HDU 1536--s-nim Game

Topic:

Description

Arthur and his sister Caroll has been playing a game called Nim for some time now. Nim is played as follows:


The starting position have 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 isn't able to make a move, loses.


Arthur and Caroll really enjoyed playing this simple game until they recently learned a easy-to-always be-able to fin D 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 would be 1 as 2 XOR 4 XOR 7 = 1).

If The xor-sum is 0, too bad, you'll lose.

Otherwise, move such that the xor-sum becomes 0. This was always possible.


It's quite easy-to-convince oneself that's this works. Consider these facts:

The player is takes the last bead wins.

After the winning player's last move the xor-sum would be 0.

The xor-sum would change after every move.


Which means that if you do sure that the xor-sum always was 0 when you had made your move, your opponent would never be a Ble to win, and, thus, you'll win.

Understandibly It is no 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, which seemed to solve this problem. Each player was now only allowed to remove a number of beads in some predefined set S, e.g. if we had S = (2, 5) each Playe Allowed to remove 2 or 5 beads. Now it isn't 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 this determines if a position of S-nim is a losing or a winning position. A position is a winning position if there are at least one move to a losing position. A position is a losing position if there be no moves to a losing position. This means, as expected, which 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 was 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 a ' L '. Print a newline after all 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 Test instructions:.. I don't understand. Analysis: S-nim game, the main study of the method of the SG to play the table

1#include <iostream>2#include <cstring>3#include <algorithm>4 using namespacestd;5 ints[ the],sg[10001];6 BOOLmex[10001];7 voidGET_SG (intTintN)8 {9     inti,j;Tenmemset (SG,0,sizeof(SG)); One      for(i=1; i<=n;i++) A     { -Memset (MEX,0,sizeof(MEX)); -          for(j=1; j<=t&&s[j]<=i;j++) themex[sg[i-s[j]]]=1; -          for(j=0; j<=n;j++) -             if(!Mex[j]) -              Break; +sg[i]=J; -     } + } A intMain () at { -     intK; -      while(cin>>k,k) -     { -          for(intI=1; i<=k;i++) -Cin>>S[i]; inSort (s+1, s+k+1); -Get_sg (k,10001); to         intm,n,ans,t; +Cin>>m; -          while(m--) the         { *Cin>>N; $ans=0;Panax Notoginseng              for(intI=0; i<n;i++) -             { theCin>>T; +ans^=Sg[t]; A             } the             if(ANS) +cout<<'W'; -             Else $cout<<'L'; $         } -cout<<Endl; -     } the     return 0; -}

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HDU 1536--s-nim Game