SG 函数 S-Nim
http://poj.org/problem?id=2960
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 3464 | Accepted: 1829 |
Description
- 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
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
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

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 105
#define M 10005 int s[N], sn;
int sg[M]; void getsg(int n)
{
int mk[M];
sg[] = ;//主要是让终止状态的sg为0
memset(mk, -, sizeof(mk));
for(int i = ; i < M; i++)//预处理sg函数
{
for(int j = ; j < n && s[j] <= i; j++)
mk[sg[i-s[j]]]=i;//将所有后继的sg标记为i,然后找到后继的sg没有出现过的最小正整数
//优化:注意这儿是标记成了i,刚开始标记成了1,这样每次需初始化mk memset,而标记成i就不需要了
int j = ;
while(mk[j] == i) j++;
sg[i] = j;
}
} int main()
{
while(~scanf("%d", &sn), sn)
{
for(int i = ; i < sn; i++) scanf("%d", &s[i]);
sort(s, s+sn);//排序算一个优化,求sg的时候会用到
getsg(sn);
int m;
scanf("%d", &m);
char ans[N];
for(int c = ; c < m; c++)
{
int n, tm;
scanf("%d", &n);
int res = ;
for(int i = ; i < n; i++)
{
scanf("%d", &tm);
res ^= sg[tm];
}
if(res == ) ans[c] = 'L';
else ans[c] = 'W';
}
ans[m]=;
printf("%s\n", ans);
}
return ;
}
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