递推dp（uva10404）

最新推荐文章于 2020-01-17 17:44:54 发布

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本文探讨了Bachet's Game的一种变化形式，通过动态规划的方法来确定两个玩家Stan和Ollie在完美策略下谁能赢得游戏。游戏规则允许每次从n块石头中移除特定集合内的任意数量的石头。

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Bachet's Game

Time Limit:6666MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

Submit Status

Description

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：两个人做游戏，有n块石字，m堆，每个人每次只能取m堆数量中的个数，问谁会赢

思路：dp[i]表示还剩i个是S能不能赢，那么当dp[i-a[j]]为偶数，S就可以赢

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=1000100;
int dp[maxn];
int a[maxn];
int N,M;
int main()
{
    while(scanf("%d",&N)!=EOF)
    {
        scanf("%d",&M);
        for(int i=1;i<=M;i++)scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=N;i++)
            for(int j=1;j<=M;j++)
                if(i>=a[j]&&dp[i-a[j]]==0)dp[i]=1;
        if(dp[N])printf("Stan wins\n");
        else printf("Ollie wins\n");
    }
    return 0;
}