hdu1356&hdu1944 博弈论的SG值（王道）

S-Nim
Problem 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 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.

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

用dfs搜索sg值

 #include <iostream>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 int a[];
 int t;
 int b[];
 int dfs(int x)
 {
     if(b[x]!=-)return b[x];
     int i;
     if(x-a[]<)return ;
     int c[];
     memset(c,,sizeof(c));
     for(i=;i<t;i++)
     {
         if(x-a[i]<)
         {
             break;
         }
         c[dfs(x-a[i])]=;
     }
     for(i=;i<;i++)
     if(c[i]==)
     {
         b[x]=i;
         break;
     }
     return b[x];
 }
 int main()
 {
     while(cin>>t&&t){
     memset(a,,sizeof(a));
     memset(b,-,sizeof(b));
     int j;
     for(j=;j<t;j++)
     cin>>a[j];
     sort(a,a+t);
     int i;
     b[]=;
     int n;
     cin>>n;
     for(i=;i<n;i++)
     {
         int m;
         int sum=,x;
         cin>>m;
         for(j=;j<m;j++)
         {
             cin>>x;
             sum^=dfs(x);
         }
         if(sum)
         cout<<"W";
         else cout<<"L";
     }
     cout<<endl;
     }
 }