S-Nim

 

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：

　　给你k个数 s[i]

　　再给你m个询问，每次询问是一个nim游戏，但是相比nim不同的是，每次只能从各个堆中选取 s[i]的值除去

题解：

　　SG函数的应用

　　对于给定的k个数，我们预处理出sg[i]

　　那么就简单了

　　

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define ls i<<1#define rs ls | 1#define mid ((ll+rr)>>1)#define pii pair
           
            #define MP make_pairtypedef long long LL;const long long INF = 1e18+1LL;const double Pi = acos(-1.0);const int N = 5e5+10, M = 2e5+20, mod = 1e9+7, inf = 2e9;int k,sg[N],s[N],vis[N];char A[N];int main() { while(scanf("%d",&k)!=EOF) { if(k == 0) break; for(int i = 1; i <= k; ++i) scanf("%d",&s[i]); sg[0] = 0; for(int i = 1; i <= 10000; ++i) { for(int j = 0; j <= 100; ++j) vis[j] = 0; for(int j = 1; j <= k; ++j) { if(i >= s[j] && sg[i - s[j]] <= 100) vis[sg[i - s[j]]] = 1; } for(int j = 0; j <= 100; ++j) { if(!vis[j]) { sg[i] = j; break; } } } int q,cnt = 0; scanf("%d",&q); while(q--) { int x,y,ans = 0; scanf("%d",&x); while(x--) { scanf("%d",&y); ans ^= sg[y]; } if(ans) printf("W"); else printf("L"); } printf("\n"); } return 0;}