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Recursion

Theory

Alternating Fibonacci numbers

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Given the small integer n (0 <= n <= 40) you need to find the n-th number of the alternating Fibonacci sequence.

The sequence starts with 0, 1, -1, 2, -3, 5, -8, 13, -21, ...

So, fib(0) = 0, fib(1) = 1 => fib(2) = -1.

Think of the recurrence relation and implement the method named fib in a recursive way. It's not efficient in the general but works well for small n.

Sample Input 1: 

2

Sample Output 1: 

-1

Sample Input 2: 

3

Sample Output 2: 

2
Write a program in Java 17
import java.util.Scanner;

public class Main {

public static long fib(long n) {
// write your code here
}

/* Do not change code below */
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
System.out.println(fib(n));
}
}
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