java - 用java实现摊销程序

Question

我需要计算一个人每年支付的贷款利息金额，并显示每年贷款余额的递减情况。我知道如何在纸上获取数字，但我不知道如何为其编写代码，而且其他论坛根本没有帮助。

它应该是什么样子的一个例子是:

Enter loan amount: 90000
Enter loan duration in years: 15
Enter interest rate as a percent: 6.75


For a 15 year loan of $90000.00 at 6.75% interest

Monthly payment =   $   796.42
Total interest =    $   53355.33

Yearly Balances
Year    Interest    Loan Balance
 1      5965.23     86408.21   
 2      5715.14     82566.33    
 3      5447.64     78456.94    
 4      5161.51     74061.43    
 5      4855.46     69359.87    
 6      4528.10     64330.94    
 7      4177.95     58951.87    
 8      3803.41     53198.26    
 9      3402.80     47044.03    
 10     2974.29     40461.31    
 11     2515.95     33420.24    
 12     2025.70     25888.91    
 13     1501.31     17833.19    
 14     940.40      9216.58    
 15     340.45      -0.00

What I currently have:

import java.util.*;
public class Loan
{
  public static void main(String[] args)
  {
    Scanner keyboard = new Scanner(System.in);
    int loanAmount, loanDurationYears, month;
    double interestRate, monthlyPayment, totalInterest;

    System.out.print("Enter loan amount:");
    loanAmount = keyboard.nextInt();

        System.out.print("Enter loan duration in years:");
    loanDurationYears = keyboard.nextInt();

        System.out.print("Enter interest rate as a percent:");
    interestRate = keyboard.nextDouble();
    keyboard.close();

    System.out.println("Loan amount: $" + loanAmount);
    System.out.println("Loan duration: " + loanDurationYears + " years");
    System.out.println("Interest rate: " + interestRate + "%");

    monthlyPayment = payment(loanAmount, loanDurationYears, interestRate);
    totalInterest = ((monthlyPayment * (loanDurationYears*12))-loanAmount);
    printTotals(loanAmount, loanDurationYears, interestRate, monthlyPayment, totalInterest);
  }
  public static double payment(int loanAmount, int loanDurationYears, double interestRate)
  {
      int a = loanAmount;
      int n = (loanDurationYears*12);
      double i = ((interestRate*.01)/12);
      double monthlyPayment = a*(Math.pow((1+i),n)*i)/((Math.pow((1+i),n))-1);
      return monthlyPayment;
  }
  public static void printTotals(int loanAmount, int loanDurationYears, double interestRate, double monthlyPayment, double totalInterest)
  {
      System.out.println("For a " + loanDurationYears + " year loan of $" + loanAmount + " at " + interestRate + "% interest:");
      System.out.printf("Monthly payment = $"+ "%.2f", monthlyPayment);
      System.out.println();
      System.out.printf("Total interest = $"+ "%.2f", totalInterest);
      System.out.println();
  }
}

Answer 1

浮点型和 double 型不应用于货币，因为小数部分并不总是精确的。

但是，由于您必须使用 double ，因此我建议您将小数点向右移动一点，直到获得所需的所有精度。

例如。 100.26 不会随时存储在 double 型上，但 100260.00 会存储。这样，您就不会丢失重要的十进制数字“.26”的精度，并在末尾添加精确的额外数字“0”以进行精确舍入(如果需要，否则停止在“26”)。

因为 100.26 绝不能存储在 double 上(否则您将面临失去小数精度的风险)，因此您将以字符串或 BigDecimal 类型读取用户输入，并在将数字传递给 double 之前“乘以”1000。如果您使用字符串，您将找到该点，将其删除，并将其放置在右侧 2/3 个点处，然后再转换为 double 。

因为您将乘以 1000，所以您的双最大值将受到更多限制，但如果您不打算使用巨大的数字，这可能不是问题。

您将像平常一样使用 x*1000 进行所有操作，在打印之前，您将把 double 值传递给您的一种精确数据类型、格式以删除 *1000 并打印。

对于这种样式，您不需要专门使用 double ，例如可以使用 long，但是这样，您将获得 double 的精确值。