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[101];
int t;
int b[10001];
int dfs(int x)
{
    if(b[x]!=-1)return b[x];
    int i;
    if(x-a[0]<0)return 0;
    int c[101];
    memset(c,0,sizeof(c));
    for(i=0;i<t;i++)
    {
        if(x-a[i]<0)
        {
            break;
        }
        c[dfs(x-a[i])]=1;
    }
    for(i=0;i<101;i++)
    if(c[i]==0)
    {
        b[x]=i;
        break;
    }
    return b[x];
}
int main()
{
    while(cin>>t&&t){
    memset(a,0,sizeof(a));
    memset(b,-1,sizeof(b));
    int j;
    for(j=0;j<t;j++)
    cin>>a[j];
    sort(a,a+t);
    int i;
    b[0]=0;
    int n;
    cin>>n;
    for(i=0;i<n;i++)
    {
        int m;
        int sum=0,x;
        cin>>m;
        for(j=0;j<m;j++)
        {
            cin>>x;
            sum^=dfs(x);
        }
        if(sum)
        cout<<"W";
        else cout<<"L";
    }
    cout<<endl;
    }
}