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 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.

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.

LWW

WWL

#include <iostream>

#include <cstring>

#include <algorithm>

using namespace std;

#define N 102

int n,m,l,s[N],sg[],ans,a;

bool vis[];

void c()

{

    sg[]=;

    for(int i=;i<=;i++)

    {

        memset(vis,,sizeof(vis));

        for(int j=;j<n&&s[j]<=i;j++)

        vis[sg[i-s[j]]]=;

        for(int x=;x<=;x++)

            if(!vis[x])

        {

            sg[i]=x;

            break;

        }

    }

}

int main()

{

    while(cin>>n&&n)

    {

        for(int i=;i<n;i++)

            cin>>s[i];

        sort(s,s+n);

        c();

        cin>>m;

        while(m--)

        {

            ans=;

            cin>>l;

            while(l--)

            {

                cin>>a;

                ans^=sg[a];

            }

            if(ans)

                cout<<"W";

            else

                cout<<"L";

        }

        cout<<endl;

    }

    return ;

}

#include <iostream>

#include <cstring>

#include <algorithm>

using namespace std;

#define N 102

int n,m,l,s[N],sg[],ans,a;

bool vis[];

void c()

{

    sg[]=;

    for(int i=;i<=;i++)

    {

        memset(vis,,sizeof(vis));

        for(int j=;j<n&&s[j]<=i;j++)

        vis[sg[i-s[j]]]=; //里面的sg[]是sg[i]的后继 vis[sg[]] 指的是求mex

        for(int x=;x<=;x++)

            if(!vis[x])

        {

            sg[i]=x;

            break;

        }

    }

}

int main()

{

    while(cin>>n&&n)

    {

        for(int i=;i<n;i++)

            cin>>s[i];

        sort(s,s+n);

        c();

        for(int i=;i<;i++)

        cout<<"i: "<<i<<"  sg[i]: "<<sg[i]<<endl;

        /*cin>>m;

        while(m--)

        {

            ans=0;

            cin>>l;

            while(l--)

            {

                cin>>a;

                ans^=sg[a];

            }

            if(ans)

                cout<<"W";

            else

                cout<<"L";

        }

        cout<<endl;*/

    }

    return ;

}