HDU 1536 - S-Nim（SG）

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4585    Accepted Submission(s): 1989

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

======================

题意：每次只能取固定数量的nim。若取子之后所有堆石子数异或值为0，则获胜。

多组输入。

第一行 输入可取石子的个数 以及 可取的数量；

第二行 询问次数。接下来的每行询问 输入堆数以及每堆的石子数，对于每次询问 胜W负L。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
int sg[11111];
int s[111],h[111];
int k,m,l;

int dfs(int x){
    if(sg[x]!=-1) return sg[x];
    bool vis[11111];//局部变量，每次递归都有不同的vis数组
    memset(vis,0,sizeof(vis));
    for(int i=0;i<k;i++){
        if(x>=s[i]){
            int tmp=x-s[i];
            vis[dfs(tmp)]=1;
        }
    }
    for(int i=0;;i++){
        if(!vis[i]) return sg[x]=i;
    }
}

int main()
{
    while(~scanf("%d",&k)){
        if(k==0) break;
        for(int i=0;i<k;i++){
            scanf("%d",&s[i]);
        }
        memset(sg,-1,sizeof(sg));
        sg[0]=0;
        scanf("%d",&m);
        while(m--){
            int ans=0;
            scanf("%d",&l);
            for(int i=0;i<l;i++){
                scanf("%d",&h[i]);
                ans^=dfs(h[i]);
            }
            if(ans==0) printf("L");
            else printf("W");
        }
        printf("\n");
    }
    return 0;
}