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。
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服务器之家。
原文链接：https://HdhCmsTestjianshu测试数据/p/31b783e3eb46
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