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.
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:
It is quite easy to convince oneself that this works. Consider these facts:
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
//还是SG值套模版函数就好了;
#include
#include
#include
using namespace std;
int k,ki[101];
int sg[10001];
int mex(int n)
{
}
int main()
{
}