HDU1536 S-nim 尼姆博弈简单题

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4004    Accepted Submission(s): 1732

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 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120

 

Sample Output

LWWWWL

#include<stdio.h>  #include<string.h>  #include<algorithm>  using namespace std;  //注意 S数组要按从小到大排序 SG函数要初始化为-1 对于每个集合只需初始化1遍  //n是集合s的大小 S[i]是定义的特殊取法规则的数组  int s[110],sg[10010],n;  int SG_dfs(int x)  {      int i;      if(sg[x]!=-1)          return sg[x];      bool vis[110];      memset(vis,0,sizeof(vis));      for(i=0;i<n;i++)      {          if(x>=s[i])          {               vis[SG_dfs(x-s[i])]=1; // SG_dfs(x-s[i]);      //vis[sg[x-s[i]]]=1;           }      }      int e;      for(i=0;;i++)          if(!vis[i])          {         return sg[x]=i;  //   e=i;          //   break;          }     // return sg[x]=e;  }  int main()  {      int i,m,t,num;      while(scanf("%d",&n)&&n)      {          for(i=0;i<n;i++)              scanf("%d",&s[i]);          memset(sg,-1,sizeof(sg));          sort(s,s+n);          scanf("%d",&m);          while(m--)          {              scanf("%d",&t);              int ans=0;              while(t--)              {                  scanf("%d",&num);                  ans^=SG_dfs(num);              }              if(ans==0)                  printf("L");              else                  printf("W");          }          printf("\n");      }      return 0;  }