uva 10404 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

題目大意：給出石頭的總數sum， 和取石頭的方法（一次能取多少石頭）n，接下來是n種取石頭的方式。在雙方都想贏的狀況下，問誰能最後拿走全部石頭（先手還是後手）。

解題思路：主體是完全背包的思路。dp數組有{0， 1}兩種狀況，1代表當前i個石頭先取者必勝，0代表當前i個石頭先取者必敗，狀態轉移方程是：dp[i?num[j]]==0??>dp[i]=1。

#include 
#include 
#include 
#include 
#include 
#define N 1000005
#define M 15
using namespace std;
typedef long long ll;
int num[M], dp[N];
int main() {
    int sum, n; 
    while (scanf("%d %d", &sum, &n) == 2) {
        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < n; i++) {
            scanf("%d", &num[i]);
        }
        for (int i = 1; i <= sum; i++) {
            for (int j = 0; j < n; j++) {
                if (i - num[j] >= 0 && !dp[i - num[j]]) { //該取石頭的方式可行 且 取完石頭之後剩余的石頭不能一次取完。
                    dp[i] = 1;
                    break;
                }
            }
        }
        if (dp[sum]) printf("Stan wins\n");
        else printf("Ollie wins\n");
    }
    return 0;
}