S-Nim(POJ-2960)

题面

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.

输入

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.

输出

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.

样例输入

 12 2 5
 23
 32 5 12
 43 2 4 7
 54 2 3 7 12
 65 1 2 3 4 5
 73
 82 5 12
 93 2 4 7
104 2 3 7 12
110

样例输出

1LWW
2WWL

提示

无

思路

SG函数打表，最后运用SG定理每堆石子异或一下即可。

代码

 1using namespace std;
 2int SG[10000], f[10000];
 3
 4void getSg(int n, int m){
 5    for(int i=1; i<=n; i++){
 6        bool S[10000]={0};
 7        for(int j=0; j<m && f[j]<=i; j++){
 8            S[SG[i-f[j]]] = 1;
 9        }
10        int mex=0;
11        while(S[mex]) mex++;
12        SG[i]=mex;
13    }
14}
15
16int main()
17{
18    int k;
19    while(~scanf("%d", &k) && k)
20    {
21        for(int i=0; i<k; i++){
22            scanf("%d", &f[i]);
23        }
24        sort(f, f+k);
25        getSg(10000, k);
26
27        int m; scanf("%d", &m);
28        while(m--){
29            int l; scanf("%d", &l);
30            int nim = 0;
31            while(l--){
32                int h; scanf("%d", &h);
33                nim ^= SG[h];
34            }
35            if(nim){
36                printf("W");
37            }else{
38                printf("L");
39            }
40        }
41        printf("\n");
42    }
43    return 0;
44}