HDU 1016 S-Nim ----SG求值

S-Nim

Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

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：第二次做题，题意完全忘记。
前面都是背景，告诉你Nim是赢和输的规则。
后面改成了：选的数字是规定的。

数字是改变的，用打表划不来！
数字大小到10000，所以Hash只要到100就可以了。

SG的求法有两种，
1.是打表的。 
参考http://www.cnblogs.com/tom987690183/archive/2013/05/30/3108564.html
2.是单点求取的。和记忆化搜索很相似。
这一题是单点的。
*/


#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;


int SG[10003];
int arry[103];


int make_GetSG(int n)//求单点的。
{
    int i,tmp,Hash[101]={0};//后继的大小开sqrt(N);
    for(i=1;i<=arry[0];i++)
    {
        if(arry[i]>n)
        break;
        tmp=n-arry[i];
        if(SG[tmp]==-1)
        SG[tmp]=make_GetSG(tmp);
        Hash[SG[tmp]]=1;
    }
    for(i=0;;i++)
    if(Hash[i]==0)
    return i;
}

void make_ini(int m)
{
    int i,j,k,n,x;
    memset(SG,-1,sizeof(SG));
    while(m--)
    {
        scanf("%d",&n);
        k=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            k=k^make_GetSG(x);
        }
        if(k==0)printf("L");
        else printf("W");
    }
    printf("
");
}

int main()
{
    int k,m,i;
    while(scanf("%d",&k)>0)
    {
        if(k==0)break;
        for(i=1;i<=k;i++)
        scanf("%d",&arry[i]);
        arry[0]=k;
        sort(arry+1,arry+1+k);
        scanf("%d",&m);
        make_ini(m);
    }
    return 0;
}