hdu2.1.2 How many prime numbers(计算素数的个数)

给出一系列的数，统计素数的个数

思路：因为数据的范围为2^31-1，先计算出[2,sqrt(2^31-1]之间的素数primes。在判断给出的数num是否是素数时，只需要判断num是否能被primes集合中的数据整除

代码如下：

import java.io.FileInputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.StreamTokenizer;
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;public class Main implements Runnable{private Scanner cin;private PrintWriter cout;private static final boolean DEBUG = false;private static final int N = 46340;private List<Integer> primes = new ArrayList<>();private List<Integer> nums = new ArrayList<>();private boolean[] vis = new boolean[N];private void preprocess(){for (int i = 2; i < N; i++){if (!vis[i]){primes.add(i);}int size = primes.size();for (int j = 0; j < size && primes.get(j) * i < N; j++){vis[primes.get(j) * i] = true;if (i % primes.get(j) == 0) break;}}}private void init(){try{if (DEBUG){cin = new Scanner(new BufferedReader(new InputStreamReader(new FileInputStream("e:\\program\\idea\\spoj\\spoj.txt"))));}else{cin = new Scanner(new BufferedReader(new InputStreamReader(System.in)));}preprocess();cout = new PrintWriter(System.out);}catch (IOException e){e.printStackTrace();;}}private boolean input(){if (!cin.hasNextInt()) return false;int n = cin.nextInt();nums.clear();for (int i = 0; i < n; i++){int num = cin.nextInt();nums.add(num);}return true;}private void solve(){int ans = 0;for (int i = 0; i < nums.size(); i++){boolean isPrime = true;for (int j = 0; j < primes.size() && nums.get(i) > primes.get(j); j++){if (nums.get(i) % primes.get(j) == 0) {isPrime = false;break;}}if (isPrime) ans++;}cout.println(ans);cout.flush();}public void run() {init();while (input()){solve();}}public static void main(String[] args){new Thread(new Main()).start();}
}

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hdu2.1.2 How many prime numbers(计算素数的个数)

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