【HDU 2138】How many prime numbers

【题目】 

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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

【分析】

题目大意：给你 n 个数，统计这 n 个数中素数的个数，这些数在 int 范围内

其实呢，这道题用暴力（即用  的时间处理每个数）是可以 A 的，但是抱着认真严谨（实际上是找不到板题）的态度，我还是决定用  素数测试写一下这道题

然后就是模板了~

【代码】

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int prime[10]={2,3,5,7,11,13,17,19,23,29};
int Quick_Multiply(int a,int b,int c)
{
    long long ans=0,res=a;
    while(b)
    {
        if(b&1)
          ans=(ans+res)%c;
        res=(res+res)%c;
        b>>=1;
    }
    return (int)ans;
}
int Quick_Power(int a,int b,int c)
{
    int ans=1,res=a;
    while(b)
    {
        if(b&1)
          ans=Quick_Multiply(ans,res,c);
        res=Quick_Multiply(res,res,c);
        b>>=1;
    }
    return ans;
}
bool Miller_Rabin(int x)
{
    int i,j,k;
    int s=0,t=x-1;
    if(x==2)  return true;
    if(x<2||!(x&1))  return false;
    while(!(t&1))
    {
        s++;
        t>>=1;
    }
    for(i=0;i<10&&prime[i]<x;++i)
    {
        int a=prime[i];
        int b=Quick_Power(a,t,x);
        for(j=1;j<=s;++j)
        {
            k=Quick_Multiply(b,b,x);
            if(k==1&&b!=1&&b!=x-1)
              return false;
            b=k;
        }
        if(b!=1)  return false;
    }
    return true;
}
int main()
{
    int n,i,x,ans;
    while(~scanf("%d",&n))
    {
        ans=0;
        for(i=1;i<=n;++i)
        {
            scanf("%d",&x);
            if(Miller_Rabin(x))
              ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}