组合博弈1536-S-Nim

Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:


  The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.

  The players take turns chosing a heap and removing a positive number of beads from it.

  The first player not able to make a move, loses.


Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:


  Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).

  If the xor-sum is 0, too bad, you will lose.

  Otherwise, move such that the xor-sum becomes 0. This is always possible.


It is quite easy to convince oneself that this works. Consider these facts:

  The player that takes the last bead wins.

  After the winning player's last move the xor-sum will be 0.

  The xor-sum will change after every move.


Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.

Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?

your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.

 

Input

Input consists of a number of test cases. For each test case: The first line contains a number k (0 < k ≤ 100 describing the size of S, followed by k numbers si (0 < si ≤ 10000) describing S. The second line contains a number m (0 < m ≤ 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l ≤ 100) describing the number of heaps and l numbers hi (0 ≤ hi ≤ 10000) describing the number of beads in the heaps. The last test case is followed by a 0 on a line of its own.

 

Output

For each position: If the described position is a winning position print a 'W'.If the described position is a losing position print an 'L'. Print a newline after each test case.

 

Sample Input

2 2 5 

3 

2 5 12 

3 2 4 7 

4 2 3 7 12 

5 1 2 3 4 5 

3 

2 5 12 

3 2 4 7 

4 2 3 7 12 

0

Sample Output

LWW 

WWL

//还是SG值套模版函数就好了;

#include

#include

#include

using namespace std;

int k,ki[101];

int sg[10001];

int mex(int n)

{

    int i,temp,judge[101] = {0};

    for(i = 0;i < k;i++)

    {

          temp = n - ki[i];

          if(temp < 0)

                   break;

          if(sg[temp] == -1)

                   sg[temp] = mex(temp);

          judge[sg[temp]] = 1;

    }

    for(i = 0;;i++)

            if(judge[i] == 0)

                 return i;

}

int main()

{

    int i,j,l,m,hi;

    while(cin >> k && k)

    {

            string s;

            for(i = 0;i < k;i++)

                  cin >> ki[i];

            sort(ki,ki+k);

            cin >> m;

           memset(sg,-1,sizeof(sg));

            while(m-- && cin >> l)

            {

                      int s_g = 0;

                      for(i = 0;i < l;i++)

                      {

                            cin >> hi;

                            s_g ^= mex(hi);

                      }

                      if(s_g == 0)

                          s += "L";

                      else

                          s += "W";

            }

            cout << s << endl;

    }

    return 0;

}