Encuentre el N-ésimo término de la serie donde cada término f[i] = f[i – 1] – f[i – 2]

Dados tres enteros X , Y y N , la tarea es encontrar el ^término N de la serie f[i] = f[i – 1] – f[i – 2] , i > 1 donde f[0] = X yf [1] = Y.
Ejemplos: 
 

Entrada: X = 2, Y = 3, N = 3 
Salida: -2 
La serie será 2 3 1 -2 -3 -1 2 y f[3] = -2
Entrada: X = 3, Y = 7, N = 8 
Salida: 4 
 

Enfoque: Una observación importante aquí es que habrá como máximo 6 términos distintos, antes de que la secuencia comience a repetirse. Por lo tanto, encuentre los primeros 6 términos de la serie y luego el N ^-ésimo término sería el mismo que el (N % 6) ^-ésimo término.
A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the Nth term
// of the given series
int findNthTerm(int x, int y, int n)
{
    int f[6];
 
    // First and second term of the series
    f[0] = x;
    f[1] = y;
 
    // Find first 6 terms
    for (int i = 2; i <= 5; i++)
        f[i] = f[i - 1] - f[i - 2];
 
    // Return the Nth term
    return f[n % 6];
}
 
// Driver code
int main()
{
    int x = 2, y = 3, n = 3;
    cout << findNthTerm(x, y, n);
 
    return 0;
}

// Java implementation of the approach
import java.io.*;
 
class GFG
{
     
    // Function to find the nth term of series
    static int findNthTerm(int x, int y, int n)
    {    
        int[] f = new int[6];
         
        f[0] = x;
        f[1] = y;
         
        // Loop to add numbers
        for (int i = 2; i <= 5; i++)
            f[i] = f[i - 1] - f[i - 2];
         
        return f[n % 6];
    }
 
     
    // Driver code
    public static void main(String args[])
    {
        int x = 2, y = 3, n = 3;
        System.out.println(findNthTerm(x, y, n));
    }
}
 
// This code is contributed by mohit kumar 29

# Python3 implementation of the approach
 
# Function to return the Nth term
# of the given series
def findNthTerm(x, y, n):
 
    f = [0] * 6
 
    # First and second term of
    # the series
    f[0] = x
    f[1] = y
 
    # Find first 6 terms
    for i in range(2, 6):
        f[i] = f[i - 1] - f[i - 2]
 
    # Return the Nth term
    return f[n % 6]
 
# Driver code
if __name__ == "__main__":
 
    x, y, n = 2, 3, 3
    print(findNthTerm(x, y, n))
 
# This code is contributed by
# Rituraj Jain

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to find the nth term of series
    static int findNthTerm(int x, int y, int n)
    {
        int[] f = new int[6];
         
        f[0] = x;
        f[1] = y;
         
        // Loop to add numbers
        for (int i = 2; i <= 5; i++)
            f[i] = f[i - 1] - f[i - 2];
         
        return f[n % 6];
    }
 
    // Driver code
    public static void Main()
    {
        int x = 2, y = 3, n = 3;
        Console.WriteLine(findNthTerm(x, y, n));
    }
}
 
// This code is contributed by Ryuga

<?php
 
//PHP implementation of the approach
// Function to find the nth term of series
function findNthTerm($x, $y, $n)
{
    $f = array(6);
         
    $f[0] = $x;
    $f[1] = $y;
         
    // Loop to add numbers
    for ($i = 2; $i <= 5; $i++)
        $f[$i] = $f[$i - 1] - $f[$i - 2];
         
    return $f[$n % 6];
}
 
// Driver code
$x = 2; $y = 3; $n = 3;
echo(findNthTerm($x, $y, $n));
 
// This code is contributed by Code_Mech.
?>

<script>
 
// JavaScript implementation of the approach
 
    // Function to return the Nth term
    // of the given series
    function findNthTerm(x, y, n)
    {
        let f = new Array(6);
   
        // First and second term of the series
        f[0] = x;
        f[1] = y;
   
        // Find first 6 terms
        for (let i = 2; i <= 5; i++)
            f[i] = f[i - 1] - f[i - 2];
   
        // Return the Nth term
        return f[n % 6];
    }
   
    // Driver code
 
    let x = 2, y = 3, n = 3;
    document.write(findNthTerm(x, y, n));
   
// This code is contributed by Surbhi Tyagi
 
</script>

-2

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