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

Javascript

 
// 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));
 

输出：

2

时间复杂度： O(1)

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。