HDU 2138 How many prime numbers

How many prime numbers

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 2138
64-bit integer IO format: %I64d      Java class name: Main

 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

Source

HDU 2007-11 Programming Contest_WarmUp

 

解题：miller - rabbin 素性测试

其实我们判的是伪素数，伪素数就是这个数很有可能是素数，但是不排除不是素数的可能，一旦miller rabin判断某个数不是素数，那么这个数一定不是素数，如果判断这个数是素数，则这个数是伪素数。由费马定理

$a^{p-1} equiv 1(mod quad p)$

 那么我们可以把p-1表示成

$p - 1 = d*2^{r}$

那么很明显

　$a^{p-1} = a^{d*2^{r}}$

如果

$a^{d} equiv 1 (mod quad p)$ 

那么

$a^{d}$ $a^{2d} dots$ $a^{d*2^{r}}$

都能模p余1

所以p是伪素数，如果

$a^{d}   otequiv 1(modquad p)$

那么我们只需要看看

$a^{d*2^{i}} modquad p == p-1qquad i < r$

么，存在，那么

$a^{p-1} equiv 1(modquad p)$

成立，否则，这个数一定是合数

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL quickPow(LL a,LL d,LL n){
    LL ret = 1;
    while(d){
        if(d&1) ret = (ret*a)%n;
        d >>= 1;
        a = (a*a)%n;
    }
    return ret;
}
bool check(int a,int d,int n){
    if(n == a) return true;
    while(~d&1) d >>= 1;
    LL t = quickPow(a,d,n);
    while(d < n-1 && t != 1 && t != n-1){
        t = t*t%n;
        d <<= 1;
    }
    return (d&1) || t == n-1;
}
bool isP(int n){
    if(n == 2) return true;
    if(n < 2 || 0 == (n&1)) return false;
    static int p[3] = {2,3,61};
    for(int i = 0; i < 3; ++i)
        if(!check(p[i],n-1,n)) return false;
    return true;
}
int main(){
    int n,cnt,tmp;
    while(~scanf("%d",&n)){
        cnt = 0;
        while(n--){
            scanf("%d",&tmp);
            cnt += isP(tmp);
        }
        printf("%d
",cnt);
    }
    return 0;
}