poj 1811 Prime Test

 这个题用到Miller_rabin与pallord算法：

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#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#include<cstring>

#include<ctime>

#include<vector>

#define LL long long 

using namespace std;

LL p[10] = { 2,3,5,7,11,13,17,19,23,29 };

const LL Max = ( LL )1<<62;

LL Multi( LL a ,LL b , LL n )//计算a*b 

{

   LL sum = 0;

   while( b )

   {

        if( ( b&1 ) ) sum = ( sum + a )%n;

        a<<=1;

        b>>=1;

        if( a >= n ) a %= n;        

   }

   return sum;    

}

LL Quick_mod( LL a ,LL b ,LL n )//计算当d位奇数时a^b 

{

   LL sum = 1;

   while( b )

   {

       if( ( b&1 ) ) sum = Multi( sum , a ,n );

       a = Multi( a , a ,n );

       b >>= 1;        

   }

   return sum;    

}

LL Miller_rabin( LL n )//米勒罗宾测试 

{

       LL m = n -1,d,L;

    int k=0;

    if( n == 2 ) return true;

    if( n < 2 || !( n&1 )) return false;

    while( !( m&1 ) )//求n-1的2的次方 

    {

           k++ ;  m >>= 1;

    } 

    for( int i = 0 ; i < 9 ; i ++ )

    { 

         if( p[i] >= n ) return  true;

         d = Quick_mod( p[i] , m ,n );

         if( d == 1 ) continue;

         for( int j =  0; j < k ; j ++ )

         {

              L = Multi( d , d ,n );

              if( L == 1 && d != 1 && d != ( n- 1 ) )

                  return false;

              d = L;        

         }

         if( d != 1 ) return false;

    }

    return true;

}

LL Gcd( LL a ,LL b )

{

     return b ==0 ? a : Gcd( b , a%b );    

}

LL Pallord( LL n )

{

   LL x,y,c=0,d,i=1,k=2;

   while(c==0 || c==2 ) c = abs( rand())%(n-1) + 1;

   x = y = (rand( )%( n-1 )) + 1;

   do

   {

      i++;

      d = Gcd( y + n - x, n );

      if( d >1 && d < n )

          return d; 

      if( i == k )

      {

         y = x; k <<= 1;        

      }    

      x = (Multi( x , x, n )+n-c )%n;

    }while( x != y );

    return n; 

}

LL Pallord_min( LL n )

{

    LL p,a,b;

    if( n == 1 ) return Max;

    if( Miller_rabin( n ) ) return n;//如果n为素数就返回 

    p = Pallord( n );//如果不是继续分解 

    a = Pallord_min( p );//继续分解p； 

    b = Pallord_min( n / p );

    return a < b ? a:b; //返回最小的质因子 

}

int main(  )

{

    LL num;

    int Case;

    while( scanf( "%d",&Case )==1 )

    {

       while( Case-- )

       {

        scanf( "%I64d",&num );

        if( Miller_rabin( num ) )

            puts( "Prime" );        

        else

           printf( "%I64d\n",Pallord_min( num )  );

       }

    }

    //system( "pause" );

    return 0;

}