HDU 1536（sg博弈） S-Nim

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 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.

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.

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

LWW
WWL

LL   |   We have carefully selected several similar problems for you:  1404 1517 1524 1729 1079 

#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;
}