C#实现斐波那契数列的几种方法整理

public static long CalcA(int n){  if (n <= 0) return 0;  if (n <= 2) return 1;  return checked(CalcA(n - 2) + CalcA(n - 1));}

public static long CalcB(int n){  if (n <= 0) return 0;  var a = 1L;  var b = 1L;  var result = 1L;  for (var i = 3; i <= n; i++)  {    result = checked(a + b);    a = b;    b = result;  }  return result;}

public static long CalcC(int n){  if (n <= 0) return 0;  var a = 1L;  var b = 1L;  for (var i = 3; i <= n; i++)  {    b = checked(a + b);    a = b - a;  }  return b;}

/// <summary>/// F(n)=(1/√5)*{[(1+√5)/2]^n - [(1-√5)/2]^n}/// </summary>/// <param name="n"></param>/// <returns></returns>public static long CalcD(int n){  if (n <= 0) return 0;  if (n <= 2) return 1; //加上，可减少运算。  var a = 1 / Math.Sqrt(5);  var b = Math.Pow((1 + Math.Sqrt(5)) / 2, n);  var c = Math.Pow((1 - Math.Sqrt(5)) / 2, n);  return checked((long)(a * (b - c)));}

using System;using System.Diagnostics;namespace Fibonacci{  class Program  {    static void Main(string[] args)    {      ulong result;      int number = 10;      Console.WriteLine("************* number={0} *************", number);      Stopwatch watch1 = new Stopwatch();      watch1.Start();      result = F1(number);      watch1.Stop();      Console.WriteLine("F1({0})=" + result + " 耗时：" + watch1.Elapsed, number);      Stopwatch watch2 = new Stopwatch();      watch2.Start();      result = F2(number);      watch2.Stop();      Console.WriteLine("F2({0})=" + result + " 耗时：" + watch2.Elapsed, number);      Stopwatch watch3 = new Stopwatch();      watch3.Start();      result = F3(number);      watch3.Stop();      Console.WriteLine("F3({0})=" + result + " 耗时：" + watch3.Elapsed, number);      Stopwatch watch4 = new Stopwatch();      watch4.Start();      double result4 = F4(number);      watch4.Stop();      Console.WriteLine("F4({0})=" + result4 + " 耗时：" + watch4.Elapsed, number);      Console.WriteLine();      Console.WriteLine("结束");      Console.ReadKey();    }    /// <summary>    /// 迭代法    /// </summary>    /// <param name="number"></param>    /// <returns></returns>    private static ulong F1(int number)    {      if (number == 1 || number == 2)      {        return 1;      }      else      {        return F1(number - 1) + F1(number - 2);      }          }    /// <summary>    /// 直接法    /// </summary>    /// <param name="number"></param>    /// <returns></returns>    private static ulong F2(int number)    {      ulong a = 1, b = 1;      if (number == 1 || number == 2)      {        return 1;      }      else      {        for (int i = 3; i <= number; i++)        {          ulong c = a + b;          b = a;          a = c;        }        return a;      }    }    /// <summary>    /// 矩阵法    /// </summary>    /// <param name="n"></param>    /// <returns></returns>    static ulong F3(int n)    {      ulong[,] a = new ulong[2, 2] { { 1, 1 }, { 1, 0 } };      ulong[,] b = MatirxPower(a, n);      return b[1, 0];    }    #region F3    static ulong[,] MatirxPower(ulong[,] a, int n)    {      if (n == 1) { return a; }      else if (n == 2) { return MatirxMultiplication(a, a); }      else if (n % 2 == 0)      {        ulong[,] temp = MatirxPower(a, n / 2);        return MatirxMultiplication(temp, temp);      }      else      {        ulong[,] temp = MatirxPower(a, n / 2);        return MatirxMultiplication(MatirxMultiplication(temp, temp), a);      }    }    static ulong[,] MatirxMultiplication(ulong[,] a, ulong[,] b)    {      ulong[,] c = new ulong[2, 2];      for (int i = 0; i < 2; i++)      {        for (int j = 0; j < 2; j++)        {          for (int k = 0; k < 2; k++)          {            c[i, j] += a[i, k] * b[k, j];          }        }      }      return c;    }    #endregion    /// <summary>    /// 通项公式法    /// </summary>    /// <param name="n"></param>    /// <returns></returns>    static double F4(int n)    {      double sqrt5 = Math.Sqrt(5);      return (1/sqrt5*(Math.Pow((1+sqrt5)/2,n)-Math.Pow((1-sqrt5)/2,n)));    }  }}