HDOJ(HDU) 2138 How many prime numbers(素数-快速筛选没用上、)
【摘要】 Problem Description Give you a lot of positive integers, just to find out how many prime numbers there are.
Input There are a lot of cases. In each case, there is an integer N represe...
Problem Description
Give you a lot of positive integers, just to find out how many prime numbers there are.
Input
There are a lot of cases. In each case, there is an integer N representing the number of integers to find. Each integer won’t exceed 32-bit signed integer, and each of them won’t be less than 2.
Output
For each case, print the number of prime numbers you have found out.
Sample Input
3
2 3 4
Sample Output
2
这个题目就是让你求一组的素数有多少个。
这个素数范围的数字有点大,所以不能用打表。
测试数据很水。。。直接判断就能过了。
不过判断的时候,有一个地方需要注意的,我在那个判断素数的方法注释了。
import java.util.Arrays;
import java.util.Scanner;
public class Main { public static void main(String[] args) { //boolean db[] = new boolean[2147483647]; //数组太大,不能打表! //dabiao(db); Scanner sc = new Scanner(System.in); while(sc.hasNext()){ int n = sc.nextInt(); long sum = 0; int m; for(int i=0;i<n;i++){ m=sc.nextInt(); if(prime(m)){ sum++; } } System.out.println(sum); } } //直接判断能过,说明数据比较水。 private static boolean prime(int m) { for(int i=2;i<=Math.sqrt(m);i++){ //***** 注意:i*i<=m 是会超时的,因为i*i每次都要计算 if(m%i==0){ return false; } } return true; } //素数筛选打表应该会超时 private static void dabiao(boolean[] db) { Arrays.fill(db, true); for(int i=2;i<=Math.sqrt(db.length);i++){ for(int j=i+i;j<db.length;j+=i){ if(db[j]){ db[j]=!db[j]; } } } }
}
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文章来源: chenhx.blog.csdn.net,作者:谙忆,版权归原作者所有,如需转载,请联系作者。
原文链接:chenhx.blog.csdn.net/article/details/51319430
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