/* * Copyright (c) 2000 David Flanagan.  All rights reserved. * This code is from the book Java Examples in a Nutshell, 2nd Edition. * It is provided AS-IS, WITHOUT ANY WARRANTY either expressed or implied. * You may study, use, and modify it for any non-commercial purpose. * You may distribute it non-commercially as long as you retain this notice. * For a commercial use license, or to purchase the book (recommended), * visit http://www.davidflanagan.com/javaexamples2. */package com.davidflanagan.examples.basics;/** * This program computes prime numbers using the Sieve of Eratosthenes * algorithm: rule out multiples of all lower prime numbers, and anything * remaining is a prime.  It prints out the largest prime number less than * or equal to the supplied command-line argument. **/public class Sieve {    public static void main(String[] args) {        // We will compute all primes less than the value specified on the        // command line, or, if no argument, all primes less than 100.        int max = 100;                           // Assign a default value        try { max = Integer.parseInt(args[0]); } // Parse user-supplied arg        catch (Exception e) {}                   // Silently ignore exceptions.        // Create an array that specifies whether each number is prime or not.        boolean[] isprime = new boolean[max+1];        // Assume that all numbers are primes, until proven otherwise.        for(int i = 0; i <= max; i++) isprime[i] = true;        // However, we know that 0 and 1 are not primes.  Make a note of it.        isprime[0] = isprime[1] = false;        // To compute all primes less than max, we need to rule out        // multiples of all integers less than the square root of max.        int n = (int) Math.ceil(Math.sqrt(max));  // See java.lang.Math class        // Now, for each integer i from 0 to n:        //   If i is a prime, then none of its multiples are primes,        //   so indicate this in the array.  If i is not a prime, then        //   its multiples have already been ruled out by one of the        //   prime factors of i, so we can skip this case.        for(int i = 0; i <= n; i++) {            if (isprime[i])                          // If i is a prime,                 for(int j = 2*i; j <= max; j = j + i) // loop through multiples                    isprime[j] = false;               // they are not prime.        }        // Now go look for the largest prime:        int largest;        for(largest = max; !isprime[largest]; largest--) ;  // empty loop body                // Output the result        System.out.println("The largest prime less than or equal to " + max +			   " is " + largest);    }}