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
0Sample 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
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define ll long long
#define re register
#define gc getchar()
inline int read()
{
re int x(0),f(1);re char ch(gc);
while(ch<'0'||ch>'9') {if(ch=='-')f=-1; ch=gc;}
while(ch>='0'&&ch<='9') x=(x*10)+(ch^48),ch=gc;
return x*f;
}
const int N=1e2,M=1e5;
int s[N+10],SG[M+10],k,vis[N+10];
void prepare()
{
memset(SG,0,sizeof(SG));
for(int i=1;i<=M;++i)
{
memset(vis,0,sizeof(vis));
for(int j=1;j<=k;++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(k=read(),k)
{
for(int i=1;i<=k;++i)
s[i]=read();
sort(s+1,s+1+k);
prepare();
int n=read();
for(int i=1;i<=n;++i)
{
int m=read(),ans=0;
for(int i=1;i<=m;++i)
ans^=SG[read()];
cout<<(!ans?"L":"W");
}
cout<<endl;
}
return 0;
}