Encuentra el término N de la serie 0, 1, 1, 2, 5, 29, 841…

Dado un entero positivo N , la tarea es encontrar el término N de la serie 0, 1, 1, 2, 5, 29, 841 …

Ejemplos:

Entrada: N = 6
Salida: 29
Explicación: El sexto término de la serie dada es 29.

Entrada: N = 1
Salida: 1

Entrada: N = 8
Salida: 750797

Enfoque: El problema dado es un problema matemático básico donde A _i = A _i-1 ² + A _i-2 ² . Por lo tanto, cree las variables a = 0 y b = 1 . Itere usando la variable i en el rango [2, N] , y para cada i , calcule el i ^-ésimo término y actualice el valor de a y b a i – 1 ^th e i – 2 ^th término respectivamente.

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find N-th term
// of the given series
int getNthTerm(int N)
{
    if (N < 3)
        return N - 1;
 
    // Initialize Variables repre-
    // senting 1st and 2nd term
    long int a = 0, b = 1;
 
    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {
 
        // pow((i - 2)th term, 2) +
        // pow((i - 1)th term, 2)
        long int c = a * a + b * b;
 
        // Update a and b
        a = b;
        b = c;
    }
 
    // Return Answer
    return b;
}
 
// Driver Code
int main()
{
    int N = 8;
    cout << getNthTerm(N);
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG{
 
  // Function to find N-th term
  // of the given series
  static int getNthTerm(int N)
  {
    if (N < 3)
      return N - 1;
 
    // Initialize Variables repre-
    // senting 1st and 2nd term
    int a = 0, b = 1;
 
    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {
 
      // Math.pow((i - 2)th term, 2) +
      // Math.pow((i - 1)th term, 2)
      int c = a * a + b * b;
 
      // Update a and b
      a = b;
      b = c;
    }
 
    // Return Answer
    return b;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 8;
    System.out.print(getNthTerm(N));
  }
}
 
// This code is contributed by 29AjayKumar

# Python code for the above approach
 
# Function to find N-th term
# of the given series
def getNthTerm(N):
    if (N < 3):
        return N - 1
       
    # Initialize Variables repre-
    # senting 1st and 2nd term
    a = 0
    b = 1
     
    # Loop to iterate through the
    # range [3, N] using variable i
    for i in range(3, N + 1):
       
        # pow((i - 2)th term, 2) +
        # pow((i - 1)th term, 2)
        c = a * a + b * b
         
        # Update a and b
        a = b
        b = c
         
    # Return Answer
    return b
 
# Driver Code
N = 8
print(getNthTerm(N))
 
# This code is contributed by Saurabh Jaiswal

// C# program for the above approach
using System;
class GFG
{
   
// Function to find N-th term
// of the given series
static int getNthTerm(int N)
{
    if (N < 3)
        return N - 1;
 
    // Initialize Variables repre-
    // senting 1st and 2nd term
    long a = 0, b = 1;
 
    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {
 
        // pow((i - 2)th term, 2) +
        // pow((i - 1)th term, 2)
        long c = a * a + b * b;
 
        // Update a and b
        a = b;
        b = c;
    }
 
    // Return Answer
    return (int)b;
}
 
// Driver Code
public static void Main()
{
    int N = 8;
    Console.Write(getNthTerm(N));
 
}
}
 
// This code is contributed by Samim Hossain Mondal.

<script>
  // JavaScript code for the above approach
 
 
  // Function to find N-th term
  // of the given series
  function getNthTerm(N) {
    if (N < 3)
      return N - 1;
 
    // Initialize Variables repre-
    // senting 1st and 2nd term
    let a = 0, b = 1;
 
    // Loop to iterate through the
    // range [3, N] using variable i
    for (let i = 3; i <= N; i++) {
 
      // pow((i - 2)th term, 2) +
      // pow((i - 1)th term, 2)
      let c = a * a + b * b;
 
      // Update a and b
      a = b;
      b = c;
    }
 
    // Return Answer
    return b;
  }
 
  // Driver Code
 
  let N = 8;
  document.write(getNthTerm(N));
 
 
// This code is contributed by Potta Lokesh
</script>

750797

Complejidad temporal: O(N)
Espacio auxiliar: O(1)