POJ2960 S-Nim （sg博弈模板）

S-Nim

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4655		Accepted: 2438

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 5
3
2 5 12
3 2 4 7
4 2 3 7 12
5 1 2 3 4 5
3
2 5 12
3 2 4 7
4 2 3 7 12
0

Sample Output

LWW
WWL

Source

Svenskt Mästerskap i Programmering/Norgesmesterskapet 2004
题目链接：http://poj.org/problem?id=2960

【题意】有k中取珠子的方法，每次只能够取si个，输入一个m，表示测试的组数，输入l，表示当前有多少堆，hi表示当前第i堆有多少个石子。当前选手不能按照si中的数进行取珠子则失败。或者为W，失败为L，最后输出m组的结果字符串。

【思路】sg函数打表算出每个点的sg值，异或求解即可。

【代码如下】

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;

int k,s[110],m,l,x,sg[11000],mex[11000];

void getSG(){
int cnt = 1;
memset(mex,0,sizeof(mex));
for(int i = 1; i <= 10000; i ++){
for(int j = 0; j < k && i >= s[j]; j ++){
mex[sg[i - s[j]]] = cnt; //避免多次memset()，节省时间
}
for(int j = 0; ; j ++){
if(mex[j] != cnt){
sg[i] = j; break;
}
}
cnt ++;
}
}

int main(){
while(~scanf("%d",&k) && k){
memset(sg,0,sizeof(sg));
for(int i = 0; i < k; i ++) scanf("%d",&s[i]);
sort(s,s+k);
getSG();
scanf("%d",&m);
vector<char>vct;
for(int i = 0; i < m; i ++){
scanf("%d",&l);
int ans = 0;
while(l --){
scanf("%d",&x);
ans ^= sg[x];
}
if(ans) vct.push_back('W');
else vct.push_back('L');
}
for(int i = 0; i < vct.size(); i ++){
printf("%c",vct[i]);
}
printf("\n");
}
return 0;
}