「POJ2960」S – Nim



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 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.


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

Sample Output


每次可取的石子数x满足x属于集合S(k) = {s1,s2,s3…sk},问先拿者是否有必胜策略?

  易得:SG(x) = mex(SG[x-s[i]])(1<i<k);
3.后面就是SG函数的应用,根据Sprague-Grundy Therem:g(G)=g(G1)^g(G2)^g(G3)^…^g(Gn)
  SG(G) = SG(x1)^SG(x2)^…^SG(xn),故若SG(G)=0那么必输