# [POJ2960]S-Nim

**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 theyrecently 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 = f2, 5g 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.

Arthur and his sister Caroll 玩nim游戏玩腻了（因为他们都知道了如何计算必胜策略），所以Arthur把这个游戏改了一下规则，Arthur定义了一个有限集合S，每次从一堆石子中取的石子数目必须在S中。现在，Arthur想知道先手是否有必胜策略。有多组测试数据。

**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.

每行输入首先给出一个数k，代表集合S的大小，接下来紧跟着k个数，表示集合S里的数。接下来一行数为m代表有m个游戏，后面m行每行第一个数字为n代表有n堆石子，后面紧跟着n个数代表每堆石子的个数。多组数据，做到0结束

**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.

对于每组数据，我们要输出n个字母，第i个字母为“W”代表第i个游戏先手必胜，“L”代表第i个游戏先手必败，做完一组数据后换行。

**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

Nim游戏的一种，只是有集合的限制。我们开始预处理出集合对SG值的影响，然后普通Nim即可。详解参考浅谈算法——博弈论中的例2与例4

```
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x>=10) print(x/10);
putchar(x%10+'0');
}
const int N=1e2,M=1e4;
int S[N+10],sg[M+10];
bool vis[N+10];
int size;
void prepare(){
memset(sg,0,sizeof(sg));
for (int i=1;i<=M;i++){
memset(vis,0,sizeof(vis));
for (int j=1;j<=size;j++){
if (i<S[j]) break;
vis[sg[i-S[j]]]=1;
}
for (int j=0;j<=N;j++) if (!vis[j]){sg[i]=j;break;}
}
}
int main(){
while (true){
size=read();
if (!size) break;
for (int i=1;i<=size;i++) S[i]=read();
sort(S+1,S+1+size);
prepare();
int n=read();
for (int i=1;i<=n;i++){
int m=read(),ans=0;
for (int j=1;j<=m;j++) ans^=sg[read()];
printf(!ans?"L":"W");
}
putchar('\n');
}
return 0;
}
```