什么是斐波那契數列？經典數學問題之一；斐波那契數列，又稱黃金分割數列，指的是這樣一個數列：1、1、2、3、5、8、13、21、……想必看到這個數列大家很容易的就推算出來后面好幾項的值，那么到底有什么規律，簡單說，就是前兩項的和是第三項的值，用遞歸算法計第50位多少。
這個數列從第3項開始，每一項都等于前兩項之和。
斐波那契數列：｛1,1,2,3,5,8,13,21...｝
遞歸算法，耗時最長的算法，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));}通過循環來實現
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;}通過循環的改進寫法
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，可減少運算。 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 + " 耗時：" + watch1.Elapsed, number); Stopwatch watch2 = new Stopwatch(); watch2.Start(); result = F2(number); watch2.Stop(); Console.WriteLine("F2({0})=" + result + " 耗時：" + watch2.Elapsed, number); Stopwatch watch3 = new Stopwatch(); watch3.Start(); result = F3(number); watch3.Stop(); Console.WriteLine("F3({0})=" + result + " 耗時：" + watch3.Elapsed, number); Stopwatch watch4 = new Stopwatch(); watch4.Start(); double result4 = F4(number); watch4.Stop(); Console.WriteLine("F4({0})=" + result4 + " 耗時：" + watch4.Elapsed, number); Console.WriteLine(); Console.WriteLine("結束"); 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> /// 矩陣法 /// </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> /// 通項公式法 /// </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，就這些了。用的long類型來存儲結果，當n>92時會內存溢出。
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