UVA 10404 Bachet's Game(dp + 博弈?)

Problem B: Bachet's Game

Bachet's game is probably known to all but probably not by this name. Initially there are  n  stones on the table. There are two players Stan and Ollie, who move alternately. Stan always starts. The legal moves consist in removing at least one but not more than  k  stones from the table. The winner is the one to take the last stone.

Here we consider a variation of this game. The number of stones that can be removed in a single move must be a member of a certain set of m numbers. Among the m numbers there is always 1 and thus the game never stalls.

Input

The input consists of a number of lines. Each line describes one game by a sequence of positive numbers. The first number is  n  <= 1000000 the number of stones on the table; the second number is  m  <= 10 giving the number of numbers that follow; the last  m  numbers on the line specify how many stones can be removed from the table in a single move.

Input

For each line of input, output one line saying either  Stan wins  or  Ollie wins  assuming that both of them play perfectly.

Sample input

20 3 1 3 8
21 3 1 3 8
22 3 1 3 8
23 3 1 3 8
1000000 10 1 23 38 11 7 5 4 8 3 13
999996 10 1 23 38 11 7 5 4 8 3 13

Output for sample input

Stan wins
Stan wins
Ollie wins
Stan wins
Stan wins
Ollie wins

题意：给定n个石头，和m种去除石头的方式，每种方式可以去除一定量的石头， 现在Stan(简称S),Ollie(简称O)，S先手，O后手，每次每个人能选择一种去除石头的方式，谁去除最后一堆谁就赢了。要求出必胜之人是谁。

思路：一开始没头绪，以为是博弈。没想出好的思路。由于n很大，去遍历状态肯定超时。之后看了别人的题解，才发现不错的dp思路:用一个dp数组记录，对于先手者能取到的记录为1，后手者为0，初始都为0，遍历1到n，如果dp[i]为0，说明上一手是后手取得，这样先手就能取，把dp[i]变为1，由于是从1 到 n，这样每个状态记录时，前面的都已经记录好了，所以是可行的。这样最后只需要判断dp[n]是1，还是0，就可以判断是先手胜还是后手胜了。

状态转移方程为：if (i - move[j] >= 0 && !dp[i - move[j]])  dp[i] = 1。

代码：

#include <stdio.h>
#include <string.h>

int n, m, move[15], dp[1000005], i, j;

int main() {
	while (~scanf("%d", &n)) {
		memset(dp, 0, sizeof(dp));
		scanf("%d", &m);
		for (i = 0; i < m; i ++) {
			scanf("%d", &move[i]);
		}
		for (i = 1; i <= n; i ++)
			for (j = 0; j < m; j ++) {
				if (i - move[j] >= 0 && !dp[i - move[j]]) {
					dp[i] = 1;
					break;
				}
			}
		if (dp[n])
			printf("Stan wins\n");
		else
			printf("Ollie wins\n");
	}
	return 0;
}