题目：
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：。。没咋懂
分析： S-nim博弈，主要学习SG打表的方法
1 #include<iostream>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5 int s[105],sg[10001];
6 bool mex[10001];
7 void get_sg(int t,int n)
8 {
9 int i,j;
10 memset(sg,0,sizeof(sg));
11 for(i=1;i<=n;i++)
12 {
13 memset(mex,0,sizeof(mex));
14 for(j=1;j<=t&&s[j]<=i;j++)
15 mex[sg[i-s[j]]]=1;
16 for(j=0;j<=n;j++)
17 if(!mex[j])
18 break;
19 sg[i]=j;
20 }
21 }
22 int main()
23 {
24 int k;
25 while(cin>>k,k)
26 {
27 for(int i=1;i<=k;i++)
28 cin>>s[i];
29 sort(s+1,s+k+1);
30 get_sg(k,10001);
31 int m,n,ans,t;
32 cin>>m;
33 while(m--)
34 {
35 cin>>n;
36 ans=0;
37 for(int i=0;i<n;i++)
38 {
39 cin>>t;
40 ans^=sg[t];
41 }
42 if(ans)
43 cout<<‘W‘;
44 else
45 cout<<‘L‘;
46 }
47 cout<<endl;
48 }
49 return 0;
50 }