11. Sieve of Eratosthenes. Using the Sieve principle, write a program that can produce all the prime numbers greater than zero and less than or equal to some largest number Max.

import java.util.*;
public class Eratosthenes
{
  int max;
  static int primes[];
 
  public static void main(String args[])
  {
    Eratosthenes erat = new Eratosthenes();
    erat.find_primes();
    for(int i=0;i<primes.length;i++)
    {
      System.out.print(primes[i]+",");
    }
  }
  public Eratosthenes()
  {
    System.out.print("Enter End Range of list:");
    Scanner scan = new Scanner(System.in);
    max=scan.nextInt();
  }
 
  public void find_primes()
  {
    int i,j,k, divisor, offset;
   
    double sqrt = Math.sqrt(max)+1;
    int tmp[];
   
    if(max  > 100)
    {
      primes = new int[max/2];
    }
    else
    {
      primes = new int[max];
    }
    primes[0] = 2;
    primes[1] = 3;
   
    for(i=2,j=5; j < max ; j+=2)
    {
      if(j % 3 != 0)
      {
        primes[i++] = j;
      }
    }
   
    for(i=2, divisor=5, offset = primes.length; divisor < sqrt ; )
    {
      j = i*i;
      tmp = new int[offset];
      offset = j;
     
      /*
       * Copy the numbers that have already been sieved to a new array.
       */
      System.arraycopy(primes,0,tmp,0,j);
      while( j < tmp.length)
      {
        k = primes[j++];
        if(k==0)
        {
          /*
           * The array may contain some zeros at the end. It's too much
           * trouble to calculate the exact size for the array. Easier
           * to pad with zeros
           */
          break;
        }
        if(k % divisor != 0)
        {
          tmp[offset++] = k;
        }
      }
      primes = null;
      primes = tmp;
      tmp = null;
      divisor = primes[i++];
    }
  }
}