hdu 2138 How many prime numbers

How many prime numbers

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7017 Accepted Submission(s): 2336

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

_____________________________
ACM STEPS里的...这题前面一道是求LCM....结果接下来就是这么一道。。。
朴素会超....筛法会爆....题目顺序真是按照难度来的？
于是想到 Miller-Rabin素数测试.......
这个方法是基于费马小定理
我的理解就是...
如果我要判断n是否为素数
只要取k个数如果满足 a^(n-1)mod n =1 那么n就很可能为素数。
证明什么的...暂时还是算了吧...论文里貌似扯了一大堆
第一次用，竟然真的A了。。。。
感觉更好的办法也许是先打一个比较小的素数表，然后每次random选取若干个进行判断...那样应该更可靠些？
本来想WA掉之后再改的。。。没想到这么写就A掉了。。。。杭电数据略水？

 1 #include <iostream>
 2 #include <cmath>
 3 #include <stdio.h>
 4 #include <cstring>
 5  
 6 using namespace std;
 7  
 8 typedef long long LL;
 9 LL power(LL m,LL n,LL k)
10 {
11     int b = 1;
12     while (n > 0)
13     {
14           if (n & 1)
15              b = (b*m)%k;
16           n = n >> 1 ;
17           m = (m*m)%k;
18     }
19     return b;
20 }
21 bool judge(LL n)
22 {
23     LL i;
24     if (n<=3) return true;
25     for (i=2;i<=ceil(sqrt(n))+1;i++)
26         if (n %i==0) return false;
27         return true;
28 }
29  
30 int main()
31 {
32     LL i,n,x;
33  
34     while (scanf("%I64d",&n)!=EOF)
35     {   LL ans=0;
36         for (i=1;i<=n;i++)
37         {
38             scanf("%I64d",&x);
39           if ((power(61,x-1,x)==1)&&(power(11,x-1,x)==1)&&(power(31,x-1,x)==1)
40                 &&(power(97,x-1,x)==1))
41                  ans++;
42         }
43           printf("%I64d
",ans);
44     }
45     return 0;
46 }
47  
48  
49  
50