HDU 1536 sg函数

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7262    Accepted Submission(s): 3074

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：m堆石子  玩家每次可以从某一堆中取出si个石子 不能取则输

题解：初步学习sg函数 sg[i]为 i的后继状 的sg值中 没有出现过的非负最小值。

sg异或值为0则后手胜

/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^    ^ ^
 O      O
******************************/
#include<bits/stdc++.h>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define  A first
#define B second
const int mod=1000000007;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef __int64 ll;
const ll MOD=1000000007;
const int INF=1000000010;
const ll MAX=1ll<<55;
const double eps=1e-8;
const double inf=~0u>>1;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned int uint;
typedef unsigned long long ull;
int k;
int sg[10005];
int a[105];
int flag[105];
int q,m,exm;
void init()
{
    sg[0]=0;
    for(int i=1;i<=10000;i++)
    {
        memset(flag,0,sizeof(flag));
        for(int j=1;j<=k;j++)
        {
            if(i-a[j]>=0)
            {
              flag[sg[i-a[j]]]=1;
            }
        }
        for(int j=0;;j++)
        {
            if(flag[j]==0){
                sg[i]=j;
            break;
                }
        }
    }
}
int main()
{
  while(scanf("%d",&k)!=EOF)
  {
      if(k==0)
        break;
      for(int i=1;i<=k;i++)
        scanf("%d",&a[i]);
       init();
       scanf("%d",&q);
       for(int i=1;i<=q;i++)
       {
           scanf("%d",&m);
           int ans=0;
           for(int j=1;j<=m;j++)
           {
               scanf("%d",&exm);
               ans^=sg[exm];
           }
           if(ans==0)
            printf("L");
           else
            printf("W");
       }
       printf("
");
  }
 return 0;
}