En la teoría de los números, un primo equilibrado es un número primo con huecos primos del mismo tamaño por encima y por debajo, de modo que es igual a la media aritmética de los primos más cercanos por encima y por debajo. O para decirlo algebraicamente, dado un número primo p n , donde n es su índice en el conjunto ordenado de números primos, los 
primeros primos balanceados son 5, 53, 157, 173……
Dado un entero positivo N . La tarea es imprimir el N-ésimo número primo balanceado.
Ejemplos:
Input : n = 2 Output : 53 Input : n = 3 Output : 157
La idea es generar números primos usando Sieve of Eratosthenes y almacenarlos en una array. Ahora itere sobre la array para verificar si es prima balanceada o no y siga contando la prima balanceada. Una vez que alcances el n-ésimo primo, devuélvelo.
A continuación se muestra la implementación de este enfoque:
C++
// CPP Program to find Nth Balanced Prime
#include<bits/stdc++.h>
#define MAX 501
using namespace std;
// Return the Nth balanced prime.
int balancedprime(int n)
{
// Sieve of Eratosthenes
bool prime[MAX+1];
memset(prime, true, sizeof(prime));
for (int p = 2; p*p <= MAX; p++)
{
// If prime[p] is not changed, then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p*2; i <= MAX; i += p)
prime[i] = false;
}
}
// storing all primes
vector<int> v;
for (int p = 3; p <= MAX; p += 2)
if (prime[p])
v.push_back(p);
int count = 0;
// Finding the Nth balanced Prime
for (int i = 1; i < v.size(); i++)
{
if (v[i] == (v[i+1] + v[i - 1])/2)
count++;
if (count == n)
return v[i];
}
}
// Driven Program
int main()
{
int n = 4;
cout << balancedprime(n) << endl;
return 0;
}
Java
// Java Program to find Nth Balanced Prime
import java.util.*;
public class GFG
{
static int MAX = 501;
// Return the Nth balanced prime.
public static int balancedprime(int n)
{
// Sieve of Eratosthenes
boolean[] prime = new boolean[MAX+1];
for(int k = 0 ; k < MAX+1; k++)
prime[k] = true;
for (int p = 2; p*p <= MAX; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p*2; i <= MAX;
i += p)
prime[i] = false;
}
}
// storing all primes
Vector<Integer> v =
new Vector<Integer>();
for (int p = 3; p <= MAX; p += 2)
if (prime[p])
v.add(p);
int count = 0;
// Finding the Nth balanced Prime
for (int i = 1; i < v.size(); i++)
{
if ((int)v.get(i) == ((int)v.get(i+1)
+ (int)v.get(i-1))/2)
count++;
if (count == n)
return (int) v.get(i);
}
return 1;
}
// Driven Program
public static void main(String[] args)
{
int n = 4;
System.out.print(balancedprime(n));
}
}
// This code is contributed by Prasad Kshirsagar
Python3
# Python3 code to find Nth Balanced Prime MAX = 501 # Return the Nth balanced prime. def balancedprime( n ): # Sieve of Eratosthenes prime = [True]*(MAX+1) p=2 while p * p <= MAX: # If prime[p] is not changed, # then it is a prime if prime[p] == True: # Update all multiples of p i = p * 2 while i <= MAX: prime[i] = False i = i + p p = p +1 # storing all primes v = list() p = 3 while p <= MAX: if prime[p]: v.append(p) p = p + 2 count = 0 # Finding the Nth balanced Prime i=1 for i in range(len(v)): if v[i] == (v[i+1] + v[i - 1])/2: count += 1 if count == n: return v[i] # Driven Program n = 4 print(balancedprime(n)) # This code is contributed by "Sharad_Bhardwaj".
PHP
<?php
// PHP Program to find Nth Balanced Prime
$MAX=501;
// Return the Nth balanced prime.
function balancedprime($n)
{
global $MAX;
// Sieve of Eratosthenes
$prime=array_fill(0,$MAX+1,true);
for ($p = 2; $p*$p <= $MAX; $p++)
{
// If prime[p] is not changed, then it is a prime
if ($prime[$p] == true)
{
// Update all multiples of p
for ($i = $p*2; $i <= $MAX; $i += $p)
$prime[$i] = false;
}
}
// storing all primes
$v=array();
for ($p = 3; $p <= $MAX; $p += 2)
if ($prime[$p])
array_push($v,$p);
$count = 0;
// Finding the Nth balanced Prime
for ($i = 1; $i < count($v); $i++)
{
if ($v[$i] == ($v[$i+1] + $v[$i - 1])/2)
$count++;
if ($count == $n)
return $v[$i];
}
}
// Driven Program
$n = 4;
echo balancedprime($n);
// this code is contributed by mits.
?>
C#
// C# Program to find Nth Balanced Prime
using System;
using System.Collections.Generic;
public class GFG
{
static int MAX = 501;
// Return the Nth balanced prime.
public static int balancedprime(int n)
{
// Sieve of Eratosthenes
bool[] prime = new bool[MAX+1];
for(int k = 0 ; k < MAX+1; k++)
prime[k] = true;
for (int p = 2; p*p <= MAX; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p*2; i <= MAX;
i += p)
prime[i] = false;
}
}
// storing all primes
List<int> v = new List<int>();
for (int p = 3; p <= MAX; p += 2)
if (prime[p])
v.Add(p);
int c = 0;
// Finding the Nth balanced Prime
for (int i = 1; i < v.Count-1; i++)
{
if ((int)v[i]==(int)(v[i+1]+v[i-1])/2)
c++;
if (c == n)
return (int) v[i];
}
return 1;
}
// Driven Program
public static void Main()
{
int n = 4;
Console.WriteLine(balancedprime(n));
}
}
// This code is contributed by mits
Javascript
<script>
// Javascript Program to find Nth Balanced Prime
var MAX = 501;
// Return the Nth balanced prime.
function balancedprime(n)
{
// Sieve of Eratosthenes
var prime = Array(MAX+1).fill(true);
for (var p = 2; p*p <= MAX; p++)
{
// If prime[p] is not changed, then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (var i = p*2; i <= MAX; i += p)
prime[i] = false;
}
}
// storing all primes
var v = [];
for (var p = 3; p <= MAX; p += 2)
if (prime[p])
v.push(p);
var count = 0;
// Finding the Nth balanced Prime
for (var i = 1; i < v.length; i++)
{
if (v[i] == (v[i+1] + v[i - 1])/2)
count++;
if (count == n)
return v[i];
}
}
// Driven Program
var n = 4;
document.write( balancedprime(n) );
</script>
Producción:
173