如果输入有效，如何处理读取三角形三边并计算面积的家庭作业程序？ (2)

Question

这是对以下问题的重新提问：

又是任务：

Create a class named MyTriangle that contains the following two methods:
/** Return true if the sum of any two sides is * greater than the third side. */
public static boolean isValid (double side1, double side2, double side3)

/** Return the area of the triangle. */ 
public static double area (double side1, double side2, double side3)
Write a test program that reads three sides for a triangle and computes the area if the input is valid. Otherwise, it displays that the input is invalid.

下面的尝试：问题：我想不通，不断重读这一章并没有突破任何障碍。问题在下面的代码中注释：

import java.util.Scanner;

public class NewClass1 {   

double area;
double side1, side2, side3;
double x1, x2, x3, y1, y2, y3;

public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        System.out.print("Enter two integers for side 1:");
        double x1 = input.nextDouble();
        double y1 = input.nextDouble();

        System.out.print("Enter two integers for side 2:");
        double x2 = input.nextDouble();
        double y2 = input.nextDouble();

        System.out.print("Enter two integers for side 3:");
        double x3 = input.nextDouble();
        double y3 = input.nextDouble();

        boolean isValid = true;

            if (isValid) {
                System.out.println("Input is invalid");
            }
                else
                    area(side1, side2, side3); //Using area does not work and I don't know how to remedy this. I've read the chapter over and over... I cannot get it to work.

    }

    public static double area(double side1, double side2, double side3) {
        double x1 = 0;
        double x2 = 0;
        double y1 = 0;
        double y2 = 0;
        double x3 = 0;
        double y3 = 0; 

            side1 = Math.pow(Math.pow(x2 - x1, 2) + Math.pow(y2 - y1, 2), 0.5);
            side2 = Math.pow(Math.pow(x3 - x1, 2) + Math.pow(y3 - y1, 2), 0.5);
            side3 = Math.pow(Math.pow(x3 - x2, 2) + Math.pow(y3 - y2, 2), 0.5);

            //Calculates the sides/angles using Heron's formula
            double s = (side1 + side2 + side3)/2;
            double area = Math.pow(s * (s - side1) * (s - side2) * (s - side3), 0.5);

            return (area);
    } 

    public static boolean isValid(double side1, double side2, double side3) {

        return (((side1 + side2) > side3) && ((side1 + side3) > side2) && ((side2 + side3) > side1));
    }
}

查看代码，有人可以解释我做错了什么，并解释可能的补救措施。一切都在那里，我根本无法连接点。谢谢。

修改后的代码

import java.util.Scanner;

public class NewClass1 {   

    public static void main(String[] args) {

            Scanner input = new Scanner(System.in);

            System.out.print("Enter side 1: ");
            double side1 = input.nextDouble();

            System.out.print("Enter side 2: ");
            double side2 = input.nextDouble();

            System.out.print("Enter side 3: ");
            double side3 = input.nextDouble();

            double a = area(side1, side2, side3);
            boolean isV = isValid(side1, side2, side3);

                    if (isV)
                        System.out.println("Inout is Invalid");
                    else
                        System.out.println("Area is: " + a);
        }

    public static boolean isValid(double side1, double side2, double side3) {

         return (((side1 + side2) > side3) && ((side1 + side3) > side2) && ((side2 + side3) > side1));
    }

    public static double area(double side1, double side2, double side3) {

                //Calculates the sides/angles using Heron's formula
        double s = (side1 + side2 + side3)/2;
        double theArea = Math.pow(s * (s - side1) * (s - side2) * (s - side3), 0.5);

            return (theArea);
    } 
}

我一直收到 NaN 作为该区域的答案。我做错了什么？

我已经处理了 7 个多小时，只是因为我不了解其中可能存在的问题。我学习这门 CompSci 课程已经 2 个月了。

Answer 1

你的方法 isValid returns 如果三角形有效但你假设相反：

double a = area(side1, side2, side3);
boolean isV = isValid(side1, side2, side3);
if (isV)
    System.out.println("Inout is Invalid");

您在 11.1、22.2 和 33.3 方面的输入数据（在评论中）进一步支持了这一点。那不是三角形，它是一系列重叠的线段。

如果您输入完全有效的 3/4/5 三角形，您将收到一条错误消息，指出它不是三角形。

因此，只需将上面的代码更改为类似（对无效数据调用 area 毫无意义）：

if (! isValid(side1, side2, side3))
    System.out.println("Inout is Invalid");
else
    System.out.println("Area is " + area(side1, side2, side3));

进行这些更改：

package test2;
import java.util.Scanner;

public class test2 {   
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.print("Enter three sides, separated by spaces: ");
        double s1 = input.nextDouble();
        double s2 = input.nextDouble();
        double s3 = input.nextDouble();
        if (isValid(s1, s2, s3))
            System.out.println("Area is: " + area(s1, s2, s3));
        else
            System.out.println("Input is Invalid");
    }

    public static boolean isValid(double s1, double s2, double s3) {
         return ((s1 + s2 > s3) && (s1 + s3 > s2) && (s2 + s3 > s1));
    }

    public static double area(double s1, double s2, double s3) {
        double s = (s1 + s2 + s3) / 2;
        double theArea = Math.pow(s * (s - s1) * (s - s2) * (s - s3), 0.5);
        return theArea;
    } 
}

并输入有效三角形，例如 3/4/5 或 5/5/8 会给您正确的结果：

Enter three sides, separated by spaces: 3 4 5
Area is: 6.0

Enter three sides, separated by spaces: 5 5 8
Area is: 12.0

Answer 2

if (isV) 应该是 if (!isV)。如果条件不正确，那么您仅执行带有潜在危险输入的 Heron 公式，可能导致 s * (s - side1) * (s - side2) * (s - side3) 为负数。 Java 基本类型未定义负数的平方根。

Answer 3

使用这个 ..Heron 的公式

double sidesSvalue = (a + b + c)/2.0d;
double productofSides = (sidesSvalue  * (sidesSvalue -a) * (sidesSvalue -b) * (sidesSvalue -c));
double Area= Math.sqrt(productofSides );
return Area;

只要您的值无效，它们就会进入该区域，请更正您的条件代码，您应该使用类似这样的东西

if(!isValid)
   print error message
else 
   calculate area

最后一件事总是在使用 DMAS 操作时尝试使用相同的原始类型。

如果输入有效，如何处理读取三角形三边并计算面积的家庭作业程序？ (2)

How to tackle a homework prgram that reads three sides for a triangle and computes the area if the input is valid? (2)

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