思路: dp,用记忆化搜索比较好实现. 实现: class Solution { public: int dfs(vector<int>& sum, int cur, int M, vector<vector<int>>& dp) { int n = sum.size(); * M) ] - sum[cur]; ) return dp[cur][M]; int ans = INT_MIN; ; i <= min( * M, n - cur); i+…

Stone Game II Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 609    Accepted Submission(s): 350 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4388 Descripton: Stone Game II comes. It needs two…

Stone Game II comes. It needs two players to play this game. There are some piles of stones on the desk at the beginning. Two players move the stones in turn. At each step of the game the player should do the following operations.   First, choose a p…

Stone Game II HDU - 4388 题目大意: 给出n堆物品,每堆物品都有若干件,现在A和B进行游戏,每人每轮操作一次,按照如下规则: 1. 任意选择一个堆,假设该堆有x个物品,从中选择k个,要保证0<k<x且0<(x^k)<k. 2. 再增加一个大小为x^k的堆(也就相当于将一个x个物品的堆变成一个k个物品的堆和一个x^k个物品的堆),另外有一个技能,可以将这个大小为x^k的堆变成(2*k)^x的堆,但是这个技能每个人只有一次机会可以使用. 现在问两人轮流操作,都采…

Stone Game II Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 531    Accepted Submission(s): 300 Problem Description Stone Game II comes. It needs two players to play this game. There are some p…

Last Stone Weight II 欢迎关注H寻梦人公众号 You are given an array of integers stones where stones[i] is the weight of the ith stone. We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have…

原题链接在这里:https://leetcode.com/problems/last-stone-weight-ii/ 题目: We have a collection of rocks, each rock has a positive integer weight. Each turn, we choose any two rocks and smash them together.  Suppose the stones have weights x and y with x <= y. …

原题链接在这里:https://leetcode.com/problems/stone-game-ii/ 题目: Alex and Lee continue their games with piles of stones.  There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].  The objective of the gam…

Description There is a stone game.At the beginning of the game the player picks n piles of stones in a circle. The goal is to merge the stones in one pile observing the following rules: At each step of the game,the player can merge two adjacent piles…

不管是否使用技能,发现操作前后所有堆二进制中1的个数之和不变.那么对于一个堆其实可以等价转换为一个k个石子的堆(k为该数二进制的个数),然后就是个nim游戏. 1 #include<bits/stdc++.h> 2 using namespace std; 3 int t,n,ans,a[101]; 4 int main(){ 5 scanf("%d",&t); 6 for(int ii=1;ii<=t;ii++){ 7 scanf("%d"…

http://acm.hdu.edu.cn/showproblem.php?pid=4388 http://blog.csdn.net/y1196645376/article/details/52143551 好久没有写题了,再这么颓下去就要被彻底踩爆了(已经被彻底踩爆了). 这道题是一道博弈论,从侧面向我们揭示了一个客观规律,取东西的博弈论(不是定理的话)大多数都是从二进制入手(虽然这道题题目很显然是和二进制有关)进行一系列的找规律. 这道题的正解也同样给了我们一种看题的思路,从最基本的条件看…

有一堆石头,每块石头的重量都是正整数. 每一回合,从中选出任意两块石头,然后将它们一起粉碎.假设石头的重量分别为 x 和 y,且 x <= y.那么粉碎的可能结果如下: 如果 x == y,那么两块石头都会被完全粉碎: 如果 x != y,那么重量为 x 的石头将会完全粉碎,而重量为 y 的石头新重量为 y-x. 最后,最多只会剩下一块石头.返回此石头最小的可能重量.如果没有石头剩下,就返回 0. 示例: 输入:[2,7,4,1,8,1] 输出:1 解释: 组合 2 和 4,得到 2,所以数组转…

2020-01-11 17:47:59 问题描述: 问题求解: 本题和另一题target sum非常类似.target sum的要求是在一个数组中随机添加正负号,使得最终得到的结果是target,这个题目被证明和背包问题是同一个问题,只是需要进行一下转化. 本题其实也是一个套壳题目,只是这次的壳套的更加隐蔽,对于本题来说,其实核心就是将其划分成两个堆,我们需要的是两堆的diff最小. 那么就需要另一个技巧了,我们不会直接使用dp去求这个最小值,而是使用dp去判断对于小堆中的和的可能性,最后在遍历…

1049. Last Stone Weight II https://leetcode.com/problems/last-stone-weight-ii/ 题意:从一堆石头里任选两个石头s1,s2,若s1==s2,则两个石头都被销毁,否则加入s1<s2,剩下一块重量为s2-s1的石头.重复上面的操作,直至只剩一块石头,或没有石头.问最后剩下的石头的重量最小为多少(0表示没有剩余石头). 解法: 每次选两块石头进行相减问最后一块石头重量最小为多少,可以看作是将所有石头分为两波,使两波石头的差值的…

原题链接在这里:https://leetcode.com/problems/last-stone-weight/ 题目: We have a collection of rocks, each rock has a positive integer weight. Each turn, we choose the two heaviest rocks and smash them together.  Suppose the stones have weights x and y with x…

原题链接在这里:https://leetcode.com/problems/stone-game/ 题目: Alex and Lee play a game with piles of stones.  There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i]. The objective of the game is to…

在介绍SG函数和SG定理之前我们先介绍介绍必胜点与必败点吧. 必胜点和必败点的概念:        P点:必败点,换而言之,就是谁处于此位置,则在双方操作正确的情况下必败.        N点:必胜点,处于此情况下,双方操作均正确的情况下必胜. 必胜点和必败点的性质:         1.所有终结点是 必败点 P .(我们以此为基本前提进行推理,换句话说,我们以此为假设)         2.从任何必胜点N 操作,至少有一种方式可以进入必败点 P.         3.无论如何操作,必败点P 都…

在介绍SG函数和SG定理之前我们先介绍介绍必胜点与必败点吧. 必胜点和必败点的概念:        P点:必败点,换而言之,就是谁处于此位置,则在双方操作正确的情况下必败.        N点:必胜点,处于此情况下,双方操作均正确的情况下必胜. 必胜点和必败点的性质:         1.所有终结点是 必败点 P .(我们以此为基本前提进行推理,换句话说,我们以此为假设)         2.从任何必胜点N 操作,至少有一种方式可以进入必败点 P.         3.无论如何操作,必败点P 都…

在介绍SG函数和SG定理之前我们先介绍介绍必胜点与必败点吧. 必胜点和必败点的概念:        P点:必败点,换而言之,就是谁处于此位置,则在双方操作正确的情况下必败.        N点:必胜点,处于此情况下,双方操作均正确的情况下必胜. 必胜点和必败点的性质:         1.所有终结点是 必败点 P .(我们以此为基本前提进行推理,换句话说,我们以此为假设)         2.从任何必胜点N 操作,至少有一种方式可以进入必败点 P.         3.无论如何操作,必败点P 都…

今天的比赛的题目相对来说比较「直白」,不像前几周都是一些特定的算法,如果你没学过不可能想出来. 做了这些周,对leetcode比赛的题目也发现了一些「规律」. 一般前两道题都很「简单」,只要有想法,直接敲代码就能解出来.更多考察的是结果是否正确,速度其次. 后两道题有些难度 ,不同场次难度不一样,也可能和不同人的水平感受不同.但是肯定比前两道要难. 一般在做后两道题的时候,只要复杂度是对的,一些细节也不用考虑太多.例如数组开的空间大小,一些线性的提前剪枝判断,写不写都可以过.最主要的是复杂度是同…

2019-09-07 16:34:48 877. Stone Game 问题描述: 问题求解: 典型的博弈问题,也是一个典型的min-max问题.通常使用算diff的方法把min-max转为求max. dp[i][j] : i ~ j 玩家 A 和 玩家 B 得分的diff. public boolean stoneGame(int[] piles) { int n = piles.length; int[][] dp = new int[n][n]; return helper(piles,…

1046. Last Stone Weight We have a collection of rocks, each rock has a positive integer weight. Each turn, we choose the two heaviest rocks and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is: I…

Hello everyone, I am a Chinese noob programmer. I have practiced questions on leetcode.com for 2 years. During this time, I studied a lot from many Great Gods' articles. After worship, I always wanted to write an article as they did, and now I take t…

2901: G-险恶逃生II 时间限制: 1 Sec  内存限制: 128 MB 提交: 44  解决: 14 题目描述     SOS!!!koha is trapped in the dangerous maze.He need your help again.     The maze is a 2D grid consisting of n rows and m columns. Each cell in the maze may have a stone or may be occup…

Problem 1538 - B - Stones IITime Limit: 1000MS Memory Limit: 65536KB Total Submit: 416 Accepted: 63 Special Judge: No DescriptionXiaoming took the flight MH370 on March 8, 2014 to China to take the ACM contest in WHU. Unfortunately, when the airplane…

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 592 Accepted Submission(s): 341 Problem Description After he has learned how to play Nim game, Mike begins to try another stone game which seems muc…

题目链接:Path Sum II | LeetCode OJ Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2],…

题目链接:Populating Next Right Pointers in Each Node II | LeetCode OJ Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only u…

原文标题:Functional Android (II): Collection operations in Kotlin 原文链接:http://antonioleiva.com/collection-operations-kotlin/ 原文作者:Antonio Leiva(http://antonioleiva.com/about/) 原文发布:2015-09-29 在简化代码方面,Lambda表达式是一个杰出的工具,而且还可以完成之前不可能完成的事.我们在这个系列文章的第一篇(Unlea…

统计分析中Type I Error与Type II Error的区别 在统计分析中,经常提到Type I Error和Type II Error.他们的基本概念是什么?有什么区别? 下面的表格显示 between truth/falseness of the null hypothesis and outcomes of the test " -------|-------|------- | Judgement of Null Hypothesis H0 | Valid | Invalid |…