HDU 2138 How many prime numbers（米勒拉宾素数测试算法）

How many prime numbers

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7120 Accepted Submission(s): 2375


[align=left]Problem Description[/align]
Give you a lot of positive integers, just to find out how many prime numbers there are.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each case, print the number of prime numbers you have found out.

[align=left]Sample Input[/align]

3
2 3 4

[align=left]Sample Output[/align]

2

[align=left]Author[/align]
wangye

[align=left]Source[/align]
HDU 2007-11 Programming Contest_WarmUp

[align=left]Recommend[/align]
威士忌

直接素数判断累加：

#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\n",sum);
}
return 0;
}

更优的解法（学长教的）：

#include<cstdio>
#include<cmath>
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\n",s);
}
return 0;
}

网上查了一下，这题要用相应的算法解：miller_rabin算法。学习下。。。
米勒拉宾素数测试，解大规模素数问题。

#include<iostream>
using namespace std ;
__int64 qpow(int a,int b,int r)
{
__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)
{
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)
{
int tab[5]={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 n;
while(~scanf("%d",&n))
{
int ans=0,a;
for(int i=0;i<n;i++)
{
scanf("%d",&a);
if(IsPrime(a))
ans++;
}
printf("%d\n",ans);
}
return 0;
}