Nth XOR Fibonacci number

Given three integers a, b and N where a and b are the first two terms of the XOR Fibonacci series and the task is to find the N^th term. 
The Nth term of the XOR Fibonacci series is defined as F(N) = F(N – 1) ^ F(N – 2) where ^ is the bitwise XOR.
Examples:

Input: a = 1, b = 2, N = 5 
Output: 3 
F(0) = 1 
F(1) = 2 
F(2) = 1 ^ 2 = 3 
F(3) = 2 ^ 3 = 1 
F(4) = 1 ^ 3 = 2 
F(5) = 1 ^ 2 = 3

Input: a = 5, b = 11, N = 1000001 
Output: 14 

Approach: Since, a ^ a = 0 and it is given that 

F(0) = a and F(1) = b
Now, F(2) = F(0) ^ F(1) = a ^ b
And, F(3) = F(1) ^ F(2) = b ^ (a ^ b) = a
F(4) = a ^ b ^ a = b
F(5) = a ^ b
F(6) = a
F(7) = b
F(8) = a ^ b
… 

It can be observed that the answer repeats itself after every 3 numbers. So the answer is F(N % 3) where F(0) = a, F(1) = b and F(2) = a ^ b.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the nth XOR Fibonacci number
int nthXorFib(int n, int a, int b)
{
    if (n == 0)
        return a;
    if (n == 1)
        return b;
    if (n == 2)
        return (a ^ b);

    return nthXorFib(n % 3, a, b);
}

// Driver code
int main()
{
    int a = 1, b = 2, n = 10;

    cout << nthXorFib(n, a, b);

    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

    // Function to return the
    // nth XOR Fibonacci number
    static int nthXorFib(int n, int a, int b)
    {
        if (n == 0)
            return a;
        if (n == 1)
            return b;
        if (n == 2)
            return (a ^ b);

        return nthXorFib(n % 3, a, b);
    }

    // Driver code
    public static void main (String[] args)
    {
        int a = 1, b = 2, n = 10;

        System.out.println(nthXorFib(n, a, b));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to return
# the nth XOR Fibonacci number
def nthXorFib(n, a, b):
    if n == 0 :
        return a
    if n == 1 :
        return b
    if n == 2 :
        return a ^ b

    return nthXorFib(n % 3, a, b)

# Driver code
a = 1
b = 2
n = 10
print(nthXorFib(n, a, b))

# This code is contributed by divyamohan123

C#

// C# implementation of the above approach
using System;

class GFG
{

    // Function to return the
    // nth XOR Fibonacci number
    static int nthXorFib(int n, int a, int b)
    {
        if (n == 0)
            return a;
        if (n == 1)
            return b;
        if (n == 2)
            return (a ^ b);

        return nthXorFib(n % 3, a, b);
    }

    // Driver code
    public static void Main (String[] args)
    {
        int a = 1, b = 2, n = 10;

        Console.WriteLine(nthXorFib(n, a, b));
    }
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript implementation of the approach

// Function to return the nth XOR Fibonacci number
function nthXorFib(n, a, b)
{
    if (n == 0)
        return a;
    if (n == 1)
        return b;
    if (n == 2)
        return (a ^ b);

    return nthXorFib(n % 3, a, b);
}

// Driver code
    let a = 1, b = 2, n = 10;

    document.write(nthXorFib(n, a, b));

</script>

Output:

2

Time Complexity: O(1)
Auxiliary Space: O(1)