S-Nim + sg函数+博弈+模板

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2898    Accepted Submission(s): 1288

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 2 532 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
 
#include <cstdio>#include <algorithm>#include <iostream>using namespace std; #define MAX 10010 int sg[MAX]; int num[110]; int v[110]; bool vis[MAX]; int k; void solve() {     int i,j;          sg[0]=0;     for(i=1;i<=10000;i++)     {         memset(vis,0,sizeof(vis));         for(j=1;j<=k;j++)         {             if(i >= num[j])             vis[sg[i-num[j]]] =1;         }         for(j=0;j<=MAX;j++)         {             if(!vis[j])             {                 sg[i]=j;                 break;             }         }     } }  int main() {     while(scanf("%d",&k))     {         if(k==0)         break;         int i,j;         for(i=1;i<=k;i++)         {             scanf("%d",&num[i]);         }         solve();                  int m;         scanf("%d",&m);         for(i=1;i<=m;i++)         {             int n;             scanf("%d",&n);             for(j=1;j<=n;j++)             scanf("%d",&v[j]);             int ans=sg[v[1]];             for(j=2;j<=n;j++)             ans = ans^sg[v[j]];             if(ans)             {                 printf("W");             }             else             {                 printf("L");             }         }         printf("\n");     } return 0;}