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
乍一看是道博弈問題，其實可以轉化為dp
從0開始倒推回去，如果目前是先手必敗，那麼推出去的任何一步都是先手必勝。而隻有之前的所有可以推出目前狀态的狀态都是先手必勝時，那麼目前狀态時先手必敗。
#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;
}