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  • C#斐波那契数列求法(比较阶乘和循环所用时间)

    using System;
    
    namespace ConsoleApp3
    {
        class Program
        {
            static void Main(string[] args)
            {
                Console.Write("你要输入多少项?");
                int a = Convert.ToInt32(Console.ReadLine());
                Console.WriteLine();
                DateTime dt1 = System.DateTime.Now;
                for (int i = 1; i <= a; i++)
                {
                    Console.Write("	{0}", J(i));
                    if (i % 5 == 0)
                    {
                        Console.WriteLine();
                    }
                }
                DateTime dt2 = System.DateTime.Now;
                TimeSpan ts = dt2.Subtract(dt1);
                Console.WriteLine("3变量循环方法用时{0}", ts.TotalMilliseconds);//3变量循环方法所用时间
    
                DateTime dt11 = System.DateTime.Now;
                for (int i = 1; i <= a; i++)
                {
                    Console.Write("	{0}", J1(i));
                    if (i % 5 == 0)
                    {
                        Console.WriteLine();
                    }
                }
                DateTime dt22 = System.DateTime.Now;
                TimeSpan ts1 = dt22.Subtract(dt11);
                Console.WriteLine("2变量循环方法用时{0}", ts1.TotalMilliseconds);//2变量循环方法所用时间
    
                DateTime dt111 = System.DateTime.Now;
                for (int i = 1; i <= a; i++)
                {
                    Console.Write("	{0}", J2(i));
                    if (i % 5 == 0)
                    {
                        Console.WriteLine();
                    }
                }
                DateTime dt222 = System.DateTime.Now;
                TimeSpan ts11 = dt222.Subtract(dt111);
                Console.WriteLine("递归方法用时{0}", ts11.TotalMilliseconds);//递归方法所用时间
            }
            /// <summary>
            /// 3个变量循环求斐波那契数列
            /// </summary>
            /// <param name="b"></param>
            /// <returns></returns>
            static int J(int b)
            {
                int x = 1;
                int y = 1;
                if (b == 1 || b == 2)
                {
                    return 1;
                }
                else
                {
                    for (int i = 3; i <= b; i++)
                    {
                        int z = x + y;
                        y = x;
                        x = z;
                    }
                    return x;
                }
            }
            /// <summary>
            /// 2个变量循环求斐波那契数列
            /// </summary>
            /// <param name="b"></param>
            /// <returns></returns>
            static int J1(int b)
            {
                int x = 1;
                int y = 1;
                if (b == 1 || b == 2)
                {
                    return 1;
                }
                else
                {
                    for (int i = 3; i <= b; i++)
                    {
                        y = x + y;
                        x = y-x;
                    }
                    return y;
                }
            }
    
            /// <summary>
            /// 阶乘求斐波那契数列
            /// </summary>
            /// <param name="b"></param>
            /// <returns></returns>
            static int J2(int b)
            {
                if (b == 1 || b == 2)
                {
                    return 1;
                }
                else
                {
                    return J2(b - 1) + J2(b - 2);
                }
            }
        }
    }

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  • 原文地址:https://www.cnblogs.com/BruceKing/p/11820763.html
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