HDU S NIM 求sg函数

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6014    Accepted Submission(s): 2568

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

Sprague-Grudy定理：

令N = {0, 1, 2, 3, ...} 为自然数的集合。Sprague-Grundy 函数给游戏中的每个状态分配了一个自然数。结点v的Grundy值等于没有在v的后继的Grundy值中出现的最小自然数.

形式上：给定一个有限子集 S ⊂ N,令mex S(最小排斥值)为没有出现在S中的最小自然数。定义mex(minimal excludant)运算，这是施加于一个集合的运算，表示最小的不属于这个集合的非负整数。例如mex{0,1,2,4}=3、mex{2,3,5}=0、mex{}=0。

对于一个给定的有向无环图，定义关于图的每个顶点的Sprague-Garundy函数g如下：g(x)=mex{ g(y) | y是x的后继 }。

5、性质：

（1）所有的终结点所对应的顶点，其SG值为0，因为它的后继集合是空集——所有终结点是必败点（P点）。

（2）对于一个g(x)=0的顶点x，它的所有后继y都满足g(y)!=0——无论如何操作，从必败点（P点）都只能进入必胜点（N点）//对手走完又只能把N留给我们。

（3）对于一个g(x)!=0的顶点，必定存在一个后继点y满足g(y)=0——从任何必胜点（N点）操作，至少有一种方法可以进入必败点（P点）//就是那种我们要走的方法。

6、应用：

（1）可选步数为1-m的连续整数，直接取模即可，SG(x) = x % (m+1); 

（2）可选步数为任意步，SG(x) = x; 

（3）可选步数为一系列不连续的数，用mex(计算每个节点的值)

ACcode：

#include <cstdio>#include <cstring>#include<algorithm>#define maxn 10010using namespace std;int a[maxn],sg[maxn];bool mex[maxn];void GS(int n){    memset(sg,0,sizeof(sg));    for(int i=0;i<maxn;++i){        memset(mex,false,sizeof(mex));        for(int j=0;j<n&&a[j]<=i;++j)            mex[sg[i-a[j]]]=1;        for(int j=0;j<=i;++j)        if(!mex[j]){            sg[i]=j;            break;        }    }}int main(){    int n,m,t,z,s;    while(scanf("%d",&n),n){        for(int i=0;i<n;++i)scanf("%d",&a[i]);        sort(a,a+n);GS(n);        scanf("%d",&m);        while(m--){            scanf("%d",&t);            s=0;            while(t--){                scanf("%d",&z);                s^=sg[z];            }            if(!s)printf("L");            else printf("W");        }        printf("\n");    }    return 0;}/*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*/