题目描述 Farmer John's cows like to play coin games so FJ has invented with a new two-player coin game called Xoinc for them. Initially a stack of N (5 <= N <= 2,000) coins sits on the ground; coin i from the top has integer value C_i (1 <= C_i <=…

题目描述 Farmer John's cows like to play coin games so FJ has invented with a new two-player coin game called Xoinc for them. Initially a stack of N (5 <= N <= 2,000) coins sits on the ground; coin i from the top has integer value C_i (1 <= C_i <=…

题目描述 Farmer John's cows like to play coin games so FJ has invented with a new two-player coin game called Xoinc for them. Initially a stack of N (5 <= N <= 2,000) coins sits on the ground; coin i from the top has integer value C_i (1 <= C_i <=…

https://daniu.luogu.org/problemnew/show/P2964 dp[i][j] 表示桌面上还剩i枚硬币时,上一次取走了j个的最大得分 枚举这一次要拿k个,转移到dp[i-k][k] dp[i][j]=max(sum[i]-dp[i-k][k]) 因为 上一次取走j个和取走j-1个 k的取值范围 只相差 2*j-1 和 2*j 所以 直接 dp[i][j]=dp[i][j-1] 然后k分别等于  2*j-1 和 2*j,转移 最后输出dp[n][1],因为先手可以拿1…

题目描述 Farmer John's cows like to play coin games so FJ has invented with a new two-player coin game called Xoinc for them. Initially a stack of N (5 <= N <= 2,000) coins sits on the ground; coin i from the top has integer value C_i (1 <= C_i <=…

题意:n颗硬币 两个人从前往后按顺序拿 如果上一个人拿了i颗 那么下一个可以拿1-2*i颗 问先手能获得的最大收益 题解:比较典型的最大最小最大最小..DP了 但是暴力做的话是n^3 所以就体现出了这个题的巧妙之处  dp[i][j]表示拿到了第i颗上一个人拿了j颗 dp[i][j]由 dp[i + k][k] k = 1,2...2 * j转移来 dp[i][j - 1]由 dp[i + k][k] k = 1,2...2 * (j - 1)转移来 有许多状态是一样的 所以dp[i][j-1]…

LINK : coin game 这道题 超级经典去年这个时候我就看过题目了 但时至今日还不会/cy 觉得在做比赛的题目的时候少写省选的题目 多做水题多做不难也不简单的题目就好了. 由于我是真的不会博弈上dp(其实我博弈都不太会...故写这道题的时候没有过多的思考只是草草想了一波状态就直接看题解了发现状态都列错了. 当我 理解题解中的做法感觉还不是特别的自然故写一篇题解来印证自己的理解. 这里我写上最初始的思路吧 题目中想让我们两个玩家都选择最优的情况下 第一个玩家最多能获得多少的钱.看起来是一…

POJ.1067 取石子游戏 (博弈论 威佐夫博弈) 题意分析 简单的威佐夫博弈 博弈论快速入门 代码总览 #include <cstdio> #include <cmath> #include <algorithm> using namespace std; int main() { int n,m; while(scanf("%d %d",&n,&m) != EOF){ if( n > m) swap(n,m); doubl…

题意:每次可以翻动一个.二个或三个硬币.(Mock Turtles游戏) 初始编号从0开始. 当N==1时,硬币为:正,先手必胜,所以sg[0]=1. 当N==2时,硬币为:反正,先手必赢,先手操作后可能为:反反或正反,方案数为2,所以sg[1]=2. 当N==3时,硬币为:反反正,先手必赢,先手操作后可能为:反反反.反正反.正反正.正正反,方案数为4,所以sg[2]=4. 位置x:0  1  2  3  4   5    6   7    8     9  10  11  12  13  14…

Daizhenyang's Coin Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 183    Accepted Submission(s): 83 Problem Description We know that Daizhenyang is chasing a girlfriend. As we all know, wheneve…