poj 2960 S-Nim（SG函数）

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 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.

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

LWW
WWL

 1 #include<cstdio>
 2 #include<cstring>
 3 #define FOR(a,b,c) for(int a=(b);a<(c);a++)
 4 using namespace std;
 5 
 6 int n,m,a[101],sg[10001];
 7 
 8 int dfs(int x) {
 9     if(sg[x]!=-1) return sg[x];
10     if(!x) return sg[x]=0;
11     int vis[10001];                    //size of [si]
12     memset(vis,0,sizeof(vis));
13     FOR(i,0,n)
14         if(x>=a[i]) vis[dfs(x-a[i])]=1;
15     for(int i=0;;i++)
16         if(!vis[i]) return sg[x]=i;
17 }
18 
19 int main() {
20     while(scanf("%d",&n)==1 && n) {
21         FOR(i,0,n) scanf("%d",&a[i]);
22         scanf("%d",&m);
23         memset(sg,-1,sizeof(sg));
24         FOR(i,0,m) {
25             int x,v,ans=0;
26             scanf("%d",&x);
27             FOR(j,0,x)
28                 scanf("%d",&v) , ans^=dfs(v);
29             if(ans) printf("W");
30             else printf("L");
31         }
32         putchar(‘\n‘);
33     }
34     return 0;
35 }