S-Nim（fromHDU）（博弈）（Sprague-Grundy定理）（nim问题加强版）
2017-04-08 09:26:25         来源：Coldfresh的博客

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

5：101
7：111
9：1001

Mex 函数值：这个函数的接受一个自然数集 N，输出自然数集中减去 N 之后最小的那个整数。例如 Mex({0,1,3,}) = 2， Mex({1,2,}) = 0, Mex({}) = 0。（所有定理不证明）

ac代码：

```import java.util.Arrays;
import java.util.HashSet;

import java.util.Scanner;

public class Main
{
static int sg[]=new int[10005];//我们用sg来存储各个状态的nim值
static int s,c[];
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
for(;;)
{
s=sc.nextInt();//能拿的种类个数
if(s==0)break;
Arrays.fill(sg, -1);//记忆化标志
c=new int[s];//存放在c数组里
for(int i=0;i0)
{
int t=sc.nextInt();
int nim=0;
for(int i=0;i set=new HashSet<>();//这里用HashSet的原因是，既然是集合，就不允许出现重复的自然数
for(int i=0;i=c[i])//获得所有可能的子状态，将其存入set中
else
break;
}
return sg[x]=Mex(set);
}
static int Mex(HashSet set)//按照定义，找出nim值
{
int x=0;
for(;;)
{
if(!set.contains(x))
break;
x++;
}
return x;
}
}```