hdu1536-SG值

S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6549 Accepted Submission(s): 2772

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

题意：就是要你求一下每种情况的SG值异或起来，0就是负态，非0就是胜态。

首先这里TLE了第一次从头到尾的暴力寻找，因为有很多重复的其实是不需要找的。

然后第二次卡住了了STL的时间，TLE第二发。

接着就改成VIS数组记忆化搜索了。

恩，感觉被游戏玩了一晚上。

还需要继续好好理解，gg。

代码：

#include <cstdio>#include <algorithm>#include <cstring>#include <iostream>#include <set>const int MAXN=10005;int k;int m;int maxx;int a[MAXN];int x[MAXN];int sg[MAXN+1];using namespace std;int getsg(int b){    if(sg[b]!=-1)        return sg[b];//搜索过了    if(b<a[0])//无解的情况    {        sg[b]=0;        return 0;    }    bool vis[100];    memset(vis,0,sizeof(vis));     //   set<int>s;        for(int j=0;j<k;j++)        {            if(a[j]<=b)            {                int c=getsg(b-a[j]);                vis[c]=1;            }        }        int ans=0;        while(vis[ans]!=0)        {            ans++;        }        sg[b]=ans;     return sg[b];}void solve()    {        int xx=0;        for(int i=0;i<m;i++)        {            scanf("%d",&x[i]);            int d=getsg(x[i]);            xx^=d;        }        if(xx!=0)        {            printf("W");            return ;        }        else        {            printf("L");            return ;        }}int main (void){    while(~scanf("%d",&k))    {        if(k==0)            break;        for(int i=0;i<k;i++)        {            scanf("%d",&a[i]);        }        sort(a,a+k);        memset(sg,-1,sizeof(sg));        int num;        scanf("%d",&num);        while(num--)        {            scanf("%d",&m);                solve();        }          printf("\n");    }    return 0;}