杭电1536 S-Nim（博弈模板）

S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4533 Accepted Submission(s): 1963



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 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[code]/*

Time:2014-8-25 16:19
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAX=10010;
int s[105],sg[MAX];//s[]取石子的特定限制 
int K;
int SG_DFS(int x){
	if(sg[x]!=-1)
		return sg[x];
	bool vis[120];//我之前把这个数组开大了，开到MAX大就一直RE……栈溢出。。。。。。。。无语
	memset(vis,0,sizeof(vis));
	
	for(int i=1;i<=K;i++){
		if(x>=s[i]){
			SG_DFS(x-s[i]);
			vis[ sg[ x-s[i] ] ]=true;
		}
	}
	for(int i=0;;i++){
		if(!vis[i]){
			sg[x]=i;
			break;
		}
	}
	return sg[x];
}
void solve(){
	
	while(scanf("%d",&K),K){
		for(int i=1;i<=K;i++)
		scanf("%d",&s[i]);
		sort(s+1,s+K+1);//记得一定要排序 
		memset(sg,-1,sizeof(sg));
		int N;
		scanf("%d",&N);
			while(N--){
				
				int M,num,ans=0;
				scanf("%d",&M);
				while(M--){
					scanf("%d",&num);
					ans^=SG_DFS(num);//位运算如果与其他一同运算记得加括号
				}
				if(ans==0)
					printf("L");
				else
					printf("W");
			}
			printf("\n");
	}
}
int main(){
	solve();
return 0;
}

[/code]