projecteuler 10001st prime (求出第10001个质数)

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

译文:

观察第一组质数可知,第6位质数为13,那么第10001位质数是多少呢?

============================

第一次code:

 1 import java.util.ArrayList;
 2 import java.util.List;
 3  
 4 public class Main
 5  {  
 6       public static void main (String[] args) 
 7       {  
 8         System.out.println(binarySearch(prime(200000),10000));
 9       } 
10       /**
11        * 对n个数进行求质数
12        * 所求出质数存到ArrayList数组中
13        * @param n
14        * @return
15        */
16       public static List prime(int n)
17       {
18           List<Integer>list = new ArrayList<Integer>();
19           for(int i=2;i<n;i++)
20           {
21               String b="YES";
22                 int [] a = new int[n]; 
23                 for(int j=2;j<i;j++)
24                 {
25                     a[j]=i % j;
26                     if(a[j] == 0)
27                     {
28                         b="NO";break;
29                     }
30                 }
31               if(b.equals("YES"))
32               {
33                   list.add(i);
34               }
35           }
36           return list;
37       }
38       /**
39        * 对ArrayList数组进行折半二查法查找
40        * @param x
41        * @param n
42        * @return
43        */
44     public static int binarySearch(List x, int n) 
45     {
46         int start = 0;
47         int end = x.size() - 1;
48         int mid = -1;
49         while (start <= end) 
50         {
51             mid = (start + end) / 2;
52             if (mid == n) 
53             {
54                 return (int) x.get(mid);
55             } 
56             else if (mid > n) 
57             {
58                 end = mid - 1;
59             } 
60             else if (mid < n) 
61             {
62                 start = mid + 1;
63             }
64         }
65         return -1;
66      }
67  }

结果为:104743

时间效率:求出200000内的所有质数后,再检索出第10001位质数,所花时间为43638毫秒。

原文地址:https://www.cnblogs.com/niithub/p/5805794.html