c#实现一元二次方程求解器示例分享
c#实现一元二次方程求解器示例分享
发布时间:2016-12-28 来源:查字典编辑
摘要:复制代码代码如下:usingSystem;usingSystem.Collections.Generic;usingSystem.Compo...

复制代码 代码如下:

using System;

using System.Collections.Generic;

using System.ComponentModel;

using System.Data;

using System.Drawing;

using System.Linq;

using System.Text;

using System.Windows.Forms;

namespace WindowsFormsApplication4

{

public partial class Form1 : Form

{

public Form1()

{

InitializeComponent();

}

private void Form1_Load(object sender, EventArgs e)

{

this.Text = "一元二次方程求解器";

}

private void button1_Click(object sender, EventArgs e)

{

A T=new A();

T.a = double.Parse(textBox1.Text);

T.b = double.Parse(textBox2.Text);

T.c = double.Parse(textBox3.Text);

if (T.a == 0)

textBox4.Text = string.Format("此为一元一次方程根为 x = {0}", (-T.c / T.b));

else

{

object box = textBox4;

T.Answer(T.a, T.b, T.c, box);

}

}

}

}

class A

{

public double a, b, c;

public double Answer(double a, double b, double c, object box)

{

double x1;

double x2;

TextBox temp = (TextBox)box;

if ((b * b - 4 * a * c) > 0)

{

x1 = ((-b + Math.Sqrt(b * b - 4 * a * c)) / (2 * a));

x2 = ((-b - Math.Sqrt(b * b - 4 * a * c)) / (2 * a));

temp.Text = String.Format("x1={0},x2={1}", x1, x2);

}

else if ((b * b - 4 * a * c) == 0)

{

x1 = x2 = ((-b + Math.Sqrt(b * b - 4 * a * c)) / (2 * a));

temp.Text = String.Format("x1={0},x2={1}", x1, x2);

}

else

temp.Text = "此参数下的一元二次方程无解";

return 0;

}

}

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