HDU1536 S-Nim（sg函数变换规则）

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9829    Accepted Submission(s): 4038

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//两种取法，只能取2或5个

3//例数

2 5 12//例一：两堆石子个数分别为5和12

3 2 4 7

4 2 3 7 12

5 1 2 3 4 5//五种取法。。。。

3

2 5 12

3 2 4 7

4 2 3 7 12

0

Sample Output

LWW

WWL

#include<iostream>
#include<string.h>
using namespace std;
const int N=10001;
int k,sg[N],fa[111];
void getsg(int n)
{
    bool mex[N];
    for(int i=1;i<=n;i++)
    {
        memset(mex,0,sizeof(mex));
        for(int j=0;j<k;j++)
            if(i>=fa[j])
                mex[sg[i-fa[j]]]=1;
        for(int j=0;;j++)
            if(!mex[j])
            {
                sg[i]=j;
                break;
            }
    }
}
int main()
{
    int m;
    while(~scanf("%d",&k)&&k)
    {
        memset(fa,0,sizeof(fa));
        for(int i=0;i<k;i++)
            scanf("%d",&fa[i]);
        getsg(N);
        char s[111];
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            int h,l,sum=0;
            scanf("%d",&l);
            while(l--)
            {
                scanf("%d",&h);
                sum^=sg[h];
            }
            if(sum)
                 s[i]='W';
            else s[i]='L';
        }
        s[m]='\0';
        printf("%s\n",s);
    }
    return 0;
}