Here is the code that is using command line parameter passing:
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
/**
*
* @author pacior
*/
public class MyPrimeNumber {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
boolean[] pm = new boolean[Integer.parseInt(args[0])];
pm[0] = true;//You rather don't need to do this, but thats the way
in article
for (int i = 1; i < pm.length; i++) {
if ( primeNumberTest(i+1) ){
crossOut(pm,i+1);
}
}
for (int i = 0; i < pm.length-1; i++) {
if ( pm[i] == false )
System.out.println("Prime number: " + (int)(i+1));
}
}
/**
* The solution depends on a statement, that if you have factors of
number, then
* the given factor f, is a sum of some integer numbers x + times. So
You can repeat adding a as long as it reach tested number
* it is done in while loop using for, for each number ( created from
count )
* @param primeNumber
* @return
* true if number is prime, false otherwise
*/
public static boolean primeNumberTest(int primeNumber){
if ( primeNumber == 1 )//1 is not a prime number
return false;
if ( primeNumber == 2 )//2 is the smallest prime number
return true;
int i = 2;//counter
int j = 1;
while ( i < Math.ceil(primeNumber/2)+1 ){//we test sum of
variable i up to primeNumber/2
j = i;
do {
j += i;
}while ( j < primeNumber );
if ( j == primeNumber ){
return false;//adding sums resulted in fact, that
primeNumber is that sum - that is not a prime number!
}
i++;
}
return true;
}
public static void crossOut(boolean[] a,int num){
int num2 = num+num;
while ( num2 < a.length ){
a[num2-1]= true;//a[3]=4
num2 += num;
}
}
}
It use static methods and it is based on this article:
http://mathforum.org/dr.math/faq/faq.prime.num.html
Pacior
On Jul 22, 3:35 pm, nguyen thanh lam <[email protected]> wrote:
> Help me plz ! How to find Prime number ?
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