HDU 356 S-Nim

S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2958 Accepted Submission(s): 1314

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

 

 

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

int f[101][101];

int arry[101];

int SG[10002];

int getsg(int num)

{

    int i,tmp,hash[101]={0};

    for(i=1;i<=arry[0];i++)

    {

        if(arry[i]>num)

            break;

        tmp=num-arry[i];

        if(SG[tmp]==-1)

            SG[tmp]=getsg(tmp);

        hash[SG[tmp]]=1;

    }

    for(i=0;;i++)

        if(hash[i]==0)

            return i;

}

void px()

{

    int i,j,k,tmp;

    for(i=1;i<arry[0];i++)

    {

        k=i;

        for(j=i+1;j<=arry[0];j++)

            if(arry[k]>arry[j])

                k=j;

        tmp=arry[k];

        arry[k]=arry[i];

        arry[i]=tmp;

    }

}

int main()

{

    int i,j,k,n,m,tmp,max;

    while(scanf("%d",&n)>0)

    {

        if(n==0)break;

        arry[0]=n;

        for(i=1;i<=n;i++)

            scanf("%d",&arry[i]);

        px();

        scanf("%d",&m);

        for(i=1,max=0;i<=m;i++)

        {

            scanf("%d",&k);

            f[i][0]=k;

            for(j=1;j<=k;j++)

            {

                scanf("%d",&f[i][j]);

                if(f[i][j]>max)

                    max=f[i][j];

            }

        }

        memset(SG,-1,sizeof(SG));

        for(i=1;i<=m;i++)

        {

            k=0;

            for(j=1;j<=f[i][0];j++)

            {

                tmp=f[i][j];

                k=k^getsg(tmp);

            }

            if(k==0) printf("L");

            else printf("W");

        }

        printf("\n");

    }

    return 0;

}