问题描述 我们可爱的KK有一个有趣的数学游戏:这个游戏需要两个人,有N\left(1\leq N\leq 5*{10}^{4} \right)N(1≤N≤5∗10​4​​)个数,每次KK都会先拿数.每次可以拿任意多个数,直到NN个数被拿完.每次获得的得分为取的数中的最小值,KK和对手的策略都是尽可能使得自己的得分减去对手的得分更大.在这样的情况下,最终KK的得分减去对手的得分会是多少? 输入描述 第一行一个数T\left( 1\leq T\leq 10\right)T(1≤T≤10),表示数据组…

KK's Number 题目链接: http://acm.hust.edu.cn/vjudge/contest/121332#problem/K Description Our lovely KK has a funny mathematical game:This game requires two people,There are numbers,every time KK will take the numbers,first.Every time you can take any num…

Play Game Alice and Bob are playing a game. There are two piles of cards. There are N cards in each pile, and each card has a score. They take turns to pick up the top or bottom card from either pile, and the score of the card will be added to his to…

Catching the Thief Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 653    Accepted Submission(s): 359 Problem Description In the Qingshui Village, there's a clever thief and a cleverer police. …

Problem D Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 7 Accepted Submission(s) : 3 Problem Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1…

本来博弈还没怎么搞懂,又和dp搞上了,哇,这真是冰火两重天,爽哉妙哉. 我自己的理解就是,博弈dp有点像对抗搜索的意思,但并不是对抗搜索,因为它是像博弈一样,大多数以当前的操作者来dp,光想是想不通的,上题练一练. POJ - 1678 I Love this Game! 题目大意:有两个人正在玩游戏,在给出的一堆数中,玩家1先在[a,b]的范围中挑选出一个x1,然后玩家2再挑选有个y1满足a≤y1-x1≤b,然后玩家1再挑选一个x2满足a≤x2-y1≤b,以此轮流操作,直到某个玩家不能再进行操…

Wireless Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5640    Accepted Submission(s): 1785 Problem Description Liyuan lives in a old apartment. One day, he suddenly found that there…

A Task Process Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1332    Accepted Submission(s): 656 Problem Description There are two kinds of tasks, namely A and B. There are N workers and the…

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面 方程: a[j]直接接在前面…

HDU 1231 题目大意以及解题思路见: HDU 1003题解,此题和HDU 1003只是记录的信息不同,处理完全相同. /* HDU 1231 最大连续子序列 --- 入门DP */ #include <cstdio> #include <cstring> ]; int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #endif int n; int maxSum…