PInvoke签名        [StructLayout(LayoutKind.Sequential)]
        public struct MARGINS
        {
            public int Left;
            public int Right;
            public int Top;
            public int Bottom;
        }

        [DllImport("dwmapi.dll", PreserveSig = false)]
        static extern void DwmExtendFrameIntoClientArea(IntPtr hwnd, ref MARGINS margins);

        [DllImport("dwmapi.dll", PreserveSig = false)]
        static extern bool DwmIsCompositionEnabled();

OnLoad        protected override void OnLoad(EventArgs e)
        {
            if (DwmIsCompositionEnabled())
            {
                margins = new MARGINS();
                margins.Right = margins.Left = margins.Top = margins.Bottom = this.Width+this.Height;
                DwmExtendFrameIntoClientArea(this.Handle, ref margins);
            }
            base.OnLoad(e);
        }

 OnPaintBackgroud   
protected override void OnPaintBackground(PaintEventArgs e)
        {
            base.OnPaintBackground(e);
            if (DwmIsCompositionEnabled())
            {
                e.Graphics.Clear(Color.Black);
            }
        }

Code
/*
 * Created by SharpDevelop.
 * User: metav
 * Date: 2008/10/30
 * Time: 11:35
 * 
 * KeyWord:FCTRL
 * Abstract:计算N!结果中0的个数
 * Analysis: n! 末尾 0的个数 取决于从 1 到 n 的各个数的因数中 2 、 5 的个数。
 *             又因为 2 的个数远多于 5 ， 5 的个数起决定作用。
 *             不断地用 n 除以 5，直到结果为 0，将中间的结果都加起来，即可得到因子 5 的个数。
 *             例，100/5 = 20，20/5 = 4，4/5 = 0。 
 *                      (20 + 4) = 24 个，
 *                      亦即 100! 末尾有 24 个 0。
 */
using System;

namespace Sphere
{

    public class Factorial
    {
        public static void Main()
        {
            int t  = int.Parse(Console.ReadLine());
            while(t--!=0)
            {
                int n = int.Parse(Console.ReadLine());
                Console.WriteLine(GetZeros(n));
            }
            //Console.ReadKey();
        }
        
        private static int GetZeros(int n)
        {
            int res = 0;
            while(n!=0)
            {
                n/=5;
                res += n;
            }
            return res;
        }
    }
}

Code
/*
 * Created by SharpDevelop.
 * User: metav
 * Date: 2008/10/30
 * Time: 18:56
 * 
 * KeyWord:ADDREV
 * Abstract:两个数的逆序数相加后将结果再逆序输出
 * Analysis: null
 */
using System;

namespace Sphere
{
    public class Adding_Reversed_Numbers
    {
        public static void Main()
        {
            int n = int.Parse(Console.ReadLine());
            while(n--!=0)
            {
                string s = Console.ReadLine();
                string[]  nums = s.Split(new char[]{' '});
                DoPlus(nums[0].TrimEnd('0'),nums[1].TrimEnd('0'));
            }
            //Console.ReadKey();
        }
        private static void DoPlus(string a,string b)
        {
            char[] na = a.ToCharArray();
            char[] nb = b.ToCharArray();
            
            int res = 0,carry = 0;
            int i,j;
            for( i= 0,j = 0;i<na.Length||j<nb.Length;i++,j++)
            {
                res *= 10;
                int t= 0;
                if(i<na.Length&&j<nb.Length)
                {
                    t = int.Parse(na[i].ToString())+int.Parse(nb[j].ToString());
                }
                if(i>=na.Length&&j<nb.Length)
                {
                    t=int.Parse(nb[j].ToString());
                }
                if(i<na.Length&&j>=nb.Length)
                {
                    t=int.Parse(na[i].ToString());
                }
                res += (t+carry)%10;
                carry =(t+carry)/10;
            }
            if(carry != 0)
            {
                res*=10;
                res+=1;
            }
            Console.WriteLine(res);
        }
    }
}