[C# Exercise] Recursively calculate f(n)=f(n-1)+f(n-2) f(0)=2 f(1)=3

Title: f(n)=f(n-1)+f(n-2)

f(0)=2; f(1)=3, find f(40)

This problem requires the use of function recursion to calculate,

We know  f(0) and  f(1) ,

Then by calculation,  f(2) is  f(0) +  f(1) = 2 + 3 = 5;

Create a function F, the input value is int n , and the return value is int;

static int F(int n)
{
}

Return the formula directly in the function:

return F(n - 1) + F(n - 2);//函数递归调用

At this time, there needs to be a condition to terminate the recursion, write  f (0) and  f(1) ;

if (n == 0) return 2;   //终止递归的条件
if (n == 1) return 3;

Call the function F in the main function, and the obtained res value is the value of f(40), and calculate the value of f(2) as a comparison:

static void Main(string[] args)
{
    int res = F(40);
    int res2 = F(2);
}

The source code is as follows:

using System;
//02
//f(n)=f(n-1)+f(n-2) f(0)=2 f(1)=3 ,用程序求得f(40)
namespace _05s
{
    class Program
    {
        static void Main(string[] args)
        {
            int res = F(40);
            int res2 = F(2);
            Console.WriteLine("f40：" + res);
            Console.WriteLine("f2：" + res2);
        }
        static int F(int n)
        {
            if (n == 0) return 2;   //终止递归的条件
            if (n == 1) return 3;
            return F(n - 1) + F(n - 2);//函数递归调用
        }
    }
}

The printed results are as follows:

Sanlian~