HDU 1024 Max Sum Plus Plus (动态规划) Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given…

A - Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a bra…

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…

Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description:  System Crawler  (2015-09-05) Description Now I think you have got an AC in Ignatius.L's "Max Sum&…

Max Sum Plus Plus     HDU - 1024 Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutiv…

快速上手leetcode动态规划题 我现在是初学的状态,在此来记录我的刷题过程,便于以后复习巩固. 我leetcode从动态规划开始刷,语言用的java. 一.了解动态规划 我上网查了一下动态规划,了解到动态规划是“带有备忘录的递归”, 而大多数用来理解动态规划的例子都是斐波那契数列,就是那个经典的递归式 f(i)=f(i-1)+f(i-2) ,f(1)=f(2)=1 那么我们就可以得到很多式子,比如求f(5): f(5)=f(4)+f(3); f(4)=f(3)+f(2); f(3)=f(2)…

1.Robberies 连接 :http://acm.hdu.edu.cn/showproblem.php?pid=2955     背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和… 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);    正确的方程是:f[j]=max(f[j],f[j-q[i].money]*q[i…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29942    Accepted Submission(s): 10516 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41033    Accepted Submission(s): 14763 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

做水题的感觉真好系列 HDU 2084 数塔 1: 12: 1 23: 1 2 34: 1 2 3 45: 1 2 3 4 5 dp[i][j]第i行第j个数取得的最大值dp[i][j] = max(dp[i-1][j-1], dp[i-1][j]) + a[i][j]. 代码: /********************************************************* Problem : 2084 ( 数塔 ) Judge Status : Accepted RunI…