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

 
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,SG 函数 S-NimSG 函数 S-Nim
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 ;
}
 
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