SG 函数 S-Nim

http://poj.org/problem?id=2960

S-Nim

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3464		Accepted: 1829

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

Source

Svenskt Mästerskap i Programmering/Norgesmesterskapet 2004

 

题意：给定数组S,接下来给出m个游戏局面。游戏局面是一些beads堆，先给出堆数，然后是每一堆中beads的数目。游戏规则是，两个人轮流取beads，每次可以选择一堆，从中取出k个beads，k ∈S，最后不能取的人输。 

 

新学的博弈，记下来，先介绍Nim 游戏

情况1 两堆石头，a,b每次可以从任意一堆中拿去任意个数个石头，最后不拿的输  ，要是有两堆石子，要是这两堆石子数目一样的话，先走的肯定会输，因为只要a先拿的话，b可以重复a上面的操作，和a在不同的堆中拿相同的个数，两个数相等的时候其异或值相等，及^想等，

而要是两堆石子的个数不同的话，先拿的可以通过第一次拿走其中个数较多的那一堆石头的一部分，使得剩余的两堆石头个数相同

情况2 n 堆石头，每堆石头有xi个石头，a和b 每次可以选取其中一堆拿取任意数目的石头,最后不能拿的人输。 设必胜态为1， 必败态为 设有某状态后面有n个后续状态，（后继状态就是a拿完后，b应该拿的时候的状态）要是所有的后继状态中有一个是0的话这个状态就是1，而要是后继状态中所有都是1 ，那么这个状态就是0，一步可以走到必败态的一定是必胜态，现在给出一个结论，将每堆石头的数目异或起来，如果等于0的话是必败态，要是不等于0为必胜态，现在假设三堆得数目是a b c , 假设a^b^c=k ;(k!=0)  那么k的第一位上的1肯定是来自于a b c 中的某一个，假设来自b ,那么b^k<b 有第一次拿b-(b^k)个石头，那么，下一个状态的异或值就是a^(b^k)^c =k^k =0 ,依次递归下去，最后三堆石头数目都是0的时候肯定是必败态，所以状态为0的肯定是必败态，而异或值不为0的可以通过一次取石头达到必败态，所以是必胜态。

 

下面介绍sg函数： 定义：为集合中没有出现过的最小的非负整数，s = {1,2,3,4,5}  则sg(s) = 0 ;  s ={0,1,2,3}  sg(s)=4 ;

 情况1： 有一堆石头 K 个，每次只能拿x个，x属于s = {2,4,5} 。做法，要寻找前驱，

集合k是k的所有前驱的sg值得集合 例： sg(8) = sg({sg(6),sg(4),sg(3)};

做法，求出sg(k)即可，要是sg(k) = 0 则为必败态，sg(1) !=0 为必胜态，递归的求 sg(0)  = 0 ;因为没有前驱，sg(1) = 0 ; sg(2) = sg{0} = 1；

sg(3) = sg({sg(1)}) = 1 ; sg(4) = sg({sg(2),sg(0)}) = 2;sg(5) = sg({sg(3),sg(1), sg(0)} = 2;

 

情况2 ： 有n 堆石头，每堆xi个石头，每次从第i 堆中只能取si = {……}中的值得个数，最后不能拿的人输

做法： 求出每一堆得sg(xi)；假设sg(xi) = k ;则它的后驱一定有k-1,k-2,......这样的话，就相当是一个Nim 游戏了，每堆都有k个石头都必须取，可以取任意多个石头，

所以将每一个sg(xi)异或起来，然后要是结果为0则必败

 

提示：一般博弈题，就把所给值异或起来一般就可以做了

 

现在给出Nim游戏用sg函数的证明， nim 游戏相当于是每次取得个数都是123……n,所以sg(0) = 0 ; sg(n) = n ;

 

 

下面是代码：

 

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 #define N 105
 #define M 10005

 int s[N], sn;
 int sg[M];

 void getsg(int n)
 {
     int mk[M];
     sg[] = ;//主要是让终止状态的sg为0
     memset(mk, -, sizeof(mk));
     for(int i = ; i < M; i++)//预处理sg函数
     {
         for(int j = ; j < n && s[j] <= i; j++)
             mk[sg[i-s[j]]]=i;//将所有后继的sg标记为i，然后找到后继的sg没有出现过的最小正整数
                              //优化：注意这儿是标记成了i，刚开始标记成了1，这样每次需初始化mk memset，而标记成i就不需要了
         int j = ;
         while(mk[j] == i) j++;
         sg[i] = j;
     }
 }

 int main()
 {
     while(~scanf("%d", &sn), sn)
     {
         for(int i = ; i < sn; i++) scanf("%d", &s[i]);
         sort(s, s+sn);//排序算一个优化，求sg的时候会用到
         getsg(sn);
         int m;
         scanf("%d", &m);
         char ans[N];
         for(int c = ; c < m; c++)
         {
             int n, tm;
             scanf("%d", &n);
             int res = ;
             for(int i = ; i < n; i++)
             {
                 scanf("%d", &tm);
                 res ^= sg[tm];
             }
             if(res == ) ans[c] = 'L';
             else ans[c] = 'W';
         }
         ans[m]=;
         printf("%s\n", ans);
     }
     return ;
 }