第N个XOR斐波那契数

给定三个整数a ， b和N ，其中a和b是XOR斐波纳契数列的前两个项，任务是找到第N^个项。
XOR斐波那契数列的第N个项定义为F(N)= F(N – 1)^ F(N – 2) ，其中^是按位XOR。

例子：

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

方法：因为， a ^ a = 0 ，所以给定

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
…

可以看出，答案每3个数字就会重复一次。因此答案是F(N％3) ，其中F(0)= a，F(1)= b和F(2)= a ^ b 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

输出：

2

时间复杂度： O(1)