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：给定n个石头，和m种去除石头的方式，每种方式可以去除一定量的石头， 现在Stan(简称S),Ollie(简称O)，S先手，O后手，每次每个人能选择一种去除石头的方式，谁去除最后一堆谁就赢了。要求出必胜之人是谁。
思路：一开始没头绪，以为是博弈。没想出好的思路。由于n很大，去遍历状态肯定超时。之后看了别人的题解，才发现不错的dp思路:用一个dp数组记录，对于先手者能取到的记录为1，后手者为0，初始都为0，遍历1到n，如果dp[i]为0，说明上一手是后手取得，这样先手就能取，把dp[i]变为1，由于是从1 到 n，这样每个状态记录时，前面的都已经记录好了，所以是可行的。这样最后只需要判断dp[n]是1，还是0，就可以判断是先手胜还是后手胜了。
状态转移方程为：if (i - move[j] >= 0 && !dp[i - move[j]]) dp[i] = 1。
代码：
#include <stdio.h>
#include <string.h>
int n, m, move[15], dp[1000005], 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 wins
");
else
printf("Ollie wins
");
}
return 0;
}