Generate N integers satisfying the given conditions

Given an integer N, the task is to generate an array of size N with the following properties:

  1. No two elements divide each other.
  2. Every odd subset has odd sum and every even subset has even sum.

Examples:

Input: N = 3
Output: 3 5 7
No two element divide each other and the sum
of all the odd subsets {3}, {5}, {7} and {3, 5, 7} is odd.
Sum of all the even subsets is even i.e. {3, 5}, {3, 7} and {5, 7}

Input: N = 6
Output: 3 5 7 11 13 17

Approach: In order to satisfy the condition when every odd subset has odd sum and even subset has even sum, every element has to be odd and in order for any two elements to not divide each other they must be prime. So, the task now is to find the first N odd prime numbers.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define MAX 1000000
  
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
bool prime[MAX + 1];
void SieveOfEratosthenes()
{
    memset(prime, true, sizeof(prime));
  
    prime[1] = false;
  
    for (int p = 2; p * p <= MAX; p++) {
  
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
  
            // Set all multiples of p to non-prime
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to find the first
// n odd prime numbers
void solve(int n)
{
    // To store the current count
    // of prime numbers
    int count = 0;
  
    // Starting with 3 as 2 is
    // an even prime number
    for (int i = 3; count < n; i++) {
  
        // If i is prime
        if (prime[i]) {
  
            // Print i and increment count
            cout << i << " ";
            count++;
        }
    }
}
  
// Driver code
int main()
{
    // Create the sieve
    SieveOfEratosthenes();
  
    int n = 6;
    solve(n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
static int MAX = 1000000;
  
// Create a boolean array "prime[0..n]" and 
// initialize all entries it as true. 
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
static boolean []prime = new boolean[MAX + 1];
static void SieveOfEratosthenes()
{
    for (int i = 0; i <= MAX; i ++)
        prime[i] = true;
  
    prime[1] = false;
  
    for (int p = 2; p * p <= MAX; p++) 
    {
  
        // If prime[p] is not changed, 
        // then it is a prime
        if (prime[p] == true
        {
  
            // Set all multiples of p to non-prime
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to find the first
// n odd prime numbers
static void solve(int n)
{
    // To store the current count
    // of prime numbers
    int count = 0;
  
    // Starting with 3 as 2 is
    // an even prime number
    for (int i = 3; count < n; i++)
    {
  
        // If i is prime
        if (prime[i]) 
        {
  
            // Print i and increment count
            System.out.print(i + " ");
            count++;
        }
    }
}
  
// Driver code
public static void main(String[] args) 
{
    // Create the sieve
    SieveOfEratosthenes();
  
    int n = 6;
    solve(n);
}
  
// This code is contributed by Rajput-Ji

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
from math import sqrt
  
MAX = 1000000
  
# Create a boolean array "prime[0..n]" and 
# initialize all entries it as true. 
# A value in prime[i] will finally be false
# if i is Not a prime, else true. 
prime = [True] * (MAX + 1); 
  
def SieveOfEratosthenes() :
  
    prime[1] = False
  
    for p in range(2, int(sqrt(MAX)) + 1) : 
  
        # If prime[p] is not changed, 
        # then it is a prime 
        if (prime[p] == True) :
  
            # Set all multiples of p to non-prime 
            for i in range(p * 2, MAX + 1, p) :
                prime[i] = False
  
# Function to find the first 
# n odd prime numbers 
def solve(n) : 
  
    # To store the current count 
    # of prime numbers 
    count = 0;
    i = 3;
  
    # Starting with 3 as 2 is 
    # an even prime number
    while count < n :
  
        # If i is prime 
        if (prime[i]) :
  
            # Print i and increment count 
            print(i, end = " "); 
            count += 1
          
        i += 1
  
# Driver code 
if __name__ == "__main__"
  
    # Create the sieve 
    SieveOfEratosthenes(); 
  
    n = 6
    solve(n); 
  
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach 
using System;
      
class GFG 
{
static int MAX = 1000000;
  
// Create a boolean array "prime[0..n]" and 
// initialize all entries it as true. 
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
static bool []prime = new bool[MAX + 1];
static void SieveOfEratosthenes()
{
    for (int i = 0; i <= MAX; i ++)
        prime[i] = true;
  
    prime[1] = false;
  
    for (int p = 2; p * p <= MAX; p++) 
    {
  
        // If prime[p] is not changed, 
        // then it is a prime
        if (prime[p] == true
        {
  
            // Set all multiples of p to non-prime
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to find the first
// n odd prime numbers
static void solve(int n)
{
    // To store the current count
    // of prime numbers
    int count = 0;
  
    // Starting with 3 as 2 is
    // an even prime number
    for (int i = 3; count < n; i++)
    {
  
        // If i is prime
        if (prime[i]) 
        {
  
            // Print i and increment count
            Console.Write(i + " ");
            count++;
        }
    }
}
  
// Driver code
public static void Main(String[] args) 
{
    // Create the sieve
    SieveOfEratosthenes();
  
    int n = 6;
    solve(n);
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

3 5 7 11 13 17


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.