poj 1811 Prime Test

 这个题用到Miller_rabin与pallord算法：

#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;
}