Compute Sieve of Eratosthenes using java.util.Bitset

Problem:

The Sieve of Eratosthenes is an array of boolean elements whose ith element is true if and
only if i is a prime number. Use the following algorithm to compute and print a sieve of size
1000:
(Precondition: p is an array of n bits.)
(Postcondition: p[i] is true if and only if i is prime.)
You're required to used a java.util.Bitset object.
1. Initialize p[0] and p[1] to be false, and all other p[i] to be true.
2. Repeat step 3 for each i from 3 to n, incrementing by 2.
3. If there is a prime the square root of i that divides i, set p[i] false.

Output:

Figure it out.

Solution:

public class TestSieve {
 private static final int SIZE=1000;
 private static BitSet isPrime = new BitSet(SIZE);
 public static void main(String[] args) {
 initializeSieve();
 ch02.pr20.TestSieve.printSieve();
 }
 private static void initializeSieve() {
 for (int i = 2; i < SIZE; i++) {
 isPrime.set(i);
 }
 for (int n = 2; 2*n < SIZE; n++) {
 if (isPrime.get(n)) {
 for (int m = n; m*n <SIZE; m++) {
 isPrime.clear(m*n);
 }
 }
 }
 }
 private static void printSieve() {
 int n=0;
 for (int i = 0; i < SIZE; i++) {
 if (isPrime.get(i)) {
 System.out.printf("%5d%s", i, ++n%16==0?"\n":"");
 }
 }
 System.out.printf("%n%d primes less than %d%n", n, SIZE);
 }
}