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 Nth 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++
#include <bits/stdc++.h>
using namespace std;
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);
}
int main()
{
int a = 1, b = 2, n = 10;
cout << nthXorFib(n, a, b);
return 0;
}
|
Java
class GFG
{
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);
}
public static void main (String[] args)
{
int a = 1 , b = 2 , n = 10 ;
System.out.println(nthXorFib(n, a, b));
}
}
|
Python3
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)
a = 1
b = 2
n = 10
print (nthXorFib(n, a, b))
|
C#
using System;
class GFG
{
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);
}
public static void Main (String[] args)
{
int a = 1, b = 2, n = 10;
Console.WriteLine(nthXorFib(n, a, b));
}
}
|
Javascript
<script>
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);
}
let a = 1, b = 2, n = 10;
document.write(nthXorFib(n, a, b));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Last Updated :
19 Oct, 2022
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