Question

所以，我制定了一个程序，在给定价格（加5％税）和用户付款时计算变更。我需要这个代码的其余要求是：

让程序循环直到用户决定退出程序输入0或任何负数（即-2）。
如果用户没有输入足够的付款来支付价格，请继续询问用户“请输入付款金额”，直到用户输入足够大的付款为止。
打印并显示正确的更改量以及每种货币的数量。最高价格，不含税，为500.00美元。

完整输出，所有可能的用户选择应如下所示：

Cost of transaction (enter 0 or negative to exit; max is $500.00): 1000
Cost of transaction (enter 0 or negative to exit; max is $500.00): 501
Cost of transaction (enter 0 or negative to exit; max is $500.00): 500
Amount due (with 5.00% tax): $525.00
Please enter payment amount: 1000.57
Change back $475.57
$100: 4  $50: 1  $20: 1  $10: 0  $5: 1  $1: 0  $0.25: 2  $0.10: 0  $0.05: 1
$0.01: 2
Cost of transaction (enter 0 or negative to exit; max is $500.00): 0
Done.

这是我的代码：

/*
* @author (Zach Daly)
* <p> (MakeChange.java)
* <p> (Project2)
* <p> (Make change)
*/

import java.util.*;

public class MakeChange
{
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        double price;
        double amountPaid;
        int paidInt;
        int priceWithTaxInt;
        double priceWithTaxDbl;
        final double TAX_RATE = .05;
        int change;
        int hundreds, fifties, twenties, tens, fives, ones, quarters, dimes, nickels, pennies;

        System.out.print("Cost of transaction (enter 0 or negative to exit; max is $500.00): ");
        price = in.nextDouble();
        priceWithTaxDbl = price + (price * TAX_RATE);
        System.out.printf("Amount due (with 5.00%% tax): $%3.2f\n", priceWithTaxDbl);
        System.out.println("Please enter payment amount: ");
        amountPaid = in.nextDouble();
        paidInt = (int) (amountPaid * 100);
        priceWithTaxInt = (int) (priceWithTaxDbl * 100);

        if (price <= 0)
        {
            System.out.println("Done.");
        }

        change = paidInt - priceWithTaxInt;
        double changePrint = (double) change / 100;

        System.out.println(changePrint);

        if (change == 0)
        {
            System.out.println("Exact change! Amazing!");
        }

        if (paidInt > priceWithTaxInt)
        {
            hundreds = change / 10000;
            if (hundreds > 0)
            {
                change %= 10000;
            }

            fifties = change / 5000;
            if (fifties > 0)
            {
                change %= 5000;
            }

            twenties = change / 2000;
            if (twenties > 0)
            {
                change %= 2000;
            }

            tens = change / 1000;
            if (tens > 0)
            {
                change %= 1000;
            }

            fives = change / 500;
            if (fives > 0)
            {
                change %= 500;
            }

            ones = change / 100;
            if (ones > 0)
            {
                change %= 100;
            }

            quarters = change / 25;
            if (quarters > 0)
            {
                change %= 25;
            }

            dimes = change / 10;
            if (dimes > 0)
            {
                change %= 10;
            }

            nickels = change / 5;
            if (nickels > 0)
            {
                change %= 5;
            }

            pennies = change;

            System.out.printf(
                    "Change back $%.2f\n$100: %d   $50: %d   $20: %d   $10: %d   $5: %d   $1: %d   $0.25: %d   $0.10: %d   $0.05: %d   $0.01: %d\n",
                    changePrint, hundreds, fifties, twenties, tens, fives, ones, quarters, dimes, nickels, pennies);
        }

        in.close();
    }
}

Answer 1

我相信你要找的是一个do-while循环。它将运行一次，并一直重复，直到不满足while条件：

do {
        System.out.print("Cost of transaction (enter 0 or negative to exit; max is $500.00): ");
        price = in.nextDouble();
        //repeating logic

        if (price > 500) {
            price = 0;
            System.out.println("Maximum price was exceeded");
        }
    } while (price >= 0);

在用户另有决定之前，如何多次保持进程？ Java的

1 个答案: