自动售货机程序（计算必须插入的数量等）

Question

我想写一个程序来模拟自动售货机，并根据支付的金额计算出零钱（必须退还给你）。考虑到成本，首先应该提示用户添加更多的钱，直到支付满足/超过成本。在

假设所有的零钱都是用硬币换的，硬币的面额是：1c，5c，10c，25c，$1

这是我的计划：

 x = eval(input("Enter the cost (in cents):\n"))
 b = 0

 for i in range(x+500):        
    if x<5 and x>=b:
        b += 1
        print("Deposit a coin or note (in cents):")
        print(1)
        diff = b-x
        for i in range(diff):
            onecents = diff//1
            new_onecents = diff - (onecents*1)
            print("Your change is:")
            if onecents != 0:
               print(onecents,"x 1c")                  
            break   


    elif x<10 and x>=b:
        b += 5
        print("Deposit a coin or note (in cents):")
        print(5)
        diff = b-x
        for i in range(diff):
            fivecents = diff//5
            new_fivecents = diff - (fivecents*5)
            onecents = new_fivecents//1
            new_onecents = new_fivecents - (onecents*1)
            print("Your change is:")
            if fivecents != 0:
                print(fivecents,"x 5c")
            if onecents != 0:
                print(onecents,"x 1c")                          
            break      

    elif x<25 and x>=b:
        b += 10
        print("Deposit a coin or note (in cents):")
        print(10)
        diff = b-x
        for i in range(diff):
            tencents = diff//10
            new_tencents = diff - (tencents*10)
            fivecents = new_tencents//5
            new_fivecents = new_tencents - (fivecents*5)
            onecents = new_fivecents//1
            new_onecents = new_fivecents - (onecents*1)
            print("Your change is:")
            if tencents !=0:
                print(tencents,"x 10c")
            if fivecents != 0:
                print(fivecents,"x 5c")                        
            if onecents != 0:
                print(onecents,"x 1c")                      
            break        

    elif x<100 and x>=b:
        b += 25
        print("Deposit a coin or note (in cents):")
        print(25)
        diff= b-x
        for i in range(diff):
            quarters = diff//25
            new_quarters = diff - (quarters*25)
            tencents = new_quarters//10
            new_tencents = new_quarters - (tencents*10)
            fivecents = new_tencents//5
            new_fivecents = new_tencents - (fivecents*5)
            onecents = new_fivecents//1
            new_onecents = new_fivecents - (onecents*1)
            print("Your change is:")
            if quarters !=0:
                print(quarters,"x 25c")     
            if tencents !=0:
                print(tencents,"x 10c")                    
            if fivecents != 0:
                print(fivecents,"x 5c")                    
            if onecents != 0:
                print(onecents,"x 1c")                  
            break

    elif x<500 and x>b:
        print("Deposit a coin or note (in cents):")
        print(100)
        b += 100
        diff = b-x
        for i in range(diff):
            quarters = diff//25
            new_quarters = diff - (quarters*25)
            tencents = new_quarters//10
            new_tencents = new_quarters - (tencents*10)
            fivecents = new_tencents//5
            new_fivecents = new_tencents - (fivecents*5)
            onecents = new_fivecents//1
            new_onecents = new_fivecents - (onecents*1)
            print("Your change is:")
            if quarters !=0:
                print(quarters,"x 25c")
            if tencents !=0:
                print(tencents,"x 10c")        
            if fivecents != 0:
                print(fivecents,"x 5c")            
            if onecents != 0:
                print(onecents,"x 1c")
            break             

    elif x<(x+500) and x>=b:
        print("Deposit a coin or note (in cents):")
        print(500)
        b += 500
        diff = b-x
        for i in range(diff):
            onedollars = diff//100
            new_onedollars = diff - (onedollars * 100)
            quarters = new_onedollars//25
            new_quarters = new_onedollars - (quarters*25)
            tencents = new_quarters//10
            new_tencents = new_quarters - (tencents*10)
            fivecents = new_tencents//5
            new_fivecents = new_tencents - (fivecents*5)
            onecents = new_fivecents//1
            new_onecents = new_fivecents - (onecents*1)
            print("Your change is:")
            if onedollars != 0:
                print(onedollars,"x $1")
            if quarters !=0:
                print(quarters,"x 25c")
            if tencents !=0:
                print(tencents,"x 10c")        
            if fivecents != 0:
                print(fivecents,"x 5c")            
            if onecents != 0:
                print(onecents,"x 1c")
            break 

当我运行此程序并按照说明操作时，它应该如下所示：

^{pr2}$

相反，我得到：

Enter the cost (in cents):
1000
Deposit a coin or note (in cents):
500
Deposit a coin or note (in cents):
500
Deposit a coin or note (in cents):
500
Your change is:
5 x $1

另一个预期产出：

Enter the cost (in cents):
3
Deposit a coin or note (in cents):
1
Deposit a coin or note (in cents):
1
Deposit a coin or note (in cents):
1

但是我得到：

Enter the cost (in cents):
3
Deposit a coin or note (in cents):
1 
Deposit a coin or note (in cents):
1
Deposit a coin or note (in cents):
1
Deposit a coin or note (in cents):
1
Your change is:
1 x 1c

其余的都是正确的。在

谢谢你们所有的帮手（尤其是@jornsharpe）。以下是解决方案（以代码形式）：

def vend():
    """Simulate a vending machine, taking user input and returning remainder."""
    total = eval(input("Enter the cost (in cents):\n"))
    inserted = 0
    while inserted < total:
        inserted += eval(input("Deposit a coin or note (in cents):\n"))
    if inserted > total:
        sum = inserted - total
        if sum != 0:
            print("Your change is:")
        dollars = sum//100
        if dollars != 0:
            print(dollars,'x $1')
        quarters = (sum - dollars*100)//25
        if quarters != 0:
            print(quarters,'x 25c')
        ten_cents = (sum - dollars*100 - quarters*25)//10
        if ten_cents != 0:
            print(ten_cents,'x 10c')
        five_cents = (sum - dollars*100 - quarters*25 - ten_cents*10)//5
        if five_cents != 0:
            print(five_cents,'x 5c')
        one_cents = (sum - dollars*100 - quarters*25 - ten_cents*10 - five_cents*5)//1
        if one_cents != 0:
            print(one_cents,'x 1c')

vend()

Answer 1

你的错误源于这样一个事实：你没有正确地处理你完全达到总数的情况——你超出了，然后不得不做出改变。但是，您的代码非常长且复杂，很难确切地知道它在每个阶段都在做什么。在

一些一般的编码建议：

1. eval不是一个好主意；最好使用int(input(...))，如果用户不输入整数，这也会给你带来麻烦。在
2. 您的外部for循环应该是while循环，而不是猜测迭代的最大数量（即使在您进行了更改之后，这些迭代都会运行！）在
3. 你有很多重复的代码和硬编码的值；每当你反复写类似的东西时，考虑把它们分成一个带参数的函数，或者某种循环。在

另外，阅读你的描述，特别是：

Given the cost, the user should first be prompted to add more money until the cost is met/exceeded by the payment.

我认为你应该允许用户使用input硬币，而不是猜测他们将输入什么。在

以下是一种可能的实现方式：

def vend():
    """Simulate a vending machine, taking user input and returning remainder."""
    total = int(input("Enter the cost (in cents): "))
    inserted = 0
    while inserted < total:
        inserted += int(input("Deposit a coin or note (in cents): "))
    if inserted > total:
        return make_change(inserted - total)

def make_change(remainder):
    """Calculate the coins required to make change equal to amount."""
    coins = ["$1", "25c", "10c", "5c", "1c"]
    amounts = [int(coin[:-1]) if coin.endswith("c") 
               else 100 * int(coin[1:]) 
               for coin in coins]
    counts = [0 for _ in coins]
    for index, amount in enumerate(amounts):
        counts[index] = remainder // amount
        remainder %= amount
    return ", ".join("{0} x {1}".format(count, coin) 
                     for count, coin in zip(counts, coins) 
                     if count)

请注意这两个函数之间的职责划分，以便于分别测试每个函数，并适当使用for和{}循环以尽量减少重复。另外，我已经使amounts和paid依赖于coins，所以你只需要在一个地方改变就可以添加新的硬币。在

用法示例：

^{pr2}$