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

///

/// F(n)=(1/√5)*{[(1+√5)/2]^n - [(1-√5)/2]^n}

///

///

///

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();

}

///

/// 迭代法

///

///

///

private static ulong F1(int number)

{

if (number == 1 || number == 2)

{

return 1;

}

else

{

return F1(number - 1) + F1(number - 2);

}

}

///

/// 直接法

///

///

///

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;

}

}

///

/// 矩阵法

///

///

///

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

///

/// 通项公式法

///

///

///

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。

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持脚本之家。