个改变计算器脚本。代码如下：

# Set up our money variables
centValueOfTenDollarBill = 1000
centValueOfFiveDollarBill = 500
centValueOfToonie = 200
centValueOfLoonie = 100
centValueOfQuarter = 25
centValueOfDime = 10
centValueOfNickel = 5

# Set up our variables
purchaseTotal = input("Enter purchase total: ")  # Purchase costs $12.50
moneyPaid = input("Enter money paid:  ")      # Customer gives cashier $20.00

# Figure out the change
change = moneyPaid - purchaseTotal

# Echo input data to user
print("""The total of the purchase is $%0.2f.
The customer paid $%0.2f.
The cashier gives $%0.2f back to the customer in the following fashion: """ %(purchaseTotal, moneyPaid, change))

#Convert dollars into cents to facilitate the computation
purchaseTotalInCents = purchaseTotal * 100
moneyPaidInCents = moneyPaid * 100
changeInCents = change * 100

# Determine # of $10 to be given back as part of the change
numberOfTenDollarBills = changeInCents // centValueOfTenDollarBill
changeInCents = changeInCents - (centValueOfTenDollarBill * numberOfTenDollarBills)

# Determine # of $5 to be given back as part of the change
numberOfFiveDollarBills = changeInCents // centValueOfFiveDollarBill
changeInCents -= (centValueOfFiveDollarBill * numberOfFiveDollarBills)

# Determine # of $2 (toonies) to be given back as part of the change
numberOfToonieCoins = changeInCents // centValueOfToonie
changeInCents -= (centValueOfToonie * numberOfToonieCoins)

# Determine # of $1 (loonies) to be given back as part of the change
numberOfLoonieCoins = changeInCents // centValueOfLoonie
changeInCents -= (centValueOfLoonie * numberOfLoonieCoins)

# Determine # of $0.25 (quarters) to be given back as part of the change
numberOfQuarterCoins = changeInCents // centValueOfQuarter
changeInCents -= (centValueOfQuarter * numberOfQuarterCoins)

# Determine # of $0.10 (dimes) to be given back as part of the change
numberOfDimeCoins = changeInCents // centValueOfDime  #<--- PROBLEM HERE IF DIMES ARE TWO
print (numberOfDimeCoins)
changeInCents -= (centValueOfDime * numberOfDimeCoins)

# At this point, changeInCents can either be
# 5 -> 1 x $0.05 (nickels) or
# 0 -> 0 x $0.05 (nickels)
numberOfNickelCoins = changeInCents // centValueOfNickel

# Output the result: change cashier needs to give back to customer
print("\t%i x $10.00" %numberOfTenDollarBills)
print("\t%i x $ 5.00" %numberOfFiveDollarBills)
print("\t%i x $ 2.00" %numberOfToonieCoins)
print("\t%i x $ 1.00" %numberOfLoonieCoins)
print("\t%i x $ 0.25" %numberOfQuarterCoins)
print("\t%i x $ 0.10" %numberOfDimeCoins)
print("\t%i x $ 0.05" %numberOfNickelCoins)

# Indicates the end of execution
print("----\n")

所有这一切都是错误的（至少从我能看到的情况）是，如果该程序应该回馈两个角钱，它会回报一分钱和一个镍，这会让客户缩短5美分。如果它应该回馈一分钱，那么就没有问题了。

示例：假设客户为$ 13.30商品支付了20美元。这个变化是6.70美元。

numberOfDimeCoins = changeInCents // centValueOfDime

上面这一行应该与2.0 = 20.0//10.0相同，而是返回1.0。

如果您支付20美元购买一分钱需要回馈的东西，一切都是正确的，如13.20美元，13.90美元或13.75美元。

以下是一些示例输出：

Erics-MacBook-Pro:Desktop eric$ python change.py
Enter purchase total: 13.75
Enter money paid:  20
The total of the purchase is $13.75.
The customer paid $20.00.
The cashier gives $6.25 back to the customer in the following fashion: 
    0 x $10.00
    1 x $ 5.00
    0 x $ 2.00
    1 x $ 1.00
    1 x $ 0.25
    0 x $ 0.10
    0 x $ 0.05
----

Erics-MacBook-Pro:Desktop eric$ python change.py
Enter purchase total: 12.8
Enter money paid:  20
The total of the purchase is $12.80.
The customer paid $20.00.
The cashier gives $7.20 back to the customer in the following fashion: 
    0 x $10.00
    1 x $ 5.00
    1 x $ 2.00
    0 x $ 1.00
    0 x $ 0.25
    1 x $ 0.10
    1 x $ 0.05
----

我有什么遗漏或做错了吗？

使用Python 2.7。

Answer 1

如果检查变量，您将看到问题

>>> changeInCents
19.999999999999886
>>> centValueOfDime
10

这是由于浮点精度有限。

您应该将初始值转换为美分。例如

numberspurchaseTotalInCents = int(purchaseTotal * 100)
moneyPaidInCents = int(moneyPaid * 100)
changeInCents = moneyPaidInCents - numberspurchaseTotalInCents

同时查看内置divmod()功能

Answer 2

这是因为浮点精度问题。试试这个：

purchaseTotal = 13.3
moneyPaid = 20

change = moneyPaid - purchaseTotal
print(repr(change))  # 6.699999999999999

您可能希望change为0.7，但实际上它是一个非常接近0.7的数字，但并不精确。最后，changeInCents会得到一个您希望为20.0的数字，但实际上要小一些。

Answer 3

我试图调试你的代码，但其他伙伴已经弄明白了。您对按照IEEE 754标准存储数字的方式存在问题（您可以了解更多信息here）。

我建议您使用十进制（阅读here）

这很直接，

#import and set precision
from decimal import *
getcontext().prec = **x** #you decide it
a = Decimal(10)
b = Decimal(0.9)
print a-b

阅读文档，它允许您设置很多关于如何舍入数字的相关问题。

Python中的'//'运算符没有返回正确的解决方案

3 个答案: