secuencia de Recaman

Dado un entero n. Imprime los primeros n elementos de la secuencia de Recaman .
Ejemplos: 
 

Input : n = 6
Output : 0, 1, 3, 6, 2, 7

Input  : n = 17
Output : 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 
         11, 22, 10, 23, 9, 24, 8

Es básicamente una función con dominio y codominio como números naturales y 0. Se define recursivamente de la siguiente manera: 
Específicamente, deje que a(n) denote el (n+1)-ésimo término. (0 ya está allí). 
La regla dice: 

a(0) = 0,
if n > 0 and the number is not 
   already included in the sequence,
     a(n) = a(n - 1) - n 
else 
     a(n) = a(n-1) + n. 

A continuación se muestra una implementación simple en la que almacenamos todos los números de secuencia de n Recaman en una array. Calculamos el siguiente número usando la fórmula recursiva mencionada anteriormente. 
 

C++

// C++ program to print n-th number in Recaman's
// sequence
#include <bits/stdc++.h>
using namespace std;
 
// Prints first n terms of Recaman sequence
int recaman(int n)
{
    // Create an array to store terms
    int arr[n];
 
    // First term of the sequence is always 0
    arr[0] = 0;
    printf("%d, ", arr[0]);
 
    // Fill remaining terms using recursive
    // formula.
    for (int i=1; i< n; i++)
    {
        int curr = arr[i-1] - i;
        int j;
        for (j = 0; j < i; j++)
        {
            // If arr[i-1] - i is negative or
            // already exists.
            if ((arr[j] == curr) || curr < 0)
            {
                curr = arr[i-1] + i;
                break;
            }
        }
 
        arr[i] = curr;
        printf("%d, ", arr[i]);
    }
}
 
// Driver code
int main()
{
    int n = 17;
    recaman(n);
    return 0;
}

Java

// Java program to print n-th number in Recaman's
// sequence
import java.io.*;
 
class GFG {
     
    // Prints first n terms of Recaman sequence
    static void recaman(int n)
    {
        // Create an array to store terms
        int arr[] = new int[n];
     
        // First term of the sequence is always 0
        arr[0] = 0;
        System.out.print(arr[0]+" ,");
     
        // Fill remaining terms using recursive
        // formula.
        for (int i = 1; i < n; i++)
        {
            int curr = arr[i - 1] - i;
            int j;
            for (j = 0; j < i; j++)
            {
                // If arr[i-1] - i is negative or
                // already exists.
                if ((arr[j] == curr) || curr < 0)
                {
                    curr = arr[i - 1] + i;
                    break;
                }
            }
     
            arr[i] = curr;
            System.out.print(arr[i]+", ");
             
        }
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 17;
        recaman(n);
 
    }
}
 
// This code is contributed by vt_m

Python 3

# Python 3 program to print n-th
# number in Recaman's sequence
 
# Prints first n terms of Recaman
# sequence
def recaman(n):
 
    # Create an array to store terms
    arr = [0] * n
 
    # First term of the sequence
    # is always 0
    arr[0] = 0
    print(arr[0], end=", ")
 
    # Fill remaining terms using
    # recursive formula.
    for i in range(1, n):
     
        curr = arr[i-1] - i
        for j in range(0, i):
         
            # If arr[i-1] - i is
            # negative or already
            # exists.
            if ((arr[j] == curr) or curr < 0):
                curr = arr[i-1] + i
                break
             
        arr[i] = curr
        print(arr[i], end=", ")
 
# Driver code
n = 17
 
recaman(n)
 
# This code is contributed by Smitha.

C#

// C# program to print n-th number in Recaman's
// sequence
using System;
 
class GFG {
     
    // Prints first n terms of Recaman sequence
    static void recaman(int n)
    {
        // Create an array to store terms
        int []arr = new int[n];
     
        // First term of the sequence is always 0
        arr[0] = 0;
        Console.Write(arr[0]+" ,");
     
        // Fill remaining terms using recursive
        // formula.
        for (int i = 1; i < n; i++)
        {
            int curr = arr[i - 1] - i;
            int j;
            for (j = 0; j < i; j++)
            {
                // If arr[i-1] - i is negative or
                // already exists.
                if ((arr[j] == curr) || curr < 0)
                {
                    curr = arr[i - 1] + i;
                    break;
                }
            }
     
            arr[i] = curr;
        Console.Write(arr[i]+", ");
             
        }
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 17;
        recaman(n);
 
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// PHP program to print n-th
// number in Recaman's sequence
 
// Prints first n terms
// of Recaman sequence
function recaman($n)
{
     
    // First term of the
    // sequence is always 0
    $arr[0] = 0;
    echo $arr[0], ", ";
 
    // Fill remaining terms
    // using recursive formula.
    for ($i = 1; $i < $n; $i++)
    {
            $curr = $arr[$i - 1] - $i;
            $j;
        for ($j = 0; $j < $i; $j++)
        {
             
            // If arr[i-1] - i
            // is negative or
            // already exists.
            if (($arr[$j] == $curr) || $curr < 0)
            {
                $curr = $arr[$i-1] + $i;
                break;
            }
        }
 
        $arr[$i] = $curr;
        echo $arr[$i], ", ";
    }
}
 
    // Driver Code
    $n = 17;
    recaman($n);
     
// This code is contributed by Ajit
?>

Javascript

<script>
 
    // Javascript program to print
    // n-th number in Recaman's sequence
     
    // Prints first n terms of Recaman sequence
    function recaman(n)
    {
        // Create an array to store terms
        let arr = new Array(n);
       
        // First term of the sequence is always 0
        arr[0] = 0;
        document.write(arr[0]+" ,");
       
        // Fill remaining terms using recursive
        // formula.
        for (let i = 1; i < n; i++)
        {
            let curr = arr[i - 1] - i;
            let j;
            for (j = 0; j < i; j++)
            {
                // If arr[i-1] - i is negative or
                // already exists.
                if ((arr[j] == curr) || curr < 0)
                {
                    curr = arr[i - 1] + i;
                    break;
                }
            }
       
            arr[i] = curr;
        document.write(arr[i]+", ");
               
        }
    }
     
      let n = 17;
      recaman(n);
     
</script>

Producción:  

0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 

Complejidad de tiempo : O(n 2Espacio
auxiliar : O(n), ya que se ha agregado n espacio extra Optimizaciones:  Podemos usar hash para almacenar valores previamente calculados y podemos hacer que este programa funcione en O(n) tiempo. 

 

C++

// C++ program to print n-th number in Recaman's
// sequence
#include <bits/stdc++.h>
using namespace std;
 
// Prints first n terms of Recaman sequence
void recaman(int n)
{
    if (n <= 0)
      return;
 
    // Print first term and store it in a hash
    printf("%d, ", 0);
    unordered_set<int> s;
    s.insert(0);
 
    // Print remaining terms using recursive
    // formula.
    int prev = 0;
    for (int i=1; i< n; i++)
    {
        int curr = prev - i;
 
        // If arr[i-1] - i is negative or
        // already exists.
        if (curr < 0 || s.find(curr) != s.end())
           curr = prev + i;
 
        s.insert(curr);
 
        printf("%d, ", curr);
        prev = curr;
    }
}
 
// Driver code
int main()
{
    int n = 17;
    recaman(n);
    return 0;
}

Java

// Java program to print n-th number
// in Recaman's sequence
import java.util.*;
 
class GFG
{
 
// Prints first n terms of Recaman sequence
static void recaman(int n)
{
    if (n <= 0)
    return;
 
    // Print first term and store it in a hash
    System.out.printf("%d, ", 0);
    HashSet<Integer> s = new HashSet<Integer>();
    s.add(0);
 
    // Print remaining terms using
    // recursive formula.
    int prev = 0;
    for (int i = 1; i< n; i++)
    {
        int curr = prev - i;
 
        // If arr[i-1] - i is negative or
        // already exists.
        if (curr < 0 || s.contains(curr))
            curr = prev + i;
 
        s.add(curr);
 
        System.out.printf("%d, ", curr);
        prev = curr;
    }
}
 
// Driver code
public static void main(String[] args)
{
    int n = 17;
    recaman(n);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to print n-th number in
# Recaman's sequence
 
# Prints first n terms of Recaman sequence
def recaman(n):
 
    if(n <= 0):
        return
 
    # Print first term and store it in a hash
    print(0, ",", end='')
    s = set([])
    s.add(0)
 
    # Print remaining terms using recursive
    # formula.
    prev = 0
    for i in range(1, n):
 
        curr = prev - i
 
        # If arr[i-1] - i is negative or
        # already exists.
        if(curr < 0 or curr in s):
            curr = prev + i
 
        s.add(curr)
 
        print(curr, ",", end='')
        prev = curr
 
# Driver code
if __name__=='__main__':
    n = 17
    recaman(n)
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to print n-th number
// in Recaman's sequence
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Prints first n terms of Recaman sequence
static void recaman(int n)
{
    if (n <= 0)
    return;
 
    // Print first term and store it in a hash
    Console.Write("{0}, ", 0);
    HashSet<int> s = new HashSet<int>();
    s.Add(0);
 
    // Print remaining terms using
    // recursive formula.
    int prev = 0;
    for (int i = 1; i < n; i++)
    {
        int curr = prev - i;
 
        // If arr[i-1] - i is negative or
        // already exists.
        if (curr < 0 || s.Contains(curr))
            curr = prev + i;
 
        s.Add(curr);
 
        Console.Write("{0}, ", curr);
        prev = curr;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 17;
    recaman(n);
}
}
 
// This code is contributed by Princi Singh

PHP

<?php
// PHP program to print n-th number in
// Recaman's sequence
 
// Prints first n terms of Recaman sequence
function recaman($n)
{
    if($n <= 0)
        return;
 
    // Print first term and store
    // it in a hash
    print("0, ");
    $s = array();
    array_push($s, 0);
 
    // Print remaining terms using recursive
    // formula.
    $prev = 0;
    for ($i = 1; $i < $n; $i++)
    {
        $curr = $prev - $i;
 
        // If arr[i-1] - i is negative or
        // already exists.
        if($curr < 0 or in_array($curr, $s))
            $curr = $prev + $i;
 
        array_push($s, $curr);
 
        print($curr.", ");
        $prev = $curr;
    }
         
}
 
// Driver code
$n = 17;
recaman($n);
 
// This code is contributed by chandan_jnu
?>

Javascript

<script>
 
//  Javascript program to print n-th number
// in Recaman's sequence
 
// Prints first n terms of Recaman sequence
function recaman(n)
{
    if (n <= 0)
    return;
  
    // Print first term and store it in a hash
    document.write(0 + ", ");
    let s = new Set();
    s.add(0);
  
    // Print remaining terms using
    // recursive formula.
    let prev = 0;
    for (let i = 1; i< n; i++)
    {
        let curr = prev - i;
  
        // If arr[i-1] - i is negative or
        // already exists.
        if (curr < 0 || s.has(curr))
            curr = prev + i;
  
        s.add(curr);
  
        document.write(curr + ", ");
        prev = curr;
    }
}
     
    // Driver code
     
    let n = 17;
    recaman(n);
     
</script>

Producción: 
 

0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 

Complejidad del tiempo: O(n) 
Espacio auxiliar: O(n)
Este artículo es una contribución de Kishlay Verma . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
 

Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

Deja una respuesta

Tu dirección de correo electrónico no será publicada. Los campos obligatorios están marcados con *