S-Nim

Problem Description

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 rst 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

 

Sample Output

 1 #include<iostream> 2 #include<math.h> 3 #include<algorithm> 4 #include<string> 5 using namespace std; 6 int a[105]; 7 int sg[10005]; 8 int k; 9 int mex(int x)10 {11     if(sg[x]!=-1) return sg[x];12     bool vis[105];13     memset(vis,0,sizeof(vis));14     for(int i=0;i<k;i++)15     {16         if(x-a[i]>=0)17         {18             mex(x-a[i]);19             vis[sg[x-a[i]]]=true;20         }21     }22     for(int i=0;i<105;i++)23         if(!vis[i])24             return sg[x]=i;25 }26 int main()27 {28     while(cin>>k&&k)29     {30         string str="";31         memset(sg,-1,sizeof(sg));32         sg[0]=0;33         for(int i=0;i<k;i++)34             cin>>a[i];35         sort(a,a+k);36         int m;37         cin>>m;38         for(;m>0;m--)39         {40             int ans=0;  41             int x,u;42             cin>>x;43             for(int i=0;i<x;i++)   44             {  45                 cin>>u;46                 ans^=mex(u);  47             }     48             if(!ans)  str+="L"; 49             else str+="W";  50         }  51         cout<<str<<endl; 52     }53     return 0;54 }