S-Nim

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3694		Accepted: 1936

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

 #include<cstdio>

 #include<cstring>

 #define FOR(a,b,c) for(int a=(b);a<(c);a++)

 using namespace std;

 int n,m,a[],sg[];

 int dfs(int x) {

     if(sg[x]!=-) return sg[x];

     if(!x) return sg[x]=;

     int vis[];                    //size of [si]

     memset(vis,,sizeof(vis));

     FOR(i,,n)

         if(x>=a[i]) vis[dfs(x-a[i])]=;

     for(int i=;;i++)

         if(!vis[i]) return sg[x]=i;

 }

 int main() {

     while(scanf("%d",&n)== && n) {

         FOR(i,,n) scanf("%d",&a[i]);

         scanf("%d",&m);

         memset(sg,-,sizeof(sg));

         FOR(i,,m) {

             int x,v,ans=;

             scanf("%d",&x);

             FOR(j,,x)

                 scanf("%d",&v) , ans^=dfs(v);

             if(ans) printf("W");

             else printf("L");

         }

         putchar('\n');

     }

     return ;

 }