hdu 2138 How many prime numbers(miller_rabin?)

最新推荐文章于 2020-02-21 21:04:13 发布

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本文介绍了一种通过预处理素数并使用这些素数来判断大范围内的整数是否为素数的方法。该方法首先生成一系列不超过32位的素数，并利用这些素数检查给定整数集合中哪些数是素数。文章提供了完整的C++实现代码，展示了如何有效地进行素数计数。

How many prime numbers

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3302 Accepted Submission(s): 1074

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

Author

wangye

Source

HDU 2007-11 Programming Contest_WarmUp

Recommend

威士忌

据说得用miller_rabin算法，但我用用了筛选，900多MS，猥琐的过了

一个数只要不能被小于其方根的质素整除即为质素

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 65537
using namespace std;
int f[N];
bool s[N];
int next=0;
void oprime()
{
  memset(s,0,sizeof(s));
  s[0]=s[1]=1;
  int i,j;
  for(i=2;i<N;i++)
  {
    if(!s[i])
    {
      f[++next]=i;//记录素数
      for(j=i+i;j<N;j+=i)
      s[i]=1;
    }
  }
}
int main()
{
  int m;
  int ans;
  oprime();
  while(scanf("%d",&m)!=EOF)
  {
    ans=0;
    for(int i=0;i<m;i++)
    {
      __int64 shu;
      scanf("%I64d",&shu);
      bool flag=0;
      if(shu==1)
      continue;
      for(int j=1;j<next&&int(sqrt(shu)+1)>f[j];j++)
      {
        if(shu%f[j]==0)
        {
          flag=1;
          break;
        }
      }
      if(!flag)
       ans++;
    }
    printf("%d\n",ans);
  }
}