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

题意： 多组测试数据 ，输入 k个集合S的元素，m种情况，m种（L堆，每堆hi个）。若存在移动某堆能到达一个必败点，则该点为必胜点，输出W。必败点指无论怎么移动都只能到达必胜点，输出L。
解题方法： SG函数，每堆看做一个子游戏，SG函数通过递归得到每种堆数的sg函数值。
关于SG函数可以看写得非常漂亮，ORZ

//HDU 1536 SG函数
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10010;
int t, s[110], sg[maxn];
bool vis[maxn];
void pre_sg(int *s, int t)//N为求解范围, s[i]为每次可取值,t是s的长度
{
    memset(sg, 0, sizeof(sg));
    int i, j;
    for(i = 1; i <= maxn; i++){
        memset(vis, 0, sizeof(vis));
        for(j = 0; j < t; j++){
            if(i - s[j] >= 0) vis[sg[i - s[j]]] = 1;
        }
        for(j = 0; j < maxn; j++){
            if(!vis[j]){
                break;
            }
        }
        sg[i] = j;
    }
}
int main()
{
    int n, m, x;
    while(scanf("%d", &t) != EOF)
    {
        if(!t) break;
        string ans = "";
        for(int i = 0; i < t ;i++) scanf("%d", &s[i]);
        pre_sg(s, t);
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            scanf("%d", &m);
            int SG = 0;
            for(int j = 0; j < m; j++){
                scanf("%d", &x);
                SG ^= sg[x];
            }
            if(SG == 0){
                ans += 'L';
            }
            else{
                ans += 'W';
            }
        }
        cout << ans << endl;
    }
    return 0;
}