UVA 10404 Bachet's Game(dp + 博弈?)
http://www.21tx.com 2014年01月20日 CSDN 帆帆帆帆帆帆帆帆帆帆

## 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```

```#include <stdio.h>
#include <string.h>

int n, m, move, dp, i, j;

int main() {
while (~scanf("%d", &n)) {
memset(dp, 0, sizeof(dp));
scanf("%d", &m);
for (i = 0; i < m; i ++) {
scanf("%d", &move[i]);
}
for (i = 1; i <= n; i ++)
for (j = 0; j < m; j ++) {
if (i - move[j] >= 0 && !dp[i - move[j]]) {
dp[i] = 1;
break;
}
}
if (dp[n])
printf("Stan winsn");
else
printf("Ollie winsn");
}
return 0;
}```