学步园

运用费马小定理和二次探测定理进行素数测试

#include<iostream>
#include<ctime>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
__int64 mod_exp(__int64 a, __int64 b, __int64 n) //计算(a^b) mod n
{
   __int64 d = 1;
   a = a % n;
   while (b) {
       if (b & 1) d = a*d%n;
        a = a*a%n;
        b = b >> 1;
      }
   return d;
}

bool Wintess(__int64 a, __int64 n) //以a为基对n进行Miller测试并实现二次探测
{
   __int64 m, x, y;
   int i, j = 0;
   m = n - 1;
   while (m % 2 == 0) //计算(n-1)=m*(2^j)中的j和m,j=0时m=n-1,不断的除以2直至n为奇数
   {
       m = m >> 1;
       j++;
   }
   x = mod_exp(a, m, n);
   for (i = 1; i <= j; i++) {
       y = mod_exp(x, 2, n);
       if ((y == 1) && (x != 1) && (x != n - 1)) //二次探测
           return true; //返回true时,n是合数
           x = y;
   }
   if (y != 1) return true;
   return false;
}

bool miller_rabin(int times, __int64 n) //对n进行times次的Miller测试
{
   __int64 a;
   int i;
   if (n == 1) return false;
   if (n == 2)  return true;
   if (n % 2 == 0) return false;
   srand(time(NULL));
   for (i = 1; i <= times; i++) {
       
       a = rand() % (n - 2) + 2;
       if (Wintess(a, n)) return false;
   }return true;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        if(miller_rabin(5,n)) cout<<n<<" is a prim"<<endl;
        else  cout<<n<<" is not a prim"<<endl;
    }return 0;
}

方法二：

#include<iostream>
#include<ctime>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
typedef __int64 L;
void power(L a,L p,L n,L& result,bool& composite)
{
    L x;
    if(p==0)
    {
         result=1;
         return;
    }
    power(a,p/2,n,x,composite);
    result=(x*x)%n;
    if(result==1&&(x!=1)&&(x!=n-1)) composite=true;
    if(p&1) result=(result*a)%n;
}
bool Miller_Rabin(L n,L times)
{
      L a,result;
      bool composite=false;
      srand(time(NULL));
      for(L i=1;i<=times;++i)
      {
          a=rand()%(n-3)+2;
          power(a,n-1,n,result,composite);
          if(composite||result!=1) return false;
      }
      return true;

}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        if(Miller_Rabin(n,5)) cout<<n<<" is a prim"<<endl;
        else  cout<<n<<" is not a prim"<<endl;
    }return 0;
}

#include <iostream>
#include <stdlib.h>
using namespace std;
typedef long long L;
void power(L a,L p,L n,L& result ,bool& flag)
{
    L x;
    if(p==0)
    {
        result=1;
        return;
    }
    power(a,p/2,n,x,flag);
    result=(x*x)%n;//二次探测
    if((result==1)&&(x!=1)&&(x!=n-1)) flag=true;
    if(p&1) result=(result*a)%n;
}
int main()
{

        int T;
        cin>>T;
        while(T--){

        int times;
        L n;
        cin>>n>>times;//n是待测的数，times是测试的次数
        if(n<=1||(n%2==0))
        {
            cout<<n<<" is not a prime "<<endl;
            continue;
        }
        if(n==2||n==3)
        {
            cout<<n<<" is  a prime "<<endl;
            continue;
        }
        bool flag=false;
        L result;
        for(int i=0;i!=times;++i)
        {
            L a=rand()%(n-3)+2;//产生2——n-2之间的随机数
            power(a,n-1,n,result ,flag);
        }
        if(flag||result!=1) cout<<n<<" is not a prime"<<endl;
        else    cout<<n<<" is a prime"<<endl;
        }return 0;
}