贴一个程序Java examples in a nutshell里面的:用的是Eratosthenes的Sieve算法: /* * 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); } }
/*
* 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);
}
}