3

我正在尝试制作一个简单的基于控制台的程序来求解二次方程。如果判别式是负数,我还没有弄清楚如何计算答案,所以我只是显示了一条消息作为占位符。那么有人可以解释如何做到这一点吗?我将发布整个程序以获取任何建议。谢谢。

正常工作的代码(谢谢,Gunther Fox):

if(discriminant < 0)
{
    double x1 = -b/(2*a);
    discriminant = -discriminant ;
    double x2 = Math.Sqrt(discriminant)/(2*a);
    double x3 = -Math.Sqrt(discriminant)/(2*a);
    Console.WriteLine(x1.ToString() + " + " + x2.ToString() + " * i ");
    Console.WriteLine(x1.ToString() + " + " + x3.ToString() + " * i ");

    Console.ReadLine();
}

旧代码:

class Program
{
    static void Main(string[] args)
    {
        int choice;
        double ans1;
        double ans2;

        Console.WriteLine("~~~~~~~~ TERMINAL CALCULATOR ~~~~~~~~");

        Console.WriteLine("1. Standard Calculator\n2. Quadratic Equation Solver\n3. Simple Interest Calculator");
        choice = int.Parse(Console.ReadLine());

        try
        {
            if (choice == 2)
            {
                Console.Write("Enter a value for 'a': ");
                double a = double.Parse(Console.ReadLine());

                Console.Write("Enter a value for 'b': ");
                double b = double.Parse(Console.ReadLine());

                Console.Write("Enter a value for 'c': ");
                double c = double.Parse(Console.ReadLine());

                //Quadratic Formula: x = (-b +- sqrt(b^2 - 4ac)) / 2a

                //Solve disriminant: (b*b) - (4*a*c)
                double discriminant = (b * b) - (4 * a * c);

                if (discriminant < 0)
                {
                    Console.WriteLine("No real solutions");
                    Console.ReadLine();
                }
                else
                {
                    double sqrt = Math.Sqrt(discriminant);
                    ans1 = (-b + sqrt) / (2 * a);
                    ans2 = (-b - sqrt) / (2 * a);
                    Console.WriteLine("{" + ans1 + " , " + ans2 + "}");
                    Console.ReadLine();
                }
            }

            if (choice == 1)
            {
                Console.Write("Enter first number: ");
                double num1 = double.Parse(Console.ReadLine());

                Console.Write("Enter second number (Enter 0 if you want to use sqrt): ");
                double num2 = double.Parse(Console.ReadLine());

                //Prompt user to choose an operation
                Console.WriteLine("Choose a math operator:\n1. +\n2. -\n3. x\n4. /\n5. ^\n6. Square root");
                int mathOpr = int.Parse(Console.ReadLine());

                if (mathOpr == 1)
                {
                    double answer = num1 + num2;
                    Console.WriteLine("\n\n" + answer);

                    Console.ReadLine();
                }

                if (mathOpr == 2)
                {
                    double answer = num1 - num2;
                    Console.WriteLine("\n\n" + answer);

                    Console.ReadLine();
                }

                if (mathOpr == 3)
                {
                    double answer = num1 * num2;
                    Console.WriteLine("\n\n" + answer);

                    Console.ReadLine();
                }

                if (mathOpr == 4)
                {
                    double answer = num1 / num2;
                    Console.WriteLine("\n\n" + answer);

                    Console.ReadLine();
                }

                if (mathOpr == 5)
                {
                    double answer = Math.Pow(num1, num2);
                    Console.WriteLine("\n\n" + answer);

                    Console.ReadLine();
                }

                if (mathOpr == 6)
                {
                    if (num1 < 0)
                    {
                        Console.WriteLine("\n\nNo real solutions");
                    }

                    else
                    {
                        double answer = Math.Sqrt(num1);
                        Console.WriteLine("\n\n" + answer);
                    }
                }

                else
                {
                    Console.WriteLine("That is not a valid option, idiot.");

                    Console.ReadLine();
                }
            }

            if (choice == 3)
            {
                Console.WriteLine("Enter initial amount: ");
                double initAmount = double.Parse(Console.ReadLine());

                Console.WriteLine("Enter interest rate (ex. 6% = .06): ");
                double rate = double.Parse(Console.ReadLine());

                Console.WriteLine("Enter time range (Year, Month, Week, etc.): ");
                string timeRange = Console.ReadLine();

                Console.WriteLine("Enter amount of " + timeRange.ToLower() + "s: ");
                int time = int.Parse(Console.ReadLine());

                for (int time2 = 1; time2 <= time; time2++)
                {
                    double totalAmount = initAmount * Math.Pow(1 + rate, time2);
                    Console.WriteLine("\n" + timeRange + " " + time2 + " ---------- " + totalAmount);
                }

                Console.ReadLine();
            }
        }
        catch 
        {
            Console.WriteLine("That is not a valid option.");
        }
    }
}
4

2 回答 2

3

如果判别式为负,你可以翻转判别式的符号,做根计算,然后将 i 添加到结果的末尾吗?

于 2012-12-20T00:17:01.767 回答
3

翻转判别并乘以 i:

double discriminate = (b*b)-(4*a*c);
if(discriminate < 0){
    double x1 = -b/(2*a);
    discriminate = -discriminate ;
    double x2 = Math.Sqrt(discriminate)/(2*a);
    double x3 = -Math.Sqrt(discriminate)/(2*a);
}
Console.WriteLine(x1.ToString() + " + " + x2.ToString() + " * i ");
Console.WriteLine(x1.ToString() + " + " + x3.ToString() + " * i ");
于 2012-12-20T00:21:58.570 回答