Question

In either Java or C#

Instead of a regular Fibonacci number, you are supposed to calculate a special one as follows:

F(n) = F(n-1)+ 2 * F(n-2) + 3 * F(n-3)

As an example, if F(0) = F(1) = F(2) = 1, then we have:

F(3) = F(2) + 2 * F(1) + 3 * F(0) = 1 + 2 + 3 = 6 F(4) = F(3) + 2 * F(2) + 3 * F(1) = 6 + 2 + 3 = 11

Given F(0), F(1), F(2), and N, your job is to calculate F(N).

Input Format

First number is F(0), second number is F(1), third number is F(2), and last number is N.

Example input: 1 1 1 4

Answer 1

import java.util.*;
class Solution{
   static int a,b,c,n;
   public static void main(String args[])
   {
       Scanner sc=new Scanner(System.in);
       a=sc.nextInt();
       b=sc.nextInt();
       c=sc.nextInt();
       n=sc.nextInt();
       sc.close();
       System.out.println(function(n));
   }
   static int function(int n1)
   { if(n1==0)
           return a;
       if(n1==1)
           return b;
       if(n1==2)
           return c;
      
       return function(n1-1)+2*function(n1-2)+3*function(n1-3);
         
   }
}

Tested Input(s):

a) 1 1 1 4 Output: 11

b)2 3 4 7 Output: 444  

The above code is written recursively such that the function calls itself every time.

The code is very simple.When even the called function parameter value becomes one of the 3 inputs a,b,c,  we return them.

else we return function(n1-1)+2*function(n1-2)+3*function(n1-3)