poj 2960 S-Nim

S-Nim

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4113		Accepted: 2158

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

 

/*
poj 2960 S-Nim

先给你一个集合，然后是类似于NIM游戏，但是你每次只能从这些石碓中取出集合中的个数，
搞出SG值然后进行计算即可
而且这题 sort什么的会RE 囧.

hhh-2016-08-02 18:16:21
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <functional>
typedef long long ll;
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std;
const int maxn = 10000+10;

int sg[maxn];
int s[maxn];
int n;

void SG(int now)
{
    if(sg[now] != -1)
        return ;
    int vis[maxn];
    memset(vis,0,sizeof(vis));
    for(int i = 0; i < n; i++)
    {
        int t = now-s[i];
        if(t < 0)
            continue;
        SG(t);
        vis[sg[t]] = 1;
    }

    for(int i = 0;; i++)
    {
        if(!vis[i])
        {
            sg[now] = i;
            break;
        }
    }
}

int main()
{
    int x,m;
    //freopen("in.txt","r",stdin);
    while(scanf("%d",&n) != EOF && n)
    {
        for(int i = 0; i < n; i++)
            scanf("%d",&s[i]);
        //sort(s,s+n);
        memset(sg,-1,sizeof(sg));
        sg[0] = 0;
        scanf("%d",&m);
        while(m--)
        {
            int ans = 0,cnt;
            scanf("%d",&cnt);
            for(int i = 0; i < cnt; i++)
            {
                scanf("%d",&x);
                if(sg[x] == -1)
                    SG(x);
                ans ^= sg[x];
            }
            if(ans)
                printf("W");
            else
                printf("L");
        }
        printf("\n");

    }
    return 0;
}