S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4770    Accepted Submission(s): 2058

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

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int s[101],k;
int sg[10001];
int n;
int seg(int num){
    if(sg[num]!=-1)return sg[num];
    bool used[101];
    memset(used,0,sizeof(used));
    for(int i=0;i<k;i++){
        if(num<s[i])break;
        used[seg(num-s[i])]=true;
    }
    for(int i=0;i<101;i++)if(!used[i])return sg[num]=i;
}
int main(){
    freopen("C:\\Users\\Administrator\\Desktop\\input.txt","w",stdout);
    while(scanf("%d",&k)==1&&k){
        for(int i=0;i<k;i++)scanf("%d",s+i);
        memset(sg,-1,sizeof(sg));
        sort(s,s+k);
        int T;
        scanf("%d",&T);
        while(T--){
            scanf("%d",&n);
            int sum=0,tmp;
            for(int i=0;i<n;i++){scanf("%d",&tmp);sum^=seg(tmp);}
            if(sum==0)putchar(‘L‘);
            else putchar(‘W‘);
        }
        putchar(‘\n‘);
    }
    return 0;
}