HDU 1536——S-nim博弈

时间：2015-10-18 22:52:34 阅读：213 评论：0 收藏：0 [点我收藏+]

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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

2 5 12

3 2 4 7

4 2 3 7 12

5 1 2 3 4 5

2 5 12

3 2 4 7

4 2 3 7 12

Sample Output

LWW

WWL

题意：。。没咋懂

分析： S-nim博弈，主要学习SG打表的方法

 1 #include<iostream>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int s[105],sg[10001];
 6 bool mex[10001];
 7 void get_sg(int t,int n)
 8 {
 9     int i,j;
10     memset(sg,0,sizeof(sg));
11     for(i=1;i<=n;i++)
12     {
13         memset(mex,0,sizeof(mex));
14         for(j=1;j<=t&&s[j]<=i;j++)
15             mex[sg[i-s[j]]]=1;
16         for(j=0;j<=n;j++)
17             if(!mex[j])
18             break;
19         sg[i]=j;
20     }
21 }
22 int main()
23 {
24     int k;
25     while(cin>>k,k)
26     {
27         for(int i=1;i<=k;i++)
28             cin>>s[i];
29         sort(s+1,s+k+1);
30         get_sg(k,10001);
31         int m,n,ans,t;
32         cin>>m;
33         while(m--)
34         {
35             cin>>n;
36             ans=0;
37             for(int i=0;i<n;i++)
38             {
39                 cin>>t;
40                 ans^=sg[t];
41             }
42             if(ans)
43                 cout<<‘W‘;
44             else
45                 cout<<‘L‘;
46         }
47         cout<<endl;
48     }
49     return 0;
50 }

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HDU 1536——S-nim博弈

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原文地址：http://www.cnblogs.com/forwin/p/4890405.html

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