Números Pentanacci

La serie de Pentanacci es una generalización de la secuencia de Fibonacci donde cada término es la suma de los cinco términos anteriores. Los primeros números de Pentanacci son los siguientes: 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464, 912, 1793, 3525, 6930, 13624, 26784 , 52656, 103519…..

El enésimo término del número de Pentanacci viene dado por:

T(n) = T(n-1) + T(n-2) + T(n-3) + T(n-4) + T(n-5) 
con T(0) = T(1) = T(2) = T(3) = 0, T(4) = 1 

Encuentre el N número de ^Pentanacci

Dado un número N. La tarea es encontrar el N-ésimo número de Pentanacci.

Ejemplos: 

Entrada: N = 7 
Salida: 2

Entrada: N = 10 
Salida: 16 

Enfoque ingenuo: la idea es seguir la recurrencia para encontrar el número y usar la recursión para resolverlo.

Relación de recurrencia: 
T(n) = T(n-1) + T(n-2) + T(n-3) + T(n-4) + T(n-5) 

A continuación se muestra la implementación del enfoque anterior: 

// A simple recursive program to print
// Nth Pentanacci number
#include<bits/stdc++.h>
using namespace std;
     
// Recursive function to find the Nth
// Pentanacci number
int printpentaRec(int n)
{
    if (n == 0 || n == 1 ||
        n == 2 || n == 3)
        return 0;
         
    else if (n == 4 || n == 5)
        return 1;
    else
        return (printpentaRec(n - 1) +
                printpentaRec(n - 2) +
                printpentaRec(n - 3)+
                printpentaRec(n - 4)+
                printpentaRec(n - 5));
}
             
// Function to print the Nth
// Pentanacci number
void printPenta(int n)
{
    cout << printpentaRec(n) << "\n";
}
             
// Driver code
int main()
{
    int n = 10;
     
    printPenta(n);
     
    return 0;
}
 
// This code is contributed by yatinagg

// A simple recursive program to print
// Nth Pentanacci number
import java.util.*;
class GFG{
     
// Recursive function to find the Nth
// Pentanacci number
static int printpentaRec(int n)
{
    if (n == 0 || n == 1 ||
        n == 2 || n == 3)
        return 0;
         
    else if (n == 4 || n == 5)
        return 1;
    else
        return (printpentaRec(n - 1) +
                printpentaRec(n - 2) +
                printpentaRec(n - 3) +
                printpentaRec(n - 4) +
                printpentaRec(n - 5));
}
             
// Function to print the Nth
// Pentanacci number
static void printPenta(int n)
{
    System.out.print(printpentaRec(n) + "\n");
}
             
// Driver code
public static void main(String[] args)
{
    int n = 10;
     
    printPenta(n);
}
}
 
// This code is contributed by gauravrajput1

# A simple recursive program to print
# Nth Pentanacci number
   
# Recursive program to find the Nth
# Pentanacci number
def printpentaRec(n) :
    if (n == 0 or n == 1 or\
        n == 2 or n == 3):
        return 0
    elif (n == 4 or n == 5):
        return 1
    else :
        return (printpentaRec(n - 1) +
                printpentaRec(n - 2) +
                printpentaRec(n - 3)+
                printpentaRec(n - 4)+
                printpentaRec(n - 5))
           
# Function to print the Nth
# Pentanacci number
def printPenta(n) :
    print(printpentaRec(n))
           
   
# Driver code
n = 10
printPenta(n)

// A simple recursive program to print
// Nth Pentanacci number
using System;
 
class GFG{
     
// Recursive function to find the Nth
// Pentanacci number
static int printpentaRec(int n)
{
    if (n == 0 || n == 1 ||
        n == 2 || n == 3)
        return 0;
          
    else if (n == 4 || n == 5)
        return 1;
    else
        return (printpentaRec(n - 1) +
                printpentaRec(n - 2) +
                printpentaRec(n - 3) +
                printpentaRec(n - 4) +
                printpentaRec(n - 5));
}
              
// Function to print the Nth
// Pentanacci number
static void printPenta(int n)
{
    Console.WriteLine(printpentaRec(n));
}
 
// Driver code
static void Main()
{
    int n = 10;
      
    printPenta(n);
}
}
 
// This code is contributed divyeshrabadiya07

<script>
    // A simple recursive program to print
    // Nth Pentanacci number
     
    // Recursive function to find the Nth
    // Pentanacci number
    function printpentaRec(n)
    {
        if (n == 0 || n == 1 ||
            n == 2 || n == 3)
            return 0;
        else if (n == 4 || n == 5)
            return 1;
        else
            return (printpentaRec(n - 1) +
                    printpentaRec(n - 2) +
                    printpentaRec(n - 3)+
                    printpentaRec(n - 4)+
                    printpentaRec(n - 5));
    }
 
    // Function to print the Nth
    // Pentanacci number
    function printPenta(n)
    {
        document.write(printpentaRec(n) + "</br>");
    }
 
    let n = 10;    
    printPenta(n);
 
// This code is contributed by divyesh072019.
</script>

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Enfoque Eficiente: La idea es usar Programación Dinámica para resolver este problema. Esa es la memorización de la solución en cuatro variables para los últimos cuatro términos de modo que el mismo subproblema no se calcule una y otra vez.

 A continuación se muestra la implementación del enfoque anterior:

// C++14 implementation to print
// Nth Pentanacci numbers.
#include<bits/stdc++.h>
using namespace std;
     
// Function to print Nth
// Pentanacci number
void printpenta(int n)
{
    if (n < 0)
        return;
     
    // Initialize first five
    // numbers to base cases
    int first = 0;
    int second = 0;
    int third = 0;
    int fourth = 0;
    int fifth = 1;
     
    // Declare a current variable
    int curr = 0;
     
    if (n == 0 || n == 1 ||
        n == 2 || n == 3)
        cout << first << "\n";
         
    else if (n == 5)
        cout << fifth << "\n";
     
    else
    {
     
        // Loop to add previous five numbers
        // for each number starting from 5
        // and then assign first, second,
        // third, fourth to second, third, fourth
        // and curr to fifth respectively
        for(int i = 5; i < n; i++)
        {
            curr = first + second +
                   third + fourth + fifth;
            first = second;
            second = third;
            third = fourth;
            fourth = fifth;
            fifth = curr;
        }
    cout << curr << "\n";
    }
}
             
// Driver code
int main()
{
    int n = 10;
     
    printpenta(n);
     
    return 0;
}
 
// This code is contributed by yatinagg

// Java implementation to print
// Nth Pentanacci numbers.
import java.util.*;
class GFG{
     
// Function to print Nth
// Pentanacci number
static void printpenta(int n)
{
  if (n < 0)
    return;
 
  // Initialize first five
  // numbers to base cases
  int first = 0;
  int second = 0;
  int third = 0;
  int fourth = 0;
  int fifth = 1;
 
  // Declare a current variable
  int curr = 0;
 
  if (n == 0 || n == 1 ||
      n == 2 || n == 3)
    System.out.print(first + "\n");
  else if (n == 5)
    System.out.print(fifth + "\n");
  else
  {
    // Loop to add previous five numbers
    // for each number starting from 5
    // and then assign first, second,
    // third, fourth to second, third,
    // fourth and curr to fifth respectively
    for(int i = 5; i < n; i++)
    {
      curr = first + second +
             third + fourth + fifth;
      first = second;
      second = third;
      third = fourth;
      fourth = fifth;
      fifth = curr;
    }
    System.out.print(curr + "\n");
  }
}
             
// Driver code
public static void main(String[] args)
{
  int n = 10;
  printpenta(n);   
}
}
 
// This code is contributed by Princi Singh

# Python3 implementation to print
# Nth Pentanacci numbers.
   
# Function to print Nth
# Pentanacci number     
def printpenta(n) :
    if (n < 0): 
        return 
   
    # Initialize first five 
    # numbers to base cases 
    first = 0 
    second = 0 
    third = 0 
    fourth = 0
    fifth = 1
   
    # declare a current variable 
    curr = 0 
   
    if (n == 0  or n == 1 or\
        n == 2 or n == 3): 
        print(first)
    elif (n == 5): 
        print(fifth) 
   
    else:
   
        # Loop to add previous five numbers 
        # for each number starting from 5 
        # and then assign first, second, 
        # third, fourth to second, third, fourth 
        # and curr to fifth respectively 
        for i in range(5, n):
            curr = first + second +\
                 third + fourth + fifth
            first = second 
            second = third 
            third = fourth 
            fourth = fifth
            fifth = curr 
           
    print(curr)  
   
# Driver code
n = 10
printpenta(n)

// C# implementation to print
// Nth Pentanacci numbers.
using System;
class GFG{
     
// Function to print Nth
// Pentanacci number
static void printpenta(int n)
{
  if (n < 0)
    return;
 
  // Initialize first five
  // numbers to base cases
  int first = 0;
  int second = 0;
  int third = 0;
  int fourth = 0;
  int fifth = 1;
 
  // Declare a current variable
  int curr = 0;
 
  if (n == 0 || n == 1 ||
      n == 2 || n == 3)
    Console.Write(first + "\n");
  else if (n == 5)
    Console.Write(fifth + "\n");
  else
  {
    // Loop to add previous five numbers
    // for each number starting from 5
    // and then assign first, second,
    // third, fourth to second, third,
    // fourth and curr to fifth respectively
    for(int i = 5; i < n; i++)
    {
      curr = first + second +
             third + fourth + fifth;
      first = second;
      second = third;
      third = fourth;
      fourth = fifth;
      fifth = curr;
    }
    Console.Write(curr + "\n");
  }
}
 
// Driver code
public static void Main(String[] args)
{
  int n = 10;
  printpenta(n);   
}
}
 
// This code is contributed by shikhasingrajput

<script>
 
    // Javascript implementation to print
    // Nth Pentanacci numbers.
     
    // Function to print Nth
    // Pentanacci number
    function printpenta(n)
    {
      if (n < 0)
        return;
 
      // Initialize first five
      // numbers to base cases
      let first = 0;
      let second = 0;
      let third = 0;
      let fourth = 0;
      let fifth = 1;
 
      // Declare a current variable
      let curr = 0;
 
      if (n == 0 || n == 1 ||
          n == 2 || n == 3)
        document.write(first + "</br>");
      else if (n == 5)
        document.write(fifth + "</br>");
      else
      {
        // Loop to add previous five numbers
        // for each number starting from 5
        // and then assign first, second,
        // third, fourth to second, third,
        // fourth and curr to fifth respectively
        for(let i = 5; i < n; i++)
        {
          curr = first + second +
                 third + fourth + fifth;
          first = second;
          second = third;
          third = fourth;
          fourth = fifth;
          fifth = curr;
        }
        document.write(curr + "</br>");
      }
    }
     
    let n = 10;
      printpenta(n); 
     
</script>

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