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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

UVA 10404 Bachet s Game(dp + 博弈?)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 winsn");
        else
            printf("Ollie winsn");
    }
    return 0;
}

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