zoukankan      html  css  js  c++  java
  • 第六章第三十九题(几何:点的位置)(Geometry: point position)

    6.39(几何:点的位置)编程练习题3.32显示如何测试一个点是否在一个有向直线的左侧、右侧,或在该直线上。使用下面的方法头编写该方法:

    public static boolean leftOfTheLine(double x0, double y0, double x1, double y1, double x2, double y2)

    public static boolean onTheSameLine(double x0, double y0, double x1, double y1, double x2, double y2)

    public static boolean onTheLineSegment(double x0, double y0, double x1, double y1, double x2, double y2)

    编写一个程序,提示用户输入三个点赋给p0、p1和p2,显示p2是否在从p0到p1的直线的左侧、右侧、直线上,或者线段上。

    下面是一些运行示例:

    Enter three points for p0, p1, and p2: 1 1 2 2 1.5 1.5

    (1.5, 1.5) is on the line segment from (1.0, 1.0) to (2.0, 2.0)

    Enter three points for p0, p1, and p2: 1 1 2 2 3 3

    (3.0, 3.0) is on the same line from (1.0, 1.0) to (2.0, 2.0)

    Enter three points for p0, p1, and p2: 1 1 2 2 1 1.5

    (1.0, 1.5) is on the left side of the line from (1.0, 1.0) to (2.0, 2.0)

    Enter three points for p0, p1, and p2: 1 1 2 2 1 –1

    (1.0, −1.0) is on the right side of the line from (1.0, 1.0) to (2.0, 2.0)

    6.39(Geometry: point position) Programming Exercise 3.32 shows how to test whether a point is on the left side of a directed line, on the right, or on the same line. Write the methods with the following headers:

    public static boolean leftOfTheLine(double x0, double y0, double x1, double y1, double x2, double y2)

    public static boolean onTheSameLine(double x0, double y0, double x1, double y1, double x2, double y2)

    public static boolean onTheLineSegment(double x0, double y0, double x1, double y1, double x2, double y2)

    Write a program that prompts the user to enter the three points for p0, p1, and p2and displays whether p2 is on the left side of the line from p0 to p1, right side, the same line, or on the line segment.

    Here are some sample runs:

    Enter three points for p0, p1, and p2: 1 1 2 2 1.5 1.5

    (1.5, 1.5) is on the line segment from (1.0, 1.0) to (2.0, 2.0)

    Enter three points for p0, p1, and p2: 1 1 2 2 3 3

    (3.0, 3.0) is on the same line from (1.0, 1.0) to (2.0, 2.0)

    Enter three points for p0, p1, and p2: 1 1 2 2 1 1.5

    (1.0, 1.5) is on the left side of the line from (1.0, 1.0) to (2.0, 2.0)

    Enter three points for p0, p1, and p2: 1 1 2 2 1 –1

    (1.0, −1.0) is on the right side of the line from (1.0, 1.0) to (2.0, 2.0)

    下面是参考答案代码:

    // https://cn.fankuiba.com
    import java.util.Scanner;
    
    public class Ans6_39_page205 {
        public static void main(String[] args) {
            Scanner input = new Scanner(System.in);
            System.out.print("Enter three points for p0, p1, and p2:");
            double x0 = input.nextDouble();
            double y0 = input.nextDouble();
            double x1 = input.nextDouble();
            double y1 = input.nextDouble();
            double x2 = input.nextDouble();
            double y2 = input.nextDouble();
    
            if (onTheSameLine(x0, y0, x1, y1, x2, y2)) {
                if (onTheLineSegment(x0, y0, x1, y1, x2, y2)) {
                    System.out.printf("(%.1f, %.1f) is on the line segment from "
                            + "(%.1f, %.1f) to (%.1f, %.1f)", x2, y2, x0, y0, x1, y1);
                } else {
                    System.out.printf("(%.1f, %.1f) is on the same line from "
                            + "(%.1f, %.1f) to (%.1f, %.1f)", x2, y2, x0, y0, x1, y1);
                }
            } else {
                if (leftOfTheLine(x0, y0, x1, y1, x2, y2)) {
                    System.out.printf("(%.1f, %.1f) is on the left side of the line from "
                            + "(%.1f, %.1f) to (%.1f, %.1f)", x2, y2, x0, y0, x1, y1);
                } else {
                    System.out.printf("(%.1f, %.1f) is on the right side of the line from "
                            + "(%.1f, %.1f) to (%.1f, %.1f)", x2, y2, x0, y0, x1, y1);
                }
            }
        }
    
        public static boolean leftOfTheLine(double x0, double y0,
                                            double x1, double y1, double x2, double y2) {
            if ((x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0) > 0)
                return true;
            else
                return false;
        }
    
        public static boolean onTheSameLine(double x0, double y0,
                                            double x1, double y1, double x2, double y2) {
            return (x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0) == 0;
        }
    
        public static boolean onTheLineSegment(double x0, double y0,
                                               double x1, double y1, double x2, double y2) {
            if ((x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0) == 0 &&
                    x2 <= (Math.max(x0, x1)) && x2 >= (Math.min(x1, x0)))
                return true;
            else
                return false;
        }
    }
    
    

    适用Java语言程序设计与数据结构(基础篇)(原书第11版)Java语言程序设计(基础篇)(原书第10/11版)

    发布在博客:(https://cn.fankuiba.com)

  • 相关阅读:
    C#面向对象五(继承、抽象类和抽象方法、多态、虚方法、is、as、new覆盖关键字)
    C#面向对象四(文件与目录操作、序列化与反序列化、XML)
    C#面向对象三(常用控件)
    C#面向对象二(集合泛型及排序)
    C#面向对象一(方法、封装、类、两种数据类型)
    如何最快找到重点 —— 小小码虫瞎想の效率与绩效篇(一)
    Android Telephony —— 手机信号实时变化源码分析过程记录
    Android 某些配置记录
    Macbook Pro 使用小记
    fhq treap 范浩强平衡树
  • 原文地址:https://www.cnblogs.com/in2013/p/13097190.html
Copyright © 2011-2022 走看看