解题报告 HDU1944 S-Nim

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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 rst 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

AC代码：

 #include<iostream>
 #include<math.h>
 #include<algorithm>
 #include<string>
 using namespace std;
 int a[];
 int sg[];
 int k;
 int mex(int x)
 {
     if(sg[x]!=-) return sg[x];
     bool vis[];
     memset(vis,,sizeof(vis));
     for(int i=;i<k;i++)
     {
         if(x-a[i]>=)
         {
             mex(x-a[i]);
             vis[sg[x-a[i]]]=true;
         }
     }
     for(int i=;i<;i++)
         if(!vis[i])
             return sg[x]=i;
 }
 int main()
 {
     while(cin>>k&&k)
     {
         string str="";
         memset(sg,-,sizeof(sg));
         sg[]=;
         for(int i=;i<k;i++)
             cin>>a[i];
         sort(a,a+k);
         int m;
         cin>>m;
         for(;m>;m--)
         {
             int ans=;
             int x,u;
             cin>>x;
             for(int i=;i<x;i++)
             {
                 cin>>u;
                 ans^=mex(u);
             }
             if(!ans)  str+="L";
             else str+="W";
         }
         cout<<str<<endl;
     }
     return ;
 }