hdu 1536 S-Nim

S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3526 Accepted Submission(s): 1550


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 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.


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


Source
Norgesmesterskapet 2004


Recommend
LL

//187MS    840K    900 B    C++
//博弈SG值的运用 
//递归用mex()函数求SG值(不打表，打表比较耗时..) 
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int ki[105],k;
int sg[10005];
int m,l,hi;
int mex(int n)
{
    int flag[105]={0}; //不用开太大，递归后数据会变小 
    for(int i=0;i<k;i++){
        int temp=n-ki[i];
        if(temp<0) break;
        if(sg[temp]==-1) sg[temp]=mex(temp);
        flag[sg[temp]]=1;
    }
    for(int i=0;;i++) 
        if(!flag[i]) return i;
}
int main(void)
{
    while(cin>>k)
    {
        string s;
        if(k==0) break;
        for(int i=0;i<k;i++)
            cin>>ki[i];
        cin>>m;
        memset(sg,-1,sizeof(sg));
        sort(ki,ki+k);
        while(m--){
            cin>>l;
            int nim=0;
            for(int i=0;i<l;i++){
                cin>>hi;
                nim^=mex(hi);
            }
            if(nim==0) s+="L";
            else s+="W";
        }
        cout<<s<<endl;
    }
    return 0;
}