Total Submissions: 4284

Accepted: 2254
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

题意：有限制的NIM
解题方法:
经典的Nim游戏中sg(x) = x，所以结果就是每一堆的状态直接xor即可，S-Nim游戏先计算每一堆的sg函数值，然后判断方法依旧是用xor。这个和之前的LA5059一样的，把那个SG打表程序套上去就可以A了。

//POJ 2960
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
int n, m, a[110], sg[10005];
bool vis[110];
void pre_init(){
    sg[0] = 0;
    for(int i = 1; i <= 10000; i++){
        memset(vis, false, sizeof(vis));
        for(int j = 0; j < n && a[j] <= i; j++) vis[sg[i-a[j]]] = 1;
        for(int j = 0;; j++) if(!vis[j]) {sg[i] = j; break;}
    }
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        string ans = "";
        if(n == 0) break;
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);
        sort(a , a + n);
        pre_init();
        scanf("%d", &m);
        while(m--){
            int SG = 0;
            int x;
            scanf("%d", &x);
            while(x--){
                int y;
                scanf("%d", &y);
                SG ^= sg[y];
            }
            if(SG == 0) ans += 'L';
            else ans += 'W';
        }
        cout << ans << endl;
    }
}