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 |
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
Problem Setter: Piotr Rudnicki
Source
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