S-Nim (hdu1536 + SG function)

S-Nim (hdu1536 + SG function)
S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 5317 Accepted Submission (s): 2288

 

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 alway S 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 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 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120

Sample Output

LWWWWL

 

 

Enter K to indicate the size of a set, and then enter a set to indicate that only the number of elements in the set can be used for the stones.

Enter m to ask about this set.

Then input n in each row in m rows, indicating that there are n heaps, and each Heap has n1 stones. Ask whether the State indicated by this row is "win" or "lose". If you win, enter W. Otherwise, L

Idea: we can divide n heap stones into n games and then combine n games.

 

Reprinted, please specify the Source: search for & STAR kids

 

Question link: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 1536

 

# Include
 
  
# Include
  
   
# Includeusing namespace std; // note that the S array should be sorted in ascending order. The SG function must be initialized to-1. For each set, only one time is required. // n is the size of the Set s. S [I] is defined. array int s [110] of special fetch rules, sg [10010], n; int SG_dfs (int x) {int I; if (sg [x]! =-1) return sg [x]; bool vis [110]; memset (vis, 0, sizeof (vis); for (I = 0; I
   
    
= S [I]) {SG_dfs (x-s [I]); vis [sg [x-s [I] = 1 ;}} int e; for (I = 0; I ++) if (! Vis [I]) {e = I; break;} return sg [x] = e;} int main () {int I, m, t, num; while (scanf (% d, & n) {for (I = 0; I