S-Nim

<span size="+0"><strong><span style="font-family:Arial;font-size:12px;color:green;FONT-WEIGHT: bold">Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4479    Accepted Submission(s): 1941</span></strong></span>

Problem 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 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120

Sample Output

LWWWWL

Source

Norgesmesterskapet 2004 

思路：尼姆博弈的变形，套用了sg函数

代码如下：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int k,a[110],f[10010];int mex(int p)//sg函数{	int i,t;	bool g[110]={0};	for(i=0;i<k;i++)	{		t=p-a[i];		if(t<0)		break;		if(f[t]==-1)		f[t]=mex(t);		g[f[t]]=1;	}	for(i=0;;i++)	if(!g[i])	return i;}int main(){	int n,i,m,j,t,s;	while(~scanf("%d",&k),k)	{		for(i=0;i<k;i++)		scanf("%d",&a[i]);		sort(a,a+k);		memset(f,-1,sizeof(f));		f[0]=0;		scanf("%d",&n);		while(n--)		{			scanf("%d",&m);			s=0;			while(m--)			{				scanf("%d",&t);				if(f[t]==-1)				f[t]=mex(t);				s^=f[t];//尼姆博弈，各个堆的数量的异或和为零的话，先手输			}			if(s==0)			printf("L");			else			printf("W");//题上的没有换行，注意细节 ，此处没有换行符		}		printf("\n");	}	return 0;}