hdu 1536 S-Nim (博弈)

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

 

Source

Norgesmesterskapet 2004

题意：先输入K 表示一个集合的大小 之后输入集合 表示对于这对石子只能去这个集合中的元素的个数

之后输入 一个m 表示接下来对于这个集合要进行m次询问 

之后m行 每行输入一个n 表示有n个堆 每堆有n1个石子 问这一行所表示的状态是赢还是输 如果赢输入W否则L

思路：sg函数的运用；

代码（搜索求sg值）：

#include<algorithm>#include<cstring>#include<iostream>using namespace std;const int N=10005;int f[105];int sg[N];int k;int dfs(int x){    if(sg[x]!=-1)        return sg[x];    int vis[105];    memset(vis,0,sizeof(vis));    for(int i=0;i<k&&x>=f[i];i++)            vis[dfs(x-f[i])]=1;    for(int i=0;;i++)        if(vis[i]==0)            return sg[x]=i;}int main(){    while(~scanf("%d",&k))    {        if(k==0) break;        for(int i=0;i<k;i++)            scanf("%d",&f[i]);        char c[105];        int T;        memset(sg,-1,sizeof(sg));        sort(f,f+k);        scanf("%d",&T);        for(int t=0;t<T;t++)        {            int n,v;            int ans=0;            scanf("%d",&n);            for(int i=0;i<n;i++)            {                scanf("%d",&v);                ans^=dfs(v);            }            if(ans) c[t]='W';            else c[t]='L';        }        for(int i=0;i<T;i++)            printf("%c",c[i]);        printf("\n");    }}