hdu 1536 S-Nim

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

【题意】

以我仅剩不多的英文水平，硬是没懂这题说啥，没办法只有搜别人的博客，总算是懂了。用我的话解释一下。

用取石子来说明：

现在你和对手约定玩取石子的游戏，但是你们不想游戏太简单，所以每次取的时候只能取特定的值，这些值放在一个集合里面，然后就是一局局的游戏，对于每一局游戏你要输出一个值表示先手是否能必胜。

还是用测试用例说话：

首先是多组输入，对于每一组输入，首先是能选择的取子数量集合的长度，然后就是能取的各个数量。然后一行是按照当前集合的游戏局数，对于每一局游戏有一行 ，第一个数据表示石子的堆数，然后就是每一堆的数量多少。

【分析】

只能说题意远比题目难，虽然是第一次做SG的题目，但是过程还是好理解的。然后就是getSG那个函数，应该就是模板了。关键有一点是能选择的石子数量得排个序，开始没注意，wa了一发（主要是测试用例给的太好，没注意）。然后就没啥了。如果不懂的话就可以去好好学学SG（主要是我比较水，讲不清楚，有时间再来更新）。

【代码】

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <list>
#include <algorithm>
#include <iostream>

using namespace std;
int len;
int canchoose[105];
int num;
int SG[10010];
int getSG(int x)
{
if(SG[x]!=-1) return SG[x];
bool flag[105];
memset(flag,false,sizeof(flag));
for(int i=0;i<len;i++)
{
if(canchoose[i]>x) break;
int tmp=x-canchoose[i];
if(SG[tmp]==-1) SG[tmp]=getSG(tmp);
flag[SG[tmp]]=true;
}
for(int i=0;;i++) if(!flag[i]) return (SG[x]=i);
}
int main() {
//    freopen("in.txt","r",stdin);
while(cin>>len&&len)
{
for(int i=0;i<len;i++)
cin>>canchoose[i];
sort(canchoose,canchoose+len);
int T;
memset(SG,-1,sizeof(SG));
cin>>T;
while(T--)
{
int dui;
cin>>dui;
int ans=0;
for(int i=0;i<dui;i++)
{
cin>>num;
ans^=getSG(num);
}
cout<<(ans?'W':'L');
}
cout<<endl;
}
return 0;
}