project euler 10

Problem 10

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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.

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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. */@Testpublic 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;}@Overridepublic 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();}}}