hdu1536(求sg)

S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6147 Accepted Submission(s): 2621

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

/* Name: Copyright: Author: Date: 16/04/16 15:09 Description: */
#include<cstdio>
const int maxn=10005;
int a[maxn];
int b[maxn];
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef __int64 ll;
#define rint(x) scanf("%d",&x);
#define rll(x) scanf("%I64d",&x);
#define rc(x) scanf("%c",&x);
#define rd(x) scanf("%lf",&x);
#define r(x) cin>>x;
#define pint(x) printf("%d\n",x);
#define pll(x) printf("%I64d\n",x);
#define pint_(x) printf("%d",x);
#define pll_(x) printf("%I64d",x);
#define pc_(x) printf("%c",x);
#define pc(x) printf("%c\n",x);
#define pxn printf("\n");


int sg[maxn];

int solve(int n,int x){
    if(sg[x]!=-1)return sg[x];
    int used[105];
    memset(used,0,sizeof(used));
    for(int i=0;i<n;++i){
        int temp=x-a[i];
        if(temp<0)break;
        ++used[solve(n,temp)];
    }
    for(int i=0;i<=10000;++i){
        if(!used[i]){
            sg[x]=i;break;
        }
    }
    return sg[x];
}
int main(){
    int n;
    while(cin>>n,n){
        for(int i=0;i<n;++i){
            rint(a[i])
        }
        sort(a,a+n);
        int m;
        rint(m)
        memset(sg,-1,sizeof(sg));
        for(int i=0;i<m;++i){
            int num;
            rint(num)
            int ans=0;
            for(int j=0;j<num;++j){
                rint(b[j])
                ans^=solve(n,b[j]);
            }
            if(!ans)
                pc_('L')
            else pc_('W')
        }
        pxn
    }
    return 0;
}

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