Question

我正在创建一个回馈变更的系统。当我运行程序时，我得到：（假设我给了100美元）

You have given 100.0 dollars.
Your change is 14.670000000000002 dollars.
You will recieve:
0.6700000000000017 dollars
0.1700000000000017 quarters
0.0700000000000017 dimes
0.020000000000001697 nickels
1.6965595595053173E-15 pennies

码

import java.util.Scanner;
public class RunGivingChange {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        double given;
        double cost = 85.33;
        double change;
        int dollars;
        int quarters;
        int dimes;
        int nickels;
        int pennies;

        Scanner input = new Scanner(System.in);

        System.out.println("Your total is " + cost + " dollars.");
        System.out.println("How much money are you giving?");
        given = input.nextDouble();
        if (given >= cost)
        {
            System.out.println("You have given " + given + " dollars.");
            change = given - cost;
            System.out.println("Your change is " + change + " dollars.");
            System.out.println("You will recieve:");

            //dollars
            change = change % 1;
            System.out.println(change + " dollars");

            //quarters
            change = change % 0.25;
            System.out.println(change + " quarters");

            //dimes
            change = change % 0.10;
            System.out.println(change + " dimes");

            //nickels
            change = change % 0.05;
            System.out.println(change + " nickels");

            //pennies
            change = change % 0.01;
            System.out.println(change + " pennies");
        }
        else if (given < cost)
        {
            System.out.println("Not enough money.");
        }
        else
        {
            System.out.println("Error.");
        }

    }

    }

Answer 1

您使用双精度浮点值来表示金钱。

不要这样做。

浮点数可以为您提供大约15个有效精度数字，但具有惊人的范围（和计算性能），这使得它们可用于科学用途。

使用固定点类型或BigDecimal作为货币值。

Answer 2

模数（％）并不像你期望的那样工作（除了精度问题）：

2.67 % 1不会导致2，但会导致0.67。

（有关更多信息，请查看此处：How do I use modulus for float/double?）

您可能想要计算Math.floor( 2.67 / 1)等并从change中减去舍入值。

或者，如果change始终为正（应该是），则可以将（并因此剪切）转换为整数：

int dollars = (int)(change / 1);
change -= dollars;

int quarters = (int)(change / 0.25);
change -= 0.25 * quarters;

...

要克服这些前提问题，请使用BigDecimal或将所有值转换为便士，即更改= 1467便士，1美元= 100便士等。

Answer 3

而不是以美元为代价，你可以用最小的部门（便士）来表达：

double cost = 85.33;// dollars
you could express it as :
int cost = 8533: .. //pennies

然后，您可以使用/和%操作序列来计算每个货币分配。例如：

int dollars = change / 100; // how many dollars fit in that pennies amount
change = change % 100; // how much pennies are left

//repeat for the other subdivisions

Answer 4

如果您使用的是双打，请确保舍入到小数位。

change = (double) Math.round(change *100) / 100;

其次，mod（％）返回语句的其余部分

14.67％1.0 = 0.67

首先要做的是找到1.0分为14的次数。

dollars = Math.floor(change / 1);
change = change % 1.0;

这会向下舍入到最接近的整数，并且会给出1.0可以进行更改的次数。

在这里使用双人的问题就像Bathsheba所说的那样。要解决此问题，每次修改更改后都必须进行舍入。

这应该会给你正确的输出。

import java.util.Scanner;

public class RunGivingChange{


    public static void main(String[] args) {


        double given;
        double cost = 85.33;
        double change;
        int dollars;
        int quarters;
        int dimes;
        int nickels;
        int pennies;

        Scanner input = new Scanner(System.in);

        System.out.println("Your total is " + cost + " dollars.");
        System.out.println("How much money are you giving?");
        given = input.nextDouble();
        if (given >= cost) {
            System.out.println("You have given " + given + " dollars.");
            change = given - cost;
            // round the change
            change = (double) Math.round(change *100) / 100;
            System.out.println("Your change is " + change + " dollars.");
            System.out.println("You will recieve:");

            //dollars     

            System.out.println(Math.floor(change / 1) + " dollars");
            change = change % 1;
            change = (double) Math.round(change *100) / 100;
            //quarters

            System.out.println(Math.floor(change / 0.25) + " quarters");
            change = change % 0.25;
            change = (double) Math.round(change *100) / 100;
            //dimes

            System.out.println(Math.floor(change / 0.1) + " dimes");
            change = change % 0.10;
            change = (double) Math.round(change *100) / 100;
            //nickels

            System.out.println(Math.floor(change / 0.05) + " nickels");
            change = change % 0.05;
            change = (double) Math.round(change *100) / 100;
            //pennies

            System.out.println(Math.floor(change / 0.01) + " pennies");

        } else if (given < cost) {
            System.out.println("Not enough money.");
        } else {
            System.out.println("Error.");
        }

    }

}

有更好的方法可以在下次搜索网站时执行此操作。这个主题有多个答案。

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4 个答案: