HDOJ 2138 How many prime numbers （数论，素数）

How many prime numbers

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

 

//确实是不解释的米勒拉宾算法啊，伟大的米勒拉宾。。。。我爱你
//耗时，0ms

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
__int64 qpow(int a,int b,int r)//快速幂 
{
    __int64 ans=1,buff=a;
    while(b)
    {
        if(b&1)ans=(ans*buff)%r;
        buff=(buff*buff)%r;
        b>>=1;
    }
    return ans;
}
bool Miller_Rabbin(int n,int a)//米勒拉宾素数测试 
{
    int r=0,s=n-1,j;
    if(!(n%a))
        return false;
    while(!(s&1)){
        s>>=1;
        r++;
    }
    __int64 k=qpow(a,s,n);
    if(k==1)
        return true;
    for(j=0;j<r;j++,k=k*k%n)
        if(k==n-1)
            return true;
    return false;
}
bool IsPrime(int n)//判断是否是素数 
{
    int tab[]={2,3,5,7};
    for(int i=0;i<4;i++)
    {
        if(n==tab[i])
            return true;
        if(!Miller_Rabbin(n,tab[i]))
            return false;
    }
    return true;
}
int main()
{
    int i,N;
    long tmp, count;
    while(scanf("%d",&N)!=EOF)
    {
        count = 0;
        for(i=0;i<N;i++)
        {
            scanf("%ld",&tmp);
            if(IsPrime(tmp))
                count++;
        }
        printf("%d
",count);
    }
}

//耗时
//625ms
#include<iostream>
#include<cmath>
using namespace std;
bool Jude(int n)
{
    int i;
    if(n==2||n==3)
        return true;
    else if(n<2)
        return false;
    else
    {
        for(i=2;i<=sqrt(1.0*n);i++)//这里sqrt(1.0*n)就算了一次,
        //如果判断条件改为i*i<=n,这里的i*i就会做sqrt(n)次,每次循环都要算一次，会超时
            if(n%i==0)
                return false;
        return true;
    }
}
int main()
{
    int t,a;
    int sum;
    while(~scanf("%d",&t))
    {
        sum=0;
        while(t--)
        {
            scanf("%d",&a);
            if(Jude(a)) 
                sum++;
        }
        printf("%d
",sum);
    }
    return 0;
}

//62ms
#include<stdio.h>
#include<math.h>
int pr[8]={4,2,4,2,4,6,2,6};
int prime(int n)
{
    int i=7,j;
    if(n<2)
        return 0;
    if(n==2||n==3||n==5)
        return 1;
    if(!(n%2&&n%3&&n%5))
        return 0;
    for(;i<=sqrt(n);)
    {
        for(j=0;j<8;j++)
        {
            if(n%i==0)
             return 0;
            i+=pr[j];
        }
        if(n%i==0)
         return 0;
    }
    return 1;
}
int main()
{
    int i,n,m,s;
    while(scanf("%d",&n)!=EOF)
    {
        s=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&m);
            if(prime(m))
                s++;
        }
        printf("%d
",s);
    }
    return 0;
}