Prime number test (Ehrlich sieve method and conventional method java)

Definition of prime numbers: Prime numbers refer to natural numbers greater than 1 that have no other factors except 1 and itself.
Ehrlich sieve method: In simple terms, it is to traverse the entire array, set each number 2 times, 3 times ... to -1 (uncrossable), and then traverse the new array to quickly find the nth prime number.
And compare its method with the conventional method

code show as below:

import java.util.Scanner;
public class Sscs {
//素数测试（找到第n个素数）
 //方法一：对一组数据中的每个数字进行鉴别是否为素数，找到对应的值
 //方法二：埃氏筛法
 static boolean sushu(int a)
 {
  
  for(int i=2;i<Math.sqrt(a);i++)//直接查询2-sqrt(a)加快速度
  {
   if(a%i==0)
      return false;
  }
  return true;
 }
 static void fa1(int n,int m)//方法一的实现
 {
  int count=0;
  for(int i=2;i<=m;i++)
  {
   if(sushu(i))
   {
    count++;
    if(count==n)
     System.out.println(i);
   }
  }
 }
 static void fa2(int n,int m)//方法二实现:埃氏筛法
 {
  int []a=new int[m];
  for(int i=2;i<a.length;i++)//遍历每一位数据，将其倍数全部置为-1
  {
   if(a[i]==-1)
    continue;
   int k=2;
   while(i*k<m)
   {
    a[i*k]=-1;
     k++;
   }
  }
  int count=0;
  for(int i=2;i<a.length;i++)//遍历查找第n个素数
  {
     if(a[i]==0)
      count++;
     if(count==n)
     { System.out.println(i);
        break;
     }  
  }
 }
 public static void main(String[] args) {
 // TODO Auto-generated method stub
  Scanner a=new Scanner(System.in);
  int n=a.nextInt();//输入需要查询的第n个数
  int b=2;
  while(b/Math.log(b)<n)//查询第n个素数时需要准备多少的数组长度
  {
   b++;
  }
  long now1=System.currentTimeMillis();//记录方法一的初始时间
  fa1(n,b);
  System.out.println(System.currentTimeMillis()-now1+"ms");//输出方法一所需要的时间
  long now2=System.currentTimeMillis();//记录方法一的初始时间 
  fa2(n,b);
  System.out.println(System.currentTimeMillis()-now2+"ms");//输出方法二所需要的时间
 }
}