hdu S-Nim （sg 函数模板题）

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8528    Accepted Submission(s): 3579

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
题意： （这里的变量 是我代码中的变量 不是题目中的变量 ）给你n个数  表示每一次你可以在某一堆中拿出a[i] 个石子，然后给你a[]   然后给你一个m 表示测试样例 然后给你一个cnt  表示 有cnt堆 石子  然后给你每一堆石子的个数。
思路： 很简单的sg函数的应用。
对于一个递增有界的图G(X, F)来说，SG函数g，是定义在X上的函数，函数值是非负整数，
用语言来描述就是：g(x)的值等于所有x的后继的SG函数中没有出现的最小非负整数。
对于递增有界的图，SG函数是唯一的、有界的。
所有的终止状态x，因为F(x)是空集，所以g(x)=0。  
x 的后继就是 x情况拿一次可以到达的状态。
这里给出  a[]  为 2 5 的情况
x:      0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
sg（x）：0 0 1 1 0 2 1 0 0 1 1  0  2  1  0  0
代码： 
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 10005using namespace std;int sg[N],a[105];int n,m,cnt;int dfs_sg(int x){int &num=sg[x];if(num!=-1) return num;int vis[110];memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++){if(x>=a[i]){if(sg[x-a[i]]!=-1){vis[sg[x-a[i]]]=1;continue;}dfs_sg(x-a[i]);vis[sg[x-a[i]]]=1;}}for(int i=0;;i++){if(!vis[i]){num=i;return num;}}}int main(){int x;while(cin>>n){if(n==0) break;for(int i=1;i<=n;i++) cin>>a[i];memset(sg,-1,sizeof(sg));sort(a+1,a+n+1);sg[0]=0;cin>>m;while(m--){cin>>cnt;int fin=0;while(cnt--){cin>>x;//printf("sg : %d\n",dfs_sg(x));fin^=dfs_sg(x);}if(fin) printf("W");else printf("L");}printf("\n");}return 0;}