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：
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 }