Tian Bogey horse racing Java solution

You must have heard the story of Tian-ji horse racing, right?
If 3 horses become 1000, King still let his horse in the order of favorably to inferior, Tian bogey can choose his horse racing in any order. Win a game, Tian Bogey can get 2002 silver, lose a game, Tian Bogey will lose 2002 silver, draw words do not lose not win. How much can Tian bogey win?

About input:
The input contains multiple sets of test data, and the first line of each set of test data is an integer n (1<=n<=1000), which means that both Tian bogey and King have n horses. The next line is n integers, which indicates the speed of the horse, and the next line is also n integers, indicating the speed of the King's Horse. The last entry ends with a 0 representation.

About output:

For each group of data, the output of an integer, that is, Tian bogey can win more than the amount of silver, if the field bogey win, on the output a negative number, that is the minimum amount of tin bogey to lose.

Example input:

3
the
about
2
the
the
2
about
the
0

Example output:
$
0
0

Package Tianjisaima;import Java.util.scanner;import Java.util.list;import java.util.arraylist;import Java.util.Collections;          public class main{public static void Main (String args[]) {int n, m;        List<integer> list1=new arraylist<integer> ();        List<integer> list2=new arraylist<integer> ();              Scanner in=new Scanner (system.in);          while (true) {n=in.nextint ();            if (n==0) break;              input data for (int i = 0; i < n; ++i) {List1.add (In.nextint ());              } for (int i = 0; i < n; ++i) {List2.add (In.nextint ());              }//Processing data collections.sort (LIST1);                    Collections.sort (LIST2);              int i=0, j=0, x=n-1, y=n-1,count=0;                    Boolean blast=true;    while (bLast) {//is the last horse if (x==i)                  Blast=false;                      if (List1.get (x) > List2.get (y)) {//If Tian bogey the best horse can win King the best horse, then a x--;                      y--;                  count+=200;                      } else if (List1.get (i) > List2.get (j)) {//If Tian Bogey's worst horse can win King's worst horse, then a                      i++;                      j + +;                  Count + = 200;                          } else {///Otherwise, the worst horse and king of the field is the best one if (List1.get (i) < List2.get (y))                      Count-= 200;                      i++;                  y--;           }} System.out.println (count); List1.clear ();
List2.clear (); }               }    }

Tian Bogey horse racing Java solution