S-Nim
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 4644   Accepted: 2431

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 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
下面展示ac代码 
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
int bz[105];
int sg[10050];
int mes[10050];

void getsg(int x)
{
	int a;
	memset(sg, 0, sizeof(sg));
	for (int s = 0; s <10010; s++)//相当于打一个表~~在允许的步骤（bz数组存储）下~从0到10010的所有数量代表的sg数先预处理出来
	{
		memset(mes, 0, sizeof(mes));//mes用来找小于s的所有sg函数的值组成的序列里，最小的没出现过整数（从0开始）
		for (a = 0; a<x; a++)
		{
			if (s - bz[a] >= 0)我曾经试着用sort先处理bz数组，这样就不用每次判断，可是忘了sort本身就很慢；
			{
				mes[sg[s - bz[a]]] = 1;
			}
		}
		for (a = 0; a < 10010; a++)
		{
			if (mes[a] == 0)
			{
				sg[s] = a;//找到最小的没出现的
				break;
			}
		}
	}
}
int main()
{
	ios::sync_with_stdio(false);
	int bzz;
	while (cin >> bzz&&bzz)
	{
		for (int s = 0; s < bzz; s++)
		{
			cin >> bz[s];
		}
		getsg(bzz);
		int te;
		cin >> te;
		queue<char>an;//用来储存答案，也有人用string存储。
		while (te--)
		{
			int jg;
			cin >> jg;
			int ans = 0;
			for (int s = 0; s < jg; s++)
			{
				int h;
				cin >> h;
				ans = ans^sg[h];
			}
			if (ans == 0)
			{
				//cout << "L";
				an.push('L');
			}
			else
			{
				an.push('W');
			}
		}
		while (!an.empty())
		{
			cout << an.front();
			an.pop();
		}
		cout << endl;
	}
	return 0;
}