hdu1536 S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5856    Accepted Submission(s): 2507

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

 

题意：给你一个集合S，再给你几个局面，每次操作某一堆只能减去S中的一个数，问当前局面是必胜局面还是必败局面。

思路：这题是sg模板题，因为终止条件是最后不能走的人输，所以当各个游戏的sg值异或和为0时为必败局面，不为0为必胜局面，我们只要处理每个游戏的sg值就行了。处理sg值的时候有两种方法，一种是每次每个局面单独求一遍，递归出它的后继游戏的sg的值，然后它的sg值就是不包含后继状态sg值的最小非负整数，还有一种是对于一个S集合求所有状态的sg值一次求出来，然后打表，那么要求的时候就可以直接用了。

代码一：递归求sg的值

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
int mex[110],sg[10010],s[110],n,m;
int get_sg(int num)
{
    int i,j;
    if(sg[num]!=-1)return sg[num];
    bool mex[110]; //mex的大小和S集合的大小差不多
    memset(mex,false,sizeof(mex));
    for(i=1;i<=n;i++){
        if(s[i]<=num){
            mex[get_sg(num-s[i])]=true;
        }
        else break;
    }
    for(i=0;;i++){
        if(!mex[i]){
            sg[num]=i;return i;
        }
    }
}
int main()
{
    int i,j,l,c,num;
    while(scanf("%d",&n)!=EOF && n!=0)
    {
        for(i=1;i<=n;i++){
            scanf("%d",&s[i]);
        }
        sort(s+1,s+1+n); //这里要排序，因为get_sg函数中进行到s[i]>num就break了。
        memset(sg,-1,sizeof(sg));
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d",&l);
            num=0;
            for(i=1;i<=l;i++){
                scanf("%d",&c);
                num=num^get_sg(c);
            }
            if(num==0)printf("L");
            else printf("W");
        }
        printf("
");
    }
    return 0;
}

代码二：先打表求出所有局面的sg函数值。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
int mex[110],sg[10010],s[110],n,m;
void get_sg( )
{
    int i,j;
    sg[0]=0;
    for(i=1;i<=10000;i++){
        memset(mex,0,sizeof(mex));
        for(j=1;j<=n;j++){
            if(s[j]<=i){
                mex[sg[i-s[j] ] ]=1;
            }
            else break;
        }
        for(j=0;;j++){
            if(!mex[j]){
                sg[i]=j;break;
            }
        }

    }




}


int main()
{
    int i,j,l,c,num;
    while(scanf("%d",&n)!=EOF && n!=0)
    {
        for(i=1;i<=n;i++){
            scanf("%d",&s[i]);
        }
        sort(s+1,s+1+n);
        get_sg();
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d",&l);
            num=0;
            for(i=1;i<=l;i++){
                scanf("%d",&c);
                num=num^sg[c];
            }
            if(num==0)printf("L");
            else printf("W");
        }
        printf("
");
    }
    return 0;
}