斐波那契问题(Fib(N)* Fib(N)– Fib(N-1)* Fib(N + 1)的值)

给定正整数N ，任务是找到Fib(N) ² –(Fib(N-1)* Fib(N + 1)) ，其中Fib(N)返回第N个斐波那契数。

例子：

Input: N = 3
Output: 1
Fib(3) * Fib(3) – Fib(2) * Fib(4) = 4 – 3 = 1

Input: N = 2
Output: -1
Fib(2) * Fib(2) – Fib(1) * Fib(3) = 1 – 2 = -1

方法：首先尝试一些测试用例就可以解决这个问题。让我们举几个例子：

For N = 1 : Fib(1) * Fib(1) – Fib(0) * Fib(2) = 1 – 0 = 1
For N = 2 : Fib(2) * Fib(2) – Fib(1) * Fib(3) = 1 – 2 = -1
For N = 3 : Fib(3) * Fib(3) – Fib(2) * Fib(4) = 4 – 3 = 1
For N = 4 : Fib(4) * Fib(4) – Fib(3) * Fib(5) = 9 – 10 = -1
We observe here that when N is even then the answer will be -1 and when the N is odd then the answer will be 1.

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
int getResult(int n)
{
    if (n & 1)
        return 1;
    return -1;
}
  
// Driver code
int main()
{
    int n = 3;
    cout << getResult(n);
}

Java

//Java implementation of the approach
  
import java.io.*;
  
class GFG {
      
static int getResult(int n)
{
    if ((n & 1)>0)
        return 1;
    return -1;
}
  
// Driver code
    public static void main (String[] args) {
      
    int n = 3;
    System.out.println(getResult(n));
    }
//This code is contributed by @Tushil.    
}

Python3

# Python 3 implementation of 
# the approach
def getResult(n):
  
    if n & 1:
        return 1
    return -1
  
# Driver code
n = 3
print(getResult(n))
  
# This code is contributed 
# by Shrikant13

C#

//C# implementation of the approach
  
using System;
  
class GFG {
      
static int getResult(int n)
{
    if ((n & 1)>0)
        return 1;
    return -1;
}
  
// Driver code
    public static void Main () {
      
    int n = 3;
    Console.WriteLine(getResult(n));
    }
//This code is contributed by anuj_67..
}

输出：

1