project euler 10

Problem
10

Summation of primes

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

素数的和

所有小于10的素数的和是2 + 3 + 5 + 7 = 17。

求所有小于两百万的素数的和。

package projecteuler;

import java.util.ArrayList;
import java.util.BitSet;
import java.util.List;

import org.junit.Test;

public class Prj10 {

/**
* The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
*
* Find the sum of all the primes below two million.
*/
@Test
public void test() {

PrimeMaster mt = new PrimeMaster().calcultePrimeList(2000000);

System.out.println(mt.toString());
System.out.println(mt.sum);

}

public static class PrimeMaster {

public List<Integer> primeList = new ArrayList<Integer>();

public Long sum = 0L;

public PrimeMaster calcultePrimeList(int upLimit ) {

primeList.clear();

BitSet bs = new BitSet(upLimit);

boolean init = true;

while (true) {
int val = getNonSetVal(init, bs, primeList, upLimit);
init = false;

for (int i = val + 1; i < upLimit; i++) {
if (i % val == 0) {
bs.set(i);
}

}

System.out.println( " val =" + val);

if (val < 0) {
break;
}
}

for (int i = 2; i < upLimit; i++) {

if (!bs.get(i)) {
primeList.add(i);
}
}

return this;
}

private int getNonSetVal(boolean init, BitSet bs,
List<Integer> primeList_, int upLimit) {

if (init) {
primeList_.add(2);
bs.set(2);
return 2;
}

for (int i = 3; i < upLimit; i++) {
if (!bs.get(i)) {
bs.set(i);
primeList_.add(i);
return i;
}
}
return -1;
}

@Override
public String toString() {

Long sum = 0L;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < primeList.size(); i++) {
if( ( i + 1 ) % 20 == 0){
sb.append("\n");
}
sb.append(primeList.get(i) + ",");

sum = sum + primeList.get(i);
}
sb.append("\n");

this.sum = sum;
return sb.toString();
}

}

}