HDU2138 & 米勒拉宾模板

题意:

　　给出n个数,判断它是不是素数.

SOL:

　　米勒拉宾裸题,思想方法略懂,并不能完全理解,所以实现只能靠背模板....

　　好在不是很长...

Code:

　　

/*==========================================================================
# Last modified: 2016-03-21 10:09
# Filename: miller-rabin.cpp
# Description:
==========================================================================*/
#define me AcrossTheSky
#include <cstdio>
#include <cmath>
#include <ctime>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm> 

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector> 

#define lowbit(x) (x)&(-x)
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)
#define ls(a,b) (((a)+(b)) << 1)
#define rs(a,b) (((a)+(b)) >> 1)
#define getlc(a) ch[(a)][0]
#define getrc(a) ch[(a)][1] 

#define maxn 100000
#define maxm 100000
#define pi 3.1415926535898
#define _e 2.718281828459
#define INF 1070000000
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 

template<class T> inline
void read(T& num) {
    bool start=false,neg=false;
    char c;
    num=0;
    while((c=getchar())!=EOF) {
        if(c=='-') start=neg=true;
        else if(c>='0' && c<='9') {
            start=true;
            num=num*10+c-'0';
        } else if(start) break;
    }
    if(neg) num=-num;
}
/*==================split line==================*/
int c[7]={2,3,5,7,11,13};
ll pow(ll a,ll k,ll mod){
	ll s=1;
	for (;k;a=(a*a)%mod,k>>=1)
		if (k&1) s*=a,s%=mod;
	return s;
}
bool check(int x,int a){
	if (!x%a) return false;
	int r=0,s=x-1;
	while (!(s&1)) s>>=1,r++;
	ll k=pow(a,s,x);
	if (k==1) return true;
	for (int j=0;j<r;j++,k=k*k%x)
		if (k==x-1) return true;
	return false;
}
bool miller_rabin(ll x){
	FORP(i,0,5) if (x==c[i]) return true;
	FORP(i,0,5){
			if (!check(x,c[i])) return false;
	}
	return true;
}
int main(){
	int n;
	while (scanf("%d",&n)!=EOF){
		 int ans=0;
		 FORP(i,1,n) {
		 	ll x; read(x);
		 	if (miller_rabin(x)) ans++;
		 }
		 printf("%d\n",ans);
	}
}