HDU 1536 S-Nim SG博弈

S-Nim

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

 

 

题意：

　　给你k个数 s[i]

　　再给你m个询问，每次询问是一个nim游戏，但是相比nim不同的是，每次只能从各个堆中选取 s[i]的值除去

题解：

　　SG函数的应用

　　对于给定的k个数，我们预处理出sg[i]

　　那么就简单了

　　

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 5e5+, M = 2e5+, mod = 1e9+, inf = 2e9;

int k,sg[N],s[N],vis[N];
char A[N];
int main() {
        while(scanf("%d",&k)!=EOF) {
            if(k == ) break;
            for(int i = ; i <= k; ++i) scanf("%d",&s[i]);
            sg[] = ;
            for(int i = ; i <= ; ++i) {
                for(int j = ; j <= ; ++j) vis[j] = ;
                for(int j = ; j <= k; ++j) {
                   if(i >= s[j] && sg[i - s[j]] <= ) vis[sg[i - s[j]]] = ;
                }
                for(int j = ; j <= ; ++j) {
                    if(!vis[j]) {
                        sg[i] = j;
                        break;
                    }
                }
            }
            int q,cnt = ;
            scanf("%d",&q);
            while(q--) {
                int x,y,ans = ;
                scanf("%d",&x);
                while(x--) {
                    scanf("%d",&y);
                    ans ^= sg[y];
                }
                if(ans) printf("W");
                else printf("L");
            }
            printf("\n");
        }
    return ;
}