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 theyrecently 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 = f2, 5g 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.

Arthur and his sister Caroll 玩nim游戏玩腻了(因为他们都知道了如何计算必胜策略),所以Arthur把这个游戏改了一下规则,Arthur定义了一个有限集合S,每次从一堆石子中取的石子数目必须在S中。现在,Arthur想知道先手是否有必胜策略。有多组测试数据。

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.

Sample Input
2 2 5
2 5 12
3 2 4 7
4 2 3 7 12
5 1 2 3 4 5
2 5 12
3 2 4 7
4 2 3 7 12

Sample Output


/*program from Wolfycz*/
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
    int x=0,f=1;char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar())  if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar())    x=(x<<1)+(x<<3)+ch-'0';
    return x*f;
inline void print(int x){
    if (x>=10)     print(x/10);
const int N=1e2,M=1e4;
int S[N+10],sg[M+10];
bool vis[N+10];
int size;
void prepare(){
    for (int i=1;i<=M;i++){
        for (int j=1;j<=size;j++){
            if (i<S[j]) break;
        for (int j=0;j<=N;j++)  if (!vis[j]){sg[i]=j;break;}
int main(){
    while (true){
        if (!size)  break;
        for (int i=1;i<=size;i++)   S[i]=read();
        int n=read();
        for (int i=1;i<=n;i++){
            int m=read(),ans=0;
            for (int j=1;j<=m;j++)  ans^=sg[read()];
    return 0;
posted @ 2018-02-08 22:52 Wolfycz 阅读(...) 评论(...) 编辑 收藏