S-Nim

Problem Description

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

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

 

Sample Output

 

Source

Norgesmesterskapet 2004

题意：m堆石子  玩家每次可以从某一堆中取出si个石子 不能取则输

题解：初步学习sg函数 sg[i]为 i的后继状 的sg值中 没有出现过的非负最小值。

sg异或值为0则后手胜

 1 /****************************** 2 code by drizzle 3 blog: www.cnblogs.com/hsd-/ 4 ^ ^    ^ ^ 5  O      O 6 ******************************/ 7 #include<bits/stdc++.h> 8 #include<map> 9 #include<set>10 #include<cmath>11 #include<queue>12 #include<bitset>13 #include<math.h>14 #include<vector>15 #include<string>16 #include<stdio.h>17 #include<cstring>18 #include<iostream>19 #include<algorithm>20 #pragma comment(linker, "/STACK:102400000,102400000")21 using namespace std;22 #define  A first23 #define B second24 const int mod=1000000007;25 const int MOD1=1000000007;26 const int MOD2=1000000009;27 const double EPS=0.00000001;28 typedef __int64 ll;29 const ll MOD=1000000007;30 const int INF=1000000010;31 const ll MAX=1ll<<55;32 const double eps=1e-8;33 const double inf=~0u>>1;34 const double pi=acos(-1.0);35 typedef double db;36 typedef unsigned int uint;37 typedef unsigned long long ull;38 int k;39 int sg[10005];40 int a[105];41 int flag[105];42 int q,m,exm;43 void init()44 {45     sg[0]=0;46     for(int i=1;i<=10000;i++)47     {48         memset(flag,0,sizeof(flag));49         for(int j=1;j<=k;j++)50         {51             if(i-a[j]>=0)52             {53               flag[sg[i-a[j]]]=1;54             }55         }56         for(int j=0;;j++)57         {58             if(flag[j]==0){59                 sg[i]=j;60             break;61                 }62         }63     }64 }65 int main()66 {67   while(scanf("%d",&k)!=EOF)68   {69       if(k==0)70         break;71       for(int i=1;i<=k;i++)72         scanf("%d",&a[i]);73        init();74        scanf("%d",&q);75        for(int i=1;i<=q;i++)76        {77            scanf("%d",&m);78            int ans=0;79            for(int j=1;j<=m;j++)80            {81                scanf("%d",&exm);82                ans^=sg[exm];83            }84            if(ans==0)85             printf("L");86            else87             printf("W");88        }89        printf("\n");90   }91  return 0;92 }