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
题目大意：给出石头的总数sum， 和取石头的方法（一次能取多少石头）n，，接下来是n种取石头的方式。在双方都想赢的状况下，问谁能最后拿走全部石头（先手还是后手）。解题思路：主体是完全背包的思路。dp数组有{0， 1}两种状况，1代表当前i个石头先取者必胜，0代表当前i个石头先取者必败，状态转移方程是：。;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]]) { //该取石头的方式可行 且 取完石头之后剩余的石头不能一次取完。dp[i] = 1;break;}}}if (dp[sum]) printf(“Stan wins\n”);else printf(“Ollie wins\n”);}return 0;}
把自己当傻瓜，不懂就问，你会学的更多
