ACM-SG函数之S-Nim——hdu1536 hdu1944 poj2960

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4091    Accepted Submission(s): 1760

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 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

Norgesmesterskapet 2004

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1536

这也是一道经典SG函数的题目。

有关于SG函数的解，能够戳这个，非常具体→http://blog.csdn.net/lttree/article/details/24886205

这道题题意：

我就按着例子格式来说吧：

先输入一个K，表示取数集合的个数。（K为0，则结束）

后面跟k个数,表示取数集合的数（就是每次仅仅能取这几个数量的物品）

然后会跟一个M,表示有M次询问。

然后接下来M行，每行先有一个N，表示有多少堆物品。

N后跟着N个数，表示每堆物品数量。

由于，OJ后台的操作，输入和输出是分开的（事实上就是将你的程序的答案存成一个TXT文件，然后和

标准答案TXT文件进行二进制的比較）

所以，我每一个N都直接输出'L'或者'W‘，

在M行结束时，换行，没实用数组来存答案。

PS：用scanf比cin快80MS

/************************************************
*************************************************
*        Author:Tree                            *
*From :http://blog.csdn.net/lttree              *
* Title : S-Nim                                 *
*Source: hdu 1536                               *
* Hint  : SG                                    *
*************************************************
*************************************************/
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 10001
int f[N],sg[N];
bool mex[N];
void get_sg(int t,int n)
{
    int i,j;
    memset(sg,0,sizeof(sg));
    for(i=1;i<=n;i++)
    {
        memset(mex,0,sizeof(mex));
        // 对于属于g(x)后继的数置1
        for( j=1 ;j<=t && f[j]<=i ;j++ )
            mex[sg[i-f[j]]]=1;
        // 找到最小不属于该集合的数
        for( j=0 ; j<=n ; j++ )
            if(!mex[j])
                break;
        sg[i] = j;
    }
}
int main()
{
    int k,m,n,i,t,temp;
    while( scanf("%d",&k) && k )
    {
        for(i=1;i<=k;++i)
            scanf("%d",&f[i]);
        sort(f+1,f+k+1);
        get_sg(k,N);
        scanf("%d",&m);
        while(m--)
        {
            temp=0;
            scanf("%d",&n);
            for(i=0;i<n;++i)
            {
                scanf("%d",&t);
                temp^=sg[t];
            }
            if( !temp )  printf("L");
            else    printf("W");
        }
        printf("\n");
    }
    return 0;
}