[2016-05-10][51nod][1106 质数检测]

• 时间:2016-05-10 12:48:09 星期二

• 题目编号:[2016-05-10][51nod][1106 质数检测]

• 题目大意:给出N个正整数，检测每个数是否为质数。如果是，输出”Yes”，否则输出”No”。

• 分析:

  1. 直接上模板
  2. 打sqrt(1E9)内的质数表,直接判断能否被所有质数整除即可

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
typedef long long ll;
const int maxp = 31622 + 10;//sqrt(1E9)
int pri[maxp];
void inipri(){
        memset(pri,0,sizeof(pri));
        for(int i = 2 ; i < maxp ; ++i){
                if(!pri[i])     pri[++pri[0]] = i;
                for(int j = 1 ; j <= pri[0] && pri[j]*i < maxp;++j){
                        pri[pri[j] * i] = 1;
                        if(i % pri[j] == 0)     break;
                }
        }
}
int main(){
        inipri();
        int n,a;
        scanf("%d",&n);
        for(int i = 0 ; i < n ; ++i){
                scanf("%d",&a);
                if((a & 1) == 0){
                        puts(a == 2 ? "Yes" : "No");
                }else {
                        int flg = 1;
                        for(int j = 1; j < pri[0] && pri[j] * pri[j] < a;++j){
                                if(a % pri[j] == 0){
                                   flg = 0;
                                   break;
                                }
                        }
                        puts(flg ? "Yes" : "No");
                }
        }
        return 0;
}

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
typedef long long ll;
ll muilt_mod(ll a, ll b ,ll c){
        a %= c;
        b %= c;
        ll ret = 0;
        ll tmp = a;
        while(b){
                if(b & 1){
                        ret += tmp;
                        if(ret > c)     ret -= c;
                }
                tmp <<= 1;
                if(tmp > c)     tmp -= c;
                b >>= 1;
        }
        return ret;
}
ll pow_mod(ll a, ll n, ll mod){
        ll ret = 1;
        ll tmp = a % mod;
        while(n){
                if(n & 1)       ret = muilt_mod(ret,tmp,mod);
                tmp = muilt_mod(tmp,tmp,mod);
                n >>= 1;
        }
        return ret;
}
bool check(ll a, ll n , ll x,ll t){
        ll ret = pow_mod(a,x,n);
        ll last = ret;
        for(int i = 1; i <= t; ++i){
                ret = muilt_mod(ret,ret,n);
                if(ret == 1 && last != 1 && last != n - 1)      return true;
                last = ret;
        }
        if(ret != 1)    return true;
        else return false;
}
bool Miller_Rabin(ll n){
        if(n < 2)       return false;
        if(n == 2)      return true;
        if((n & 1) == 0 )       return false;
        ll x = n - 1,t = 0;
        while((x & 1) == 0){
                x >>= 1;
                ++t;
        }
        srand(NULL);
        int S = 8;
        for(int i = 0 ; i < S;++i){
                ll a = rand() % (n - 1) + 1;
                if(check(a,n,x,t))      return false;
        }
        return true;
}
int main(){
        int n,a;
        scanf("%d",&n);
        for(int i = 0 ; i < n ; ++i){
                scanf("%d",&a);
                puts(Miller_Rabin(a)?"Yes":"No");
        }
        return 0;
}

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