HDOJ 2466 Cryptography Reloaded (RSA算法)

HDOJ 2466 Cryptography Reloaded (RSA算法)

给出N,e,d，3<=e<31,求出p,q;


二分给BigInteger开方很好用。
另外就是枚举(p-1)*(q-1)在比赛的时候也不失为一种很不错的方法。

import  java.math. * ;
import  java.util. * ;
import  java.io. * ;


public   class  Main
{
     public   static  BigInteger Sqrt(BigInteger x)
    {
        BigInteger l = new  BigInteger( " 0 " );
        BigInteger r = x;
         while (l.compareTo(r) < 0 )
        {
            BigInteger mid = l.add(r).divide(BigInteger.valueOf( 2 ));
             if (mid.multiply(mid).compareTo(x) < 0 )
                l = mid.add(BigInteger.ONE);
             else   if (mid.multiply(mid).compareTo(x) == 0 )
                 return  mid;
             else

                r = mid.subtract(BigInteger.ONE);
        }
         return  l;
    }
    
     public   static   boolean  Judge(BigInteger x)
    {
        BigInteger tem = new  BigInteger( " 0 " );
        tem = Sqrt(x);
         if (tem.multiply(tem).compareTo(x) == 0 )
             return   true ;
         else
             return   false ;
    }

     public   static   void  main(String[] args)
    {
        Scanner cin  =   new  Scanner ( new  BufferedInputStream(System.in));
        String sn;
        String se;
        String sd;


         int  ca = 0 ;
         while ( true )
        {
            ca ++ ;
            sn = cin.next();
            se = cin.next();
            sd = cin.next();

            BigInteger n = new  BigInteger(sn);
            BigInteger e = new  BigInteger(se);
            BigInteger d = new  BigInteger(sd);

             if (n.compareTo(BigInteger.ZERO) == 0   &&  e.compareTo(BigInteger.ZERO) == 0   && d.compareTo(BigInteger.ZERO) == 0 )
                 break ;

             for ( int  i = 1 ;i <= 100 ;i ++ )
            {
                BigInteger t = e.multiply(d).subtract(BigInteger.ONE).divide(BigInteger.valueOf(i));
                BigInteger t2 =  t.subtract(n).subtract(BigInteger.ONE).pow( 2 ).subtract(BigInteger.valueOf( 4 ).multiply(n));
                 if (Judge( t2 ) == true )
                {
                    t2 = Sqrt(t2);
                    //  System.out.println(t2);
                    BigInteger p = n.subtract(t).add(BigInteger.ONE).add(t2).divide(BigInteger.valueOf( 2 ));
                    BigInteger q = n.subtract(t).add(BigInteger.ONE).subtract(t2).divide(BigInteger.valueOf( 2 ));
                     if (q.compareTo(p) < 0 )
                    {
                        BigInteger mm = q;
                        q = p;
                        p = mm;
                    }


                    System.out.println( " Case # " + ca + " : " + "   " + p + "   " + q);
                     break ;



                }

            }


        }




    }
}