C# 4种方法计算斐波那契数列 Fibonacci

F1:　　迭代法

最慢，复杂度最高

F2:　　直接法

 

F3:　　矩阵法

参考《算法之道（The Way of Algorithm）》第38页-魔鬼序列：斐波那契序列

 

F4:　　通项公式法

 

 由于公式中包含根号5，无法取得精确的结果，数字越大误差越大

 

  1 using System;
  2 using System.Diagnostics;
  3 
  4 
  5 namespace Fibonacci
  6 {
  7     class Program
  8     {
  9         static void Main(string[] args)
 10         {
 11             ulong result;
 12 
 13             int number = 10;
 14             Console.WriteLine("************* number={0} *************", number);
 15 
 16             Stopwatch watch1 = new Stopwatch();
 17             watch1.Start();
 18             result = F1(number);
 19             watch1.Stop();
 20             Console.WriteLine("F1({0})=" + result + "  耗时：" + watch1.Elapsed, number);
 21 
 22             Stopwatch watch2 = new Stopwatch();
 23             watch2.Start();
 24             result = F2(number);
 25             watch2.Stop();
 26             Console.WriteLine("F2({0})=" + result + "  耗时：" + watch2.Elapsed, number);
 27 
 28             Stopwatch watch3 = new Stopwatch();
 29             watch3.Start();
 30             result = F3(number);
 31             watch3.Stop();
 32             Console.WriteLine("F3({0})=" + result + "  耗时：" + watch3.Elapsed, number);
 33 
 34             Stopwatch watch4 = new Stopwatch();
 35             watch4.Start();
 36             double result4 = F4(number);
 37             watch4.Stop();
 38             Console.WriteLine("F4({0})=" + result4 + "  耗时：" + watch4.Elapsed, number);
 39 
 40             Console.WriteLine();
 41 
 42             Console.WriteLine("结束");
 43             Console.ReadKey();
 44         }
 45 
 46         /// <summary>
 47         /// 迭代法
 48         /// </summary>
 49         /// <param name="number"></param>
 50         /// <returns></returns>
 51         private static ulong F1(int number)
 52         {
 53             if (number == 1 || number == 2)
 54             {
 55                 return 1;
 56             }
 57             else
 58             {
 59                 return F1(number - 1) + F1(number - 2);
 60             }
 61             
 62         }
 63 
 64         /// <summary>
 65         /// 直接法
 66         /// </summary>
 67         /// <param name="number"></param>
 68         /// <returns></returns>
 69         private static ulong F2(int number)
 70         {
 71             ulong a = 1, b = 1;
 72             if (number == 1 || number == 2)
 73             {
 74                 return 1;
 75             }
 76             else
 77             {
 78                 for (int i = 3; i <= number; i++)
 79                 {
 80                     ulong c = a + b;
 81                     b = a;
 82                     a = c;
 83                 }
 84                 return a;
 85             }
 86         }
 87 
 88         /// <summary>
 89         /// 矩阵法
 90         /// </summary>
 91         /// <param name="n"></param>
 92         /// <returns></returns>
 93         static ulong F3(int n)
 94         {
 95             ulong[,] a = new ulong[2, 2] { { 1, 1 }, { 1, 0 } };
 96             ulong[,] b = MatirxPower(a, n);
 97             return b[1, 0];
 98         }
 99 
100         #region F3
101         static ulong[,] MatirxPower(ulong[,] a, int n)
102         {
103             if (n == 1) { return a; }
104             else if (n == 2) { return MatirxMultiplication(a, a); }
105             else if (n % 2 == 0)
106             {
107                 ulong[,] temp = MatirxPower(a, n / 2);
108                 return MatirxMultiplication(temp, temp);
109             }
110             else
111             {
112                 ulong[,] temp = MatirxPower(a, n / 2);
113                 return MatirxMultiplication(MatirxMultiplication(temp, temp), a);
114             }
115         }
116 
117         static ulong[,] MatirxMultiplication(ulong[,] a, ulong[,] b)
118         {
119             ulong[,] c = new ulong[2, 2];
120             for (int i = 0; i < 2; i++)
121             {
122                 for (int j = 0; j < 2; j++)
123                 {
124                     for (int k = 0; k < 2; k++)
125                     {
126                         c[i, j] += a[i, k] * b[k, j];
127                     }
128                 }
129             }
130             return c;
131         }
132         #endregion
133 
134         /// <summary>
135         /// 通项公式法
136         /// </summary>
137         /// <param name="n"></param>
138         /// <returns></returns>
139         static double F4(int n)
140         {
141             double sqrt5 = Math.Sqrt(5);
142             return (1/sqrt5*(Math.Pow((1+sqrt5)/2,n)-Math.Pow((1-sqrt5)/2,n)));
143         }
144     }
145 }

 

n=50时

 

n=500

 

 n=5000

 

 n=50000

 

 n=5000000

 

posted @ 2018-06-14 15:55  自恋狂学长  阅读(8027)  评论(0)  编辑  收藏  举报