UVA 10404 Bachet's Game

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本文介绍了一个经典的博弈问题Bachet's Game及其变种，并通过动态规划的方法来解决这一问题。输入包括初始石头数量及每回合可移除的石头数集合，输出谁将获胜。

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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 ofm 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

乍一看是道博弈问题，其实可以转化为dp

从0开始倒推回去，如果当前是先手必败，那么推出去的任何一步都是先手必胜。而只有之前的所有可以推出当前状态的状态都是先手必胜时，那么当前状态时先手必败。

#include<iostream>  
#include<algorithm>
#include<math.h>
#include<cstdio>
#include<string>
#include<string.h>
using namespace std;
const int maxn = 1000010;
int n, m, i, j, f[maxn], a[10];

int main(){
	while (cin >> n >> m)
	{
		memset(f, 0, sizeof(f));
		for (i = 0; i < m; i++) cin >> a[i];
		for (i = 0; i <= n; i++)
		for (j = 0; j < m; j++)
		if (i - a[j] >= 0 && !f[i - a[j]]) f[i] = 1;
		if (f[n]) printf("Stan wins\n"); else printf("Ollie wins\n");
	}
	return 0;
}