poj-2960 S-Nim

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

S-Nim

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3891		Accepted: 2037

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

题目大意：给出n堆石子，两人轮流在任意一堆里可以取出s[i]个石子，直到对方不能取，你获胜，否则对方获胜，游戏结束

基础博弈，sg函数模板，但是要先计算出大概11000内的sg，否则会RE，TLE

sg函数：sg(n) = min( N – {sg(n-1), sg(n-2), …sg(n-m)} )

二维sg函数的异或sg(<n1, n2>) = sg(n1)^sg(n2)

#include<iostream>
#include<cstdio>
#include<string>
#include<map>
#include<cstring>
#include<algorithm>
using namespace std;
int s[11000], k;
int SG[11000];
int sg(int n){
    int a[11000], i, j;
    memset(a, 0, sizeof(a));
   // SG[0] = 0;
    for(i=1; i<=k; i++){
        if(n-s[i] >= 0){
            //printf("----%d\n",SG[n-s[i]]);
            a[SG[n-s[i]]] = 1;
        }
    }
    for(i=0; i<=1100; i++){
        if(!a[i]){
            //printf("  %d\n",i);
            return i;
        }
    }
}
int main(){
    int m, n, a, i, j;
    int l;
    char ch[11000];
    while(cin>>k && k){
        for(i=1; i<=k; i++)
            cin>>s[i];
        //sg[0] = 0;
        cin>>m;
        SG[0]=0;
        for(int h=1; h<=11000; h++){
            SG[h] = sg(h);//sg(n) = min( N – {sg(n-1), sg(n-2), …sg(n-m)} )
        }
        for(l=0; l<m; l++){
            cin>>a;
            int sum=0;
            for(i=0; i<a; i++){
                cin>>n;
                //printf("  %d\n",SG[n]);
                sum = sum^SG[n];//sg(5) sg(12)
            }
            if(sum==0)
                ch[l] = 'L';//异或结果为0则输
            else
                ch[l] = 'W';//异或结果不为0则赢
        }
        ch[l]='\0';
        printf("%s\n",ch);
    }
    return 0;
}