UVA10404- Bachet's Game
2014-07-08 09:23
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题意:玩取石子游戏,最后一个取到石子的获胜。题目给出石子总量n,m中取法,每次都是Stan先取,然后求胜者
思路:记录剩余i个石子时的状态,Stan为1时是获胜,0时是失败。
思路:记录剩余i个石子时的状态,Stan为1时是获胜,0时是失败。
<span style="font-size:14px;">#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int MAXN = 1000005; int n, m; int dp[MAXN], arr[MAXN]; int main() { while (scanf("%d %d", &n, &m) != EOF) { for (int i = 0; i < m; i++) scanf("%d", &arr[i]); memset(dp, 0, sizeof(dp)); for (int i = 0; i <= n; i++) for (int j = 0; j < m; j++) { if (i >= arr[j] && dp[i - arr[j]] == 0) { dp[i] = 1; break; } } if (dp ) printf("Stan wins\n"); else printf("Ollie wins\n"); } return 0; }</span>
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