HDU 1536（sg博弈） S-Nim

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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题意：给一个n，n个数。下面博弈的时候取数只能从这n个数里面取。

下面m行，每行开头是石头的堆数，后面的数字代表每堆石头的个数。然后开始博弈，先手胜的话，输出W，否则输出L。

用序列构造sg函数。然后博弈即可。

#include <cstring>#include <cmath>#include <algorithm>#include <queue>#include <iostream>#include <cstdio>using namespace std;int sg[10005];int f[110];void getsg(int n){    int i,j;    memset(sg,0,sizeof(sg));    bool hash1[10005]; //一定要用bool 定义，不然会超时。    for(i=0; i<10005; i++)    {        memset(hash1,0,sizeof(hash1));        for(j=0; j<n; j++)        {            if(i-f[j]>=0)                hash1[sg[i-f[j]]]=1;        }        for(j=0; j<10005; j++)        {            if(!hash1[j])            {                sg[i]=j;                break;            }        }    }    return ;}int main(){    int n;    while(scanf("%d",&n)!=EOF&&n!=0)    {        for(int i=0; i<n; i++)            scanf("%d",&f[i]);        sort(f,f+n);        getsg(n);        int m;//        for(int i=0; i<1000; i++)//            cout<<sg[i]<<"    ";        scanf("%d",&m);        for(int i=0; i<m; i++)        {            int t;            int pp;            int res=0;            scanf("%d",&t);            for(int j=0; j<t; j++)            {                scanf("%d",&pp);                res=res^sg[pp];            }            if(res==0)                printf("L");            else                printf("W");        }        printf("\n");    }    return 0;}