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UVa 10404. Bachet's Game

时间:2014-12-04 20:02:40      阅读:246      评论:0      收藏:0      [点我收藏+]

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题意为给出总石子数n,和m堆石子,两个人轮着取石子,每次只能从总石子中取m堆中的一堆的个数的石子,取走最后一个石子的胜。S先取,O后取。


用博弈的输赢观念+dp就可以了,由于记忆化搜索会爆栈,那就递推了,索性还好推……


不然就不会了……

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int dp[1000010];
int a[20];
int n,m;
int main()
{
	int i,j;
	while(cin>>n>>m)
	{
		for(i=0;i<m;i++)
			cin>>a[i];
		for(i=0;i<=n;i++)
		{
			dp[i]=0;
			for(j=0;j<m;j++)
				if(i-a[j]>-1&&dp[i-a[j]]==0)
				{
					dp[i]=1;
					continue;
				}
		}
		if(dp[n]==1)
			cout<<"Stan wins"<<endl;
		else
			cout<<"Ollie wins"<<endl;
	}
}

Time Limit:6666MS   Memory Limit:Unknown   64bit IO Format:%lld & %llu

Status

Description

bubuko.com,布布扣

Problem B: Bachet‘s Game

bubuko.com,布布扣 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

Problem Setter: Piotr Rudnicki

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Mathematics :: Game Theory :: Standard
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 5. Dynamic Programming
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Mathematics :: Game Theory - Standard

Root :: Prominent Problemsetters :: Piotr Rudnicki

UVa 10404. Bachet's Game

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原文地址:http://blog.csdn.net/stl112514/article/details/41728205

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