解题报告 HDU1944 S-Nim

时间：2015-01-25 16:40:27 阅读：221 评论：0 收藏：0 [点我收藏+]

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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

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

Sample Output

LWW

WWL

AC代码：

 1 #include<iostream>
 2 #include<math.h>
 3 #include<algorithm>
 4 #include<string>
 5 using namespace std;
 6 int a[105];
 7 int sg[10005];
 8 int k;
 9 int mex(int x)
10 {
11     if(sg[x]!=-1) return sg[x];
12     bool vis[105];
13     memset(vis,0,sizeof(vis));
14     for(int i=0;i<k;i++)
15     {
16         if(x-a[i]>=0)
17         {
18             mex(x-a[i]);
19             vis[sg[x-a[i]]]=true;
20         }
21     }
22     for(int i=0;i<105;i++)
23         if(!vis[i])
24             return sg[x]=i;
25 }
26 int main()
27 {
28     while(cin>>k&&k)
29     {
30         string str="";
31         memset(sg,-1,sizeof(sg));
32         sg[0]=0;
33         for(int i=0;i<k;i++)
34             cin>>a[i];
35         sort(a,a+k);
36         int m;
37         cin>>m;
38         for(;m>0;m--)
39         {
40             int ans=0;  
41             int x,u;
42             cin>>x;
43             for(int i=0;i<x;i++)   
44             {  
45                 cin>>u;
46                 ans^=mex(u);  
47             }     
48             if(!ans)  str+="L"; 
49             else str+="W";  
50         }  
51         cout<<str<<endl; 
52     }
53     return 0;
54 }

解题报告 HDU1944 S-Nim

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原文地址：http://www.cnblogs.com/verlen11/p/4248374.html

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