📜  彭塔纳奇数字

📅  最后修改于: 2021-04-21 23:53:06             🧑  作者: Mango

Pentanacci系列 是斐波那契数列的概括,其中每个项是前五个项的和。前几个Pentanacci编号如下-0、0、0、0、1、1、2、4、8、16、31、61、120、236、464、912、1793、3525、6930、13624、26784 ,52656,103519…..

彭塔纳奇数的第N项由下式给出:

找出第NPentanacci号码

给定数字N。任务是找到第N个Pentanacci编号。

例子:

天真的方法:想法是跟随递归来找到数字并使用递归来求解。

下面是上述方法的实现:

C++14
// A simple recursive program to print
// Nth Pentanacci number
#include
using namespace std;
     
// Recursive function to find the Nth
// Pentanacci number
int printpentaRec(int n)
{
    if (n == 0 || n == 1 ||
        n == 2 || n == 3 ||
        n == 4)
        return 0;
         
    else if (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


Java
// 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 ||
        n == 4)
        return 0;
         
    else if (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


Python3
# 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 or n == 4):
        return 0
    elif (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)


C#
// 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 ||
        n == 4)
        return 0;
          
    else if (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


Javascript


C++14
// C++14 implementation to print
// Nth Pentanacci numbers.
#include
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
// 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
# 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#
// 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


Javascript


输出:
16

高效的方法:想法是使用动态编程来解决此问题。那就是对最后四个术语的四个变量的解决方案的记忆,这样就不会一次又一次地计算出相同的子问题。

下面是上述方法的实现:

C++ 14

// C++14 implementation to print
// Nth Pentanacci numbers.
#include
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

// 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

# 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#

// 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

Java脚本


输出:
16