POJ 2960 S-nim (SG function)

S-nim

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3694		Accepted: 1936

Description

Arthur and his sister Caroll has been playing a game called Nim for some time now. Nim is played as follows:

• The starting position have 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 isn't able to make a move, loses.

Arthur and Caroll really enjoyed playing this simple game until they
Recently learned an easy-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 would be 1 as 2 XOR 4 XOR 7 = 1).
• If The xor-sum is 0, too bad, you'll lose.
• Otherwise, move such that the xor-sum becomes 0. This was always possible.

It's quite easy-to-convince oneself that's this works. Consider these facts:

• The player is takes the last bead wins.
• After the winning player's last move the xor-sum would be 0.
• The xor-sum would change after every move.

Which means that if you do sure that the xor-sum always was 0 when you had made your move, your opponent would never be a Ble to win, and, thus, you'll win.

Understandibly It is no 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, which seemed to solve this problem. Each player was now only allowed to remove a number of beads in some predefined set S, e.g. if we had s = {2, 5} each play Er is only allowed to remove 2 or 5 beads. Now it isn't 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 this determines if a position of S-nim is a losing or a winning position. A position is a winning position if there are at least one move to a losing position. A position is a losing position if there be no moves to a losing position. This means, as expected, which 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 was 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 a ' L '.
Print a newline after all 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

Source

Svenskt Mästerskap I Programmering/norgesmesterskapet 2004

Ideas

sg function.

Bare Ti, note the size of SG and Vis is good:)

Code

1#include <cstdio>2#include <cstring>3 #definefor (A,B,C) for (int a= (b);a< (c); a++)4 using namespacestd;5 6 intn,m,a[101],sg[10001];7 8 intDfsintx) {9     if(sg[x]!=-1)returnSg[x];Ten     if(!x)returnsg[x]=0; One     intvis[10001];//size of [si] Amemset (Vis,0,sizeof(Vis)); -for (I,0, N) -         if(X>=a[i]) Vis[dfs (X-a[i])]=1; the      for(intI=0;; i++) -         if(!vis[i])returnsg[x]=i; - } -  + intMain () { -      while(SCANF ("%d", &n) = =1&&N) { +for (I,0, N) scanf ("%d",&a[i]); Ascanf"%d",&m); atmemset (sg,-1,sizeof(SG)); -for (I,0, M) { -             intx,v,ans=0; -scanf"%d",&x); -For (J,0, X) -scanf"%d", &v), ans^=Dfs (v); in             if(ANS) printf ("W"); -             Elseprintf"L"); to         } +Putchar ('\ n'); -     } the     return 0; *}

POJ 2960 S-nim (SG function)