hdu 1536 OR poj 2960 S-Nim 博弈论，，求出SG'函数就可以解决

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4975    Accepted Submission(s): 2141

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

 

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 <cstdio>#include <string>#include <cstring>#include <iostream>#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<<ans<<endl ;}return 0 ;}