Representar el número dado como la suma de dos números compuestos

Dado un número entero N , la tarea es representar N como la suma de dos números enteros compuestos. Puede haber múltiples formas posibles, imprima cualquiera de ellas. Si no es posible representar el número como la suma de dos números compuestos, imprima -1 .
Ejemplos: 
 

Entrada: N = 13 
Salida: 4 9 
4 + 9 = 13 y tanto 4 como 9 son compuestos.
Entrada: N = 18 
Salida: 4 14 
 

Enfoque: cuando N ≤ 11 , solo 8 y 10 son los números enteros que se pueden representar como la suma de dos números enteros compuestos, es decir, 4 + 4 y 4 + 6 respectivamente. 
Cuando N > 11 entonces hay dos casos: 
 

1. Cuando N es par: N se puede representar como 4 + (N – 4) ya que ambos son compuestos.
2. Cuando N es impar: N se puede representar como 9 + (N – 9) .

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find two composite
// numbers which when added
// give sum as n
void findNums(int n)
{
 
    // Only 8 and 10 can be represented
    // as the sum of two composite integers
    if (n <= 11) {
        if (n == 8)
            cout << "4 4";
        if (n == 10)
            cout << "4 6";
        else
            cout << "-1";
        return;
    }
 
    // If n is even
    if (n % 2 == 0)
        cout << "4 " << (n - 4);
 
    // If n is odd
    else
        cout << "9 " << (n - 9);
}
 
// Driver code
int main()
{
    int n = 13;
 
    findNums(n);
 
    return 0;
}

// Java implementation of the approach
class GFG
{
     
// Function to find two composite
// numbers which when added
// give sum as n
static void findNums(int n)
{
 
    // Only 8 and 10 can be represented
    // as the sum of two composite integers
    if (n <= 11)
    {
        if (n == 8)
            System.out.print("4 4");
        if (n == 10)
            System.out.print("4 6");
        else
            System.out.print("-1");
        return;
    }
 
    // If n is even
    if (n % 2 == 0)
        System.out.print("4 " + (n - 4));
 
    // If n is odd
    else
        System.out.print("9 " + (n - 9));
}
 
// Driver code
public static void main(String args[])
{
    int n = 13;
 
    findNums(n);
}
}
 
// This code is contributed by andrew1234

# Python3 implementation of the approach
 
# Function to find two composite
# numbers which when added
# give sum as n
def findNums(n):
 
    # Only 8 and 10 can be represented
    # as the sum of two composite integers
    if (n <= 11):
        if (n == 8):
            print("4 4", end = " ")
        if (n == 10):
            print("4 6", end = " ")
        else:
            print("-1", end = " ")
 
    # If n is even
    if (n % 2 == 0):
        print("4 ", (n - 4), end = " ")
 
    # If n is odd
    else:
        print("9 ", n - 9, end = " ")
 
# Driver code
n = 13
 
findNums(n)
 
# This code is contributed by Mohit Kumar

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to find two composite
    // numbers which when added
    // give sum as n
    static void findNums(int n)
    {
     
        // Only 8 and 10 can be represented
        // as the sum of two composite integers
        if (n <= 11)
        {
            if (n == 8)
                Console.Write("4 4");
            if (n == 10)
                Console.Write("4 6");
            else
                Console.Write("-1");
            return;
        }
     
        // If n is even
        if (n % 2 == 0)
            Console.Write("4 " + (n - 4));
     
        // If n is odd
        else
            Console.Write("9 " + (n - 9));
    }
     
    // Driver code
    public static void Main()
    {
        int n = 13;
     
        findNums(n);
    }
}
 
// This code is contributed by AnkitRai01

<script>
 
// javascript implementation of the approach
 
     
// Function to find two composite
// numbers which when added
// give sum as n
function findNums(n)
{
 
    // Only 8 and 10 can be represented
    // as the sum of two composite integers
    if (n <= 11)
    {
        if (n == 8)
            document.write("4 4");
        if (n == 10)
            document.write("4 6");
        else
            document.write("-1");
        return;
    }
 
    // If n is even
    if (n % 2 == 0)
        document.write("4 " + (n - 4));
 
    // If n is odd
    else
        document.write("9 " + (n - 9));
}
 
// Driver code
 
var n = 13;
 
findNums(n);
 
// This code contributed by shikhasingrajput
 
</script>

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