什么是斐波那契数列？经典数学问题之一；斐波那契数列，又称黄金分割数列，指的是这样一个数列：1、1、2、3、5、8、13、21、……想必看到这个数列大家很容易的就推算出来后面好几项的值，那么到底有什么规律，简单说，就是前两项的和是第三项的值，用递归算法计第50位多少。

这个数列从第3项开始，每一项都等于前两项之和。

斐波那契数列：｛1,1,2,3,5,8,13,21...｝

递归算法，耗时最长的算法，效率很低。

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

通过循环来实现

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;}

通过循环的改进写法

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;}

通用公式法

/// <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)));}

其他方法

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))); } }}

OK，就这些了。用的long类型来存储结果，当n>92时会内存溢出。

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