Question

我正在创建一个程序，您可以在其中输入格式为y = mx + c

的等式

它将给出从-2到+2的y值。

用户可以输入的内容的示例是y = 2x + 5.

我该如何解决这个问题？我想输入x的整数值我不知道在哪里或如何开始。

Answer 1

如果您只想输入x值的整数，可以使用下面的方法...该方法允许您选择渐变和y截距的值。

您可以使用此方法：

    public static void rangeCalculator(int startPoint, int endPoint){

    Scanner input = new Scanner(System.in);
    System.out.print("Enter gradient:");
    double gradient = input.nextDouble();
    System.out.print("\nEnter intercept:");
    double intercept = input.nextDouble();

    for(int i=startPoint; i<=endPoint; i++){
        System.out.println("y="+gradient+"x + " +intercept+"\t" + "input:"+i + " output:" + (gradient*i + intercept));
    }
}

Answer 2

import java.util.Scanner;

public class graphTester {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter equation: ");
        String input = scanner.nextLine();
        Character equalsChar = '=';
        Character xChar = 'x';
        int iEquals = 0;
        int iX =0;
        int iPlusOrSubtract = 0;
        int[] xValues = {-2, -1, 0, 1, 2, 3, 4};
        int[] yValues = new int[7];
        int i = 0;

        for (iEquals = 0; iEquals <= input.length(); iEquals++){
            Character c1 = input.charAt(iEquals);
            if (c1 == equalsChar){
                System.out.println("Found equals at index: " + iEquals);
                break;
            }else{

            }
        }

        for (iX = 0; iX <= input.length(); iX++){
            Character c2 = input.charAt(iX);
            if (c2 == xChar){
                System.out.println("Found x at index: " + iX);
                break;
            }else{

            }
        }

        String coEfficientString = input.substring(iEquals + 1, iX);
        int coEfficient = Integer.parseInt(coEfficientString);
        System.out.println("coEfficient: " + coEfficient);

        String yInterceptString = input.substring(iX + 1, input.length());
        int yIntercept = Integer.parseInt(yInterceptString);
        System.out.println("Y-Intercept: " + yIntercept);

        for (int value : xValues){
            i++;
            System.out.println("X-Value:" + value + "     Y-Value:" + value*coEfficient);
            yValues[i] = value*coEfficient;

        }
        System.out.println(yValues);
    }

}

Answer 3

import java.util.regex.*;

public class RegexTester {

    public static void main(String[] args) {
        String str2Check = "3x+2";

        //Find x - \\d{1,}+[x-x]
        //Find y-intercept [[\\+] | [\\-]]+\\d{1}

        String regexStringCoefficient = "[[\\+] | [\\-]]+\\d{1}";

        regexChecker(regexStringCoefficient, str2Check);
    }

    public static void regexChecker(String regexString, String str2Check){

        Pattern checkRegex = Pattern.compile(regexString);
        Matcher regexMacher = checkRegex.matcher(str2Check);

        while (regexMacher.find()){
            if (regexMacher.group().length() != 0){
                System.out.println(regexMacher.group());
                System.out.println("First Index: " + regexMacher.start());
                System.out.println("Ending index: " + regexMacher.end());
            }
        }
    }

}

求解格式为y = mx + c的方程的所有y值

3 个答案: