Java——求解一元n次方程(V1.0)

主要思路

1. 通过 接口(interface) 进行求解方法声明，通过 implements 声明自己使用的接口
2. 通过字符串 equals() 方法以及 while() 循环进行选择是否退出程序
3. switch case 语句判断求解的方程次幂

代码

/*
 @author big_fw
 @function 求解一元n次方程
 @version 1.o
 */

import java.util.Map;
import java.util.Scanner;

interface Solve1 { //一元一次
    double solve1(double a, double b, double y);
}

interface Solve2 { //一元二次
    double solve2(double a, double b, double c);
}

public class SolveEquation implements Solve1, Solve2 {
    //一元一次
    public double solve1(double a, double b, double y) {
        double x;
        x = (y - b) / a;
        return x;
    }

    //一元二次
    public double solve2(double a, double b, double c) {
        double x, x1;
        x1 = b * b - 4 * a * c;
        x = Math.sqrt(x1);
        return x; //返回 b^2-4ac
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        String str1 = "q";
        System.out.println("进入程序请输入i,如要退出程序请输入 'q':");
        String str2 = scan.next();
        while (!str2.equals(str1)) {
            System.out.print("请选择输入的一元n次方程的次幂：");
            int n = scan.nextInt();

            switch (n) {
                case 1://一元一次方程
                    SolveEquation s1 = new SolveEquation();

                    System.out.println("输入一元一次方程，其形式为 ax + b = y");

                    System.out.print("a为：");
                    double a = scan.nextDouble();
                    System.out.print("b为：");
                    double b = scan.nextDouble();
                    System.out.print("y为：");
                    double y = scan.nextDouble();

                    System.out.println("方程解为：" + s1.solve1(a, b, y));
                    break;

                case 2://一元二次方程
                    SolveEquation s2 = new SolveEquation();
                    System.out.println("输入一元二次方程，其形式为 ax＾2＋bx＋c＝0(其中a != 0)");

                    System.out.print("a为：");
                    double a1 = scan.nextDouble();
                    System.out.print("b为：");
                    double b1 = scan.nextDouble();
                    System.out.print("c为：");
                    double c1 = scan.nextDouble();

                    if (b1 * b1 - 4 * a1 * c1 < 0) {
                        System.out.println("方程无实解");
                    } else if (b1 * b1 - 4 * a1 * c1 >= 0) {

                        double x3 = s2.solve2(a1, b1, c1);
                        double x1 = (-b1 + x3) / (2 * a1);
                        double x2 = (-b1 - x3) / (2 * a1);

                        System.out.println("一元二次方程的解为：");
                        System.out.println("x1 = " + x1);
                        System.out.println("x2 = " + x2);

                        break;
                    }
                default:
                    System.out.println("还没写好 ^_^");
            }
        }
    }
}