POJ2960S-Nim

        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

本来想着每组数据打一次表时间更短，没想到TLE了，哎，还是每个测试样例打一次表吧，直接按最大打表出来

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
LL x,y;
int n;
int t;
int l;
int a[10010];
int f[110];
int sg[10010];
int vis[10010];
void getSG(int MAX)
{
    memset(sg,0,sizeof(sg));
    for(int i = 1; i < MAX; i++)
    {
        memset(vis,0,sizeof(vis));
        for(int j = 0; f[j] <= i&&j<n; j++)
        {
            vis[sg[i-f[j]]] = 1;
        }
        for(int k = 0; ;k++)
        {
            if(!vis[k])
            {
                sg[i] = k;
                break;
            }
        }
    }
}
void debug()
{
    for(int i = 0; i < 20; i++)
        printf("%d
",sg[i]);
}
int main()
{
    while(scanf("%d",&n)!=EOF&&n)
    {
        for(int i = 0; i < n; i++)
            scanf("%d",&f[i]);
        sort(f,f+n);
        scanf("%d",&t);
        getSG(10010);
        while(t--)
        {
            scanf("%d",&l);
            int maxn = -1;
            for(int i = 0; i < l; i++)
            {
                scanf("%d",&a[i]);
                if(a[i]>maxn)
                    maxn = a[i];
            }
            //debug();
            LL ans = 0;
            for(int i = 0; i < l; i++)
            {
                ans^=sg[a[i]];
            }
            if(ans==0)
                printf("L");
            else
                printf("W");
        }
        printf("
");
    }
    return 0;
}