S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5637 Accepted Submission(s): 2414

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

     #include<cstdio>

     #include<algorithm>

     using namespace std;

     #define N 100+10

     int knum,mnum,lnum;

     int ans[N],si[N],hi[N],sg[];

     int mex(int x)//求x的sg值（可作为模版应用）

     {

         if(sg[x]!=-) return sg[x];

         bool vis[N];

         memset(vis,false,sizeof(vis));

         for(int i=;i<knum;i++)    {

             int temp=x-si[i];

             if(temp<) break;

             sg[temp]=mex(temp);

             vis[sg[temp]]=true;

         }

         for(int i=;;i++)    {

             if(!vis[i])    {

                 sg[x]=i; break;

             }

         }

         return sg[x];

     }

     int main()    {

         while(scanf("%d",&knum) && knum)    {

             for(int i=;i<knum;i++)

                 scanf("%d",&si[i]);

             sort(si,si+knum);

             memset(sg,-,sizeof(sg));

             sg[]=;

             memset(ans,,sizeof(ans));

             scanf("%d",&mnum);

             for(int i=;i<mnum;i++)   {

                 scanf("%d",&lnum);

                 for(int j=;j<lnum;j++) {

                     scanf("%d",&hi[i]); ans[i]^=mex(hi[i]);//尼姆博弈

                 }

             }

             for(int i=;i<mnum;i++)

             {

                 if(ans[i]==) printf("L");

                 else printf("W");

             }

             printf("\n");

         }

         return ;

     }

 #include"iostream"

 #include"algorithm"

 #include"string.h"

 using namespace std;

 int s[],sg[],k;

 int getsg(int m)

 {

     int hash[]={};

     int i;

     for(i=;i<k;i++){

         if(m-s[i]<)

             break;

         if(sg[m-s[i]]==-)

             sg[m-s[i]]=getsg(m-s[i]);

         hash[sg[m-s[i]]]=;

     }

     for(i=;;i++)

         if(hash[i]==)

             return i;

 }

 int main()

 {

     //int k;

    // freopen("game.in","r",stdin);

     //freopen("game.out","w",stdout);

     while(cin>>k,k)

     {

         int i;

         for(i=;i<k;i++)

             cin>>s[i];

         sort(s,s+k);

         memset(sg,-,sizeof(sg));

         sg[]=;

         int t;

         cin>>t;

         while(t--)

         {

             int n,m;

             cin>>n;

             int ans=;

             while(n--)

             {

                 cin>>m;

                 if(sg[m]==-)

                     sg[m]=getsg(m);

                 ans^=sg[m];

             }

             if(ans)

                 cout<<'W';

             else cout<<'L';

         }

         cout<<endl;

     }

     return ;

 }

巴特西

hdu 1536 sg (dfs实现)

S-Nim

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