S-Nim

    Problem DescriptionArthur 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.

    InputInput 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.

    OutputFor 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

    这道题的题目挺长的，大致就说，给定一个s集合，每次 行走的步数属于s集合，，，共有n堆石子，问首先出手的能否必胜

    话说看了好久的SG函数，，有一丁点儿领悟~~不敢误导大家~就直接上代码了

    代码：

/*S-NimTime Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4975    Accepted Submission(s): 2141Problem DescriptionArthur 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. InputInput 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. OutputFor 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 Input2 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 OutputLWWWWL */#include#include#include#include#define LEN 110#define MAX 10010using namespace std ;int s[LEN] , p ,sg[MAX];void getSG(int k){	bool hash[MAX] ;	memset(sg,0,sizeof(sg)) ;	for(int i = 0 ; i< MAX ; ++i)	{		memset(hash,false,sizeof(hash)) ;		for(int j = 0 ; j< k ; ++j)		{			if(i-s[j]>=0)			{				hash[sg[i-s[j]]] = true ;			}		}		for(int j = 0 ; j< MAX ; ++j)		{			if(!hash[j])			{				sg[i] = j ;				break;			}		}	}}int main(){	int k;	while(~scanf("%d",&k) && k)	{		for(int i = 0 ; i< k ; ++i)		{			scanf("%d",&s[i]) ;		}		getSG(k);		int m ; 		scanf("%d",&m) ;		string ans ;		while(m--)		{			int temp = 0 , l;			scanf("%d",&l) ;			for(int i = 0 ; i< l ; ++i)			{				scanf("%d",&p) ; 				temp = temp^sg[p] ;			}			if(temp == 0)			{				ans += "L";			}			else			{				ans += "W";			}		}		cout<