C＃Windows窗体应用程序| Cookie Clicker

Question

我想做一个饼干制造商！ 它工作得很好，除了每次点击cookie之外，只需要点击2次就可以在cookie计数中添加一个。我想让计数系统变得更快，因为如果我每秒点击2次，它只算作一个atm ......

代码：

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
using System.Threading;
namespace FormalCookie
{
    public partial class Form1 : Form
    {
        int cnum = 0;        
        int add = 1;


        public Form1()
        {
            InitializeComponent();
            label1.Text = ("Price For Mouse:" + " 50");
            textBox1.Text = ("Cookies:" + " " + cnum);
            mouse.Enabled = false;

            for (int i = 0; i< add; i++)
            { 
                if (add > 1 && add % 2 == 0)
                {
                    cnum += 1 * add;
                }
                Thread.Sleep(1000);
            }
        }

        private void textBox1_TextChanged(object sender, EventArgs e)
        {

        }

        private void button1_Click(object sender, EventArgs e)
        {

            //Mouse
            if (cnum >= add * add * 50)
            {
                cnum -= (add * 100);
                add+=2;
            }


        }

        private void button2_Click(object sender, EventArgs e)
        {
            Application.Restart();
        }

        private void button3_Click(object sender, EventArgs e)
        {
            Application.Exit();
        }

        private void pictureBox1_Click(object sender, EventArgs e)
        {

            if (cnum >= (add * add * 50))
            {
                mouse.Enabled = true;
            }
            else
            {
                mouse.Enabled = false;
            }

            cnum += (add * add);
            textBox1.Text = ("Cookies:" + " " + cnum);

            label1.Text = ("Price For Mouse:" + " " + (add * add * 50));
        }

        private void label1_Click(object sender, EventArgs e)
        {

        }

        private void pictureBox1_Click(object sender, MouseEventArgs e)
        {

        }
    }
}