HDU 1536 S-Nim（SG博弈）

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7117    Accepted Submission(s): 3012

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

 

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题目大意：

    有M堆石子，堆有hi个石子，有S个操作，每次可以选择拿si个石子，不能拿着失败。

解题思路：

    明显的SG博弈，先打一个SG表，在把各个堆的SG值异或一下就行了。

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;#define mem(a,b) memset((a),(b),sizeof(a))const int maxn=100+3;const int maxi=10000;int op[maxn],S,M,sg[maxi+3];bool used[maxi+3];void make_sg()//打SG表{    sg[0]=0;//0个石子是必负    for(int i=1;i<=maxi;++i)    {        //mem(used,0);        for(int j=0;j<S;++j)//把前面的状态的SG值加入数组            if(i-op[j]>=0)                used[sg[i-op[j]]]=true;        for(int j=0;j<maxi;++j)//找到最小的没出现的SG值            if(!used[j])            {                sg[i]=j;                break;            }        for(int j=0;j<S;++j)//恢复used数组（有时用memset快一些，具体分析决定用哪个）            if(i-op[j]>=0)                used[sg[i-op[j]]]=false;    }}int main(){    while(~scanf("%d",&S)&&S)    {        for(int i=0;i<S;++i)            scanf("%d",&op[i]);        make_sg();        scanf("%d",&M);        while(M--)        {            int num,ans=0,tmp;            scanf("%d",&num);            while(num--)            {                scanf("%d",&tmp);                ans^=sg[tmp];//各个堆异或            }            if(ans)                putchar('W');            else putchar('L');        }        putchar('\n');    }        return 0;}