Question

我有一个象牙，我被要求编写一个程序来计算在现金交易后给予客户的变更，并确定每个面额中的纸币和硬币的数量。

用户必须支付商品成本和从客户收到的金额。

我必须有一个类接受十进制参数，Cost和Chashreceived，以及整数参数：Hunderds，Fifties，Twenties，Tens，Fives，Twos，Ones，50c，10c，5c，2c和1c。 / p>

i减去Cost和Cashreceived的成本，并计算退货所需的笔记和硬币的确切数量。

我已经尝试过，但是当我必须放入硬币时会出现问题。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ChangeCalculator
{
class Program
{
    static void Main(string[] args)
    {
        clsCash Money = new clsCash();
        clsCash Paid = new clsCash();
        Console.WriteLine("What is the cost of the goods?");
        Money.Cost = Convert.ToDecimal(Console.ReadLine());
        Console.WriteLine("How much was recived?");
        Paid.CashRecieved = Convert.ToDecimal(Console.ReadLine());
        Money.GetChange(Money.Cost, Paid.CashRecieved);
        Console.Read();
    }
}

class clsCash
{


private decimal cost;
private decimal cashRecieved;
public decimal Cost
{
    get
    {
        return cost;
    }
    set
    {
        cost = value;
    }
}
public decimal CashRecieved
{
    get
    {
        return cashRecieved;
    }
    set
    {
        cashRecieved = value;
    }
}




public void GetChange(decimal Cost, decimal CashRecieved)
{

    decimal change = CashRecieved - Cost;
    int hundreds = 0;
    int fifty = 0;
    int twenty = 0;
    int ten = 0;
    int five = 0;
    int two = 0;
    int one = 0;
    int centsfifty = 0;
    int centsten = 0;
    int centsfive = 0;
    int centstwo = 0;
    int centsone = 0;
    do
   {

    if (change >= 100)
    {
         hundreds = (int)change / 100;
         change = (int)change % 100;
    } //while (change > 0);
    else if (change >= 50)
    {
        fifty = (int)change / 50;
        change = change % 50;
    }
    else if (change >= 20)
    {
        twenty = (int)change / 20;
        change = change % 20;
    }
    else if (change >= 10)
    {
        ten = (int)change / 10;
        change = change % 10;
    }
    else if (change >= 5)
    {
        five = (int)change / 5;
        change = change % 5;
    }
    else if (change >= 2)
    {
        two = (int)change / 2;
        change = change % 2;
    }

    else if (change >= 1)
    {
        one = (int)change / 1;
        change = change % 1;
    }
    else if (change > 1)
    {
        decimal fhu = change / 0.5m;
        centsfifty = (int)fhu;
        change = change % 0.5m;
        Console.WriteLine("YOUR CHANGE IS:");
    }

    } while (change >= 0);


    Console.WriteLine("YOUR CHANGE IS:");
    Console.WriteLine("---------------");
    Console.WriteLine("HUNDREDS RANDS \t:  {0}", hundreds);
    Console.WriteLine("FIFTY RANDS \t:  {0}", fifty);
    Console.WriteLine("TWENTY RANDS \t:  {0}", twenty);
    Console.WriteLine("TEN RANDS \t:  {0}", ten);
    Console.WriteLine("FIVE RANDS \t:  {0}", five);
    Console.WriteLine("TWO RANDS \t:  {0}", two);
    Console.WriteLine("ONE RANDS \t:  {0}", one);
    Console.WriteLine("50 CENTS \t:  {0}", centsfifty);

}
}
}

Answer 1

您可以使用一种模式来获取此类金额，

这是一个让你入门的小例子，你可以将它包装在一个函数或其他东西中，但它可以让你知道从哪里开始。

// number to find values from
int change = 254;

int _hundreds = 100;
int _fifty = 50;
int _twenty = 20;
int _ten = 10;
int _five = 5;
int _two = 2;
int _one = 1;

int hundreds = (int)(change / _hundreds);
int fifty = (int)((change % _hundreds) / _fifty);
int twenty = (int)(((change % _hundreds) % _fifty) / _twenty);
int ten = (int)((((change % _hundreds) % _fifty) % _twenty) / _ten);
int five = (int)(((((change % _hundreds) % _fifty) % _twenty) % _ten) / _five);
int two = (int)((((((change % _hundreds) % _fifty) % _twenty) % _ten) % _five) / _two);
int one = (int)(((((((change % _hundreds) % _fifty) % _twenty) % _ten) % _five) % _two) / _one);

返回

hundreds = 2
fifty = 1
twenty = 0
ten = 0
five = 0
two = 2
one = 0;

Answer 2

帮助简化代码并提高代码可读性的一种方法是消除条件和条件嵌套算术运算，几乎总是让代码更难理解。

我看到int被抛出太多次了，这可能意味着从一开始就使用int是最好的。尽管您将成本和cashreceived变量声明为十进制类型，但它们都被转换为int。因此，最好从头开始将它们都声明为int，这将有助于简化转换。

你可能会做这样的事情

hundreds = change / 100;
change %= 100;

fifty = change / 50;
change %= 50;

ten = change / 10;
change %= 10;

five = change / 5;
change %= 5;

two = change / 2;
change %= 2;

one = change;

Answer 3

而不是

     if (change >= 100)
     {
     hundreds = (int)change / 100;
     change = (int)change % 100;

     }

尝试使用

        if (change >= 100)
        {
            decimal rand = change / 100;
            hundred= (int)rand;
            change = change % 100;
        }


        if (change >= 50)
        {
            decimal rand = (int)change / 50;
            fifty = (int)rand;
            change = change % 50;
        }


        if (change >= 20)
        {
            decimal rand = change / 20;
            twenty = (int)rand;
            change = change % 20;
        }

我希望这会有所帮助。

改变计算器

3 个答案: