HDU 1016 S-Nim ----SG求值

S-Nim

Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

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

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

 

 /*
 题意：第二次做题，题意完全忘记。
 前面都是背景，告诉你Nim是赢和输的规则。
 后面改成了：选的数字是规定的。

 数字是改变的，用打表划不来！
 数字大小到10000，所以Hash只要到100就可以了。

 SG的求法有两种，
 1.是打表的。
 参考http://www.cnblogs.com/tom987690183/archive/2013/05/30/3108564.html
 2.是单点求取的。和记忆化搜索很相似。
 这一题是单点的。
 */

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<algorithm>
 using namespace std;

 int SG[];
 int arry[];

 int make_GetSG(int n)//求单点的。
 {
     int i,tmp,Hash[]={};//后继的大小开sqrt(N);
     for(i=;i<=arry[];i++)
     {
         if(arry[i]>n)
         break;
         tmp=n-arry[i];
         if(SG[tmp]==-)
         SG[tmp]=make_GetSG(tmp);
         Hash[SG[tmp]]=;
     }
     for(i=;;i++)
     if(Hash[i]==)
     return i;
 }

 void make_ini(int m)
 {
     int i,j,k,n,x;
     memset(SG,-,sizeof(SG));
     while(m--)
     {
         scanf("%d",&n);
         k=;
         for(i=;i<=n;i++)
         {
             scanf("%d",&x);
             k=k^make_GetSG(x);
         }
         if(k==)printf("L");
         else printf("W");
     }
     printf("\n");
 }

 int main()
 {
     int k,m,i;
     while(scanf("%d",&k)>)
     {
         if(k==)break;
         for(i=;i<=k;i++)
         scanf("%d",&arry[i]);
         arry[]=k;
         sort(arry+,arry++k);
         scanf("%d",&m);
         make_ini(m);
     }
     return ;
 }