Uva--10404（动规，博弈）

2014-08-02 16:13:18

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 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

思路：很有意思的一道题目，结合了博弈的知识。首先dp[k]表示还剩k个石头的状态，dp[k] == 1表示必胜态，dp[k] == 0表示必败态。
　　由两条博弈准则：
　　1：当前状态为必胜态，当且仅当他的后继状态中至少有一个必败态。
　　2：当前状态为必败态，当且仅当他的后继状态全部都为必胜态。
而这题只用到了 1 准则，dp[0]必为必败态（因为没有剩余的石头可取了），所以可以从dp[0]开始往起点dp[n]推即可。

/*************************************************************************
    > File Name: f.cpp
    > Author: Nature
    > Mail: 564374850@qq.com 
    > Created Time: Wed 30 Jul 2014 04:10:19 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

int n,m,v[15];
int dp[1000005];

int main(){
    while(scanf("%d%d",&n,&m) == 2){
        memset(dp,-1,sizeof(dp));
        for(int i = 0; i < m; ++i)
            scanf("%d",&v[i]);
        for(int i = 0; i <= n; ++i){
            dp[i] = 0;
            for(int j = 0; j < m; ++j){
                if(i - v[j] >= 0 && !dp[i - v[j]]){
                    dp[i] = 1;
                    break;
                }
            }

        }
        if(dp[n])
            printf("Stan wins\n");
        else
            printf("Ollie wins\n");
    }
    return 0;
}