HDU1944 S-NIM(多个NIM博弈)

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.

InputInput consists of a number of test cases. 
For each test case: The rst 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.OutputFor 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
题解：可以当成多个NIM博弈，最终答案等于每个NIMA博弈结果的异或；(注意求SG函数时，不要每次都把vis数组清空，用一个t标记即可，每次改变标记，否则会超时)
参考代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define clr(a,val) memset(a,val,sizeof a)
 const int maxn=;
 int num,f[maxn],ans;
 int l,t,cas,SG[maxn],vis[maxn];
 void GetSG(int x)
 {
     clr(SG,);
     int t=;
     for(int i=;i<=x;++i)
     {
         for(int j=;f[j]<=i&&j<=num;++j) vis[SG[i-f[j]]]=t;
         for(int j=;j<=x;j++) {if(vis[j]!=t) {SG[i]=j;break;}}
         ++t;
     }
 }

 int main()
 {
     while(~scanf("%d",&num) && num)
     {
         for(int i=;i<=num;++i)scanf("%d",&f[i]);
         sort(f+,f++num);
         GetSG(maxn-);
         scanf("%d",&cas);
         while(cas--)
         {
             scanf("%d",&l);ans=;
             for(int i=;i<=l;++i) scanf("%d",&t),ans^=SG[t];
             if(!ans) printf("L");
             else printf("W");
         }
         puts("");
     }
     return ;
 }