五角星数

Pentanacci 系列 是斐波那契数列的推广，其中每一项都是前五项的总和。前几个Pentanacci 数如下 – 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464, 912, 1793, 3525, 6930, 136284, 26 , 52656, 103519…..

Pentanacci 数的第 N 项由下式给出：

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

找到第 N^个Pentanacci 数

给定一个数字N 。任务是找到第 N 个 Pentanacci 数。

例子：

Input: N = 7 
Output: 2

Input: N = 10 
Output: 16 

朴素的方法：这个想法是按照递归找到数字并使用递归来解决它。

Recurrence relation: 
T(n) = T(n-1) + T(n-2) + T(n-3) + T(n-4) + T(n-5) 

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

    // 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 ||
            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
    function printPenta(n)
    {
        document.write(printpentaRec(n) + "</br>");
    }
 
    let n = 10;    
    printPenta(n);
 
// This code is contributed by divyesh072019.

输出：

16

Efficient Approach：思路是使用动态规划来解决这个问题。那就是将最后四项的四个变量的解决方案记忆化，这样就不会一次又一次地计算相同的子问题。

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

蟒蛇3

# 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

 
    // 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); 
     

输出：

16