poj2960（hdu也有，SG博弈）

poj2960地址：http://poj.org/problem?id=2960

hdu1536地址：http://acm.hdu.edu.cn/showproblem.php?pid=1536

hdu1944地址：http://acm.hdu.edu.cn/showproblem.php?pid=1944

S-Nim

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 2800		Accepted: 1492

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

简单SG函数应用

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int fi[110],m,sg[10010];

void init()  //SG函数
{  
   bool used[110];
   int k;
   for(int i=fi[0];i<=10000;i++)
   {
	   memset(used,0,sizeof(used));
	   for(int j=0;j<m;j++)
	   {
		   if(fi[j]>i) break;
		   int num=i-fi[j];
		   used[sg[num]]=true;
	   }
	   k=0;
	   while(used[k]) k++;
	   sg[i]=k;
   }
}  

int main()
{
	int k,n,h;
	char anst[110];
	while(scanf("%d",&m)>0,m)
	{
		memset(sg,0,sizeof(sg));
		for(int i=0;i<m;i++)
			scanf("%d",&fi[i]);
		sort(fi,fi+m);
		init();k=0;
		scanf("%d",&m);
		while(m--)
		{
			scanf("%d",&n);
			int ans=0;
			while(n--)
			{
				scanf("%d",&h);
				ans=ans^sg[h];
			}
			if(ans) anst[k++]='W';
			else anst[k++]='L';
		}
		for(int i=0;i<k;i++)
			printf("%c",anst[i]);
		puts("");
	}
	return 0;
}