【题解】POJ-2960 S-Nim

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

1
2
3
4
5
6
7
8
9
10
11
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

样例输出

1
2
LWW
WWL

提示

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

查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
using namespace std;
int SG[10000], f[10000];

void getSg(int n, int m){
for(int i=1; i<=n; i++){
bool S[10000]={0};
for(int j=0; j<m && f[j]<=i; j++){
S[SG[i-f[j]]] = 1;
}
int mex=0;
while(S[mex]) mex++;
SG[i]=mex;
}
}

int main()
{
int k;
while(~scanf("%d", &k) && k)
{
for(int i=0; i<k; i++){
scanf("%d", &f[i]);
}
sort(f, f+k);
getSg(10000, k);

int m; scanf("%d", &m);
while(m--){
int l; scanf("%d", &l);
int nim = 0;
while(l--){
int h; scanf("%d", &h);
nim ^= SG[h];
}
if(nim){
printf("W");
}else{
printf("L");
}
}
printf("\n");
}
return 0;
}
_/_/_/_/_/ EOF _/_/_/_/_/