题解是看的这里的: http://www.acmerblog.com/hdu-1024-Max-Sum-Plus-Plus-1276.html 当前这个状态是dp[i][j],i 表示当前的段,j表示前j个数组成了当前的这i个段的最大值,而且a[j]在最后一个段中 状态dp[i][j]可以从dp[i][j-1]转移过来,表示第j个数字正好可以和 i 个段的前j-1个数字相加的和是当前所在状态中最大的 状态dp[i][j]可以从dp[i-1][j-1]转移过来,表示第 j 个数字正好可以成为第 i…

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 consecutive number sequence S 1, S 2, S 3,…

给定n个数字,求其中m段的最大值(段与段之间不用连续,但是一段中要连续) 例如:2 5 1 -2 2 3 -1五个数字中选2个,选择1和2 3这两段. dp[i][j]从前j个数字中选择i段,然后根据第j个数字是否独立成一段,可以写出 状态转移方程:dp[i][j]=max(dp[i][j-1]+num[j],max(dp[i-1][k])+num[j]) 这里的max(dp[i-1][k])代表的拥有i-1段时的最大值,然后再加上num[j]独立成的一段. 但是题目中没有给出m的取值范围,有可…

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…

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…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=1024     Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23844    Accepted Submission(s): 8143 Problem Description Now I think you hav…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35988    Accepted Submission(s): 12807 Problem Description Now I think you ha…

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: 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…

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…

Max Sum Plus Plus Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1024 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 mor…

扫描线求周长: hdu1828 Picture(线段树+扫描线+矩形周长) 参考链接:https://blog.csdn.net/konghhhhh/java/article/details/78236036 假想有一条扫描线,从左往右(从右往左),或者从下往上(从上往下)扫描过整个多边形(或者说畸形..多个矩形叠加后的那个图形).如果是竖直方向上扫描,则是离散化横坐标,如果是水平方向上扫描,则是离散化纵坐标.下面的分析都是离散化横坐标的,并且从下往上扫描的. 扫描之前还需要做一个工作,就是保存…

http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Problem 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 fac…

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 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23697    Accepted Submission(s): 8094 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

子串和再续 时间限制:1000 ms  |  内存限制:65535 KB 难度:4 描述 给你一个序列 S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). 我们定义 sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).现在给你一个 m(8>m>0&&m<n)你的任务是计算 sum(i1, j1) + sum(i2, j2) + sum(i3,…

Problem 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 consecutive number sequ…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 42478    Accepted Submission(s): 15335 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): 25639    Accepted Submission(s): 8884 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

传送门:Max Sum Plus Plus 题意:从n个数中选出m段不相交的连续子段,求这个和最大. 分析:经典dp,dp[i][j][0]表示不取第i个数且前i个数分成j段达到的最优值,dp[i][j][1]表示取了第i个数且前i个数分成j段达到的最优值. 那么有: dp[i][j][0]=max(dp[i-1][j][0],dp[i-1][j][1]). dp[i][j][1]=max(dp[i-1][j-1][0]+a[i],max(dp[i-1][j-1][1],dp[i][j][1])…

题目: Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 27299    Accepted Submission(s): 9499 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" prob…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 36673    Accepted Submission(s): 13069 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" proble…

操作1 的时候标记deng[rt]表示以下一段数都是与当前节点的值同样 下次操作2时直接对有deng标记的节点gcd更新 (可能还能够更简单) #include <stdio.h> #include <string.h> #include <stdlib.h> #include <limits.h> #include <malloc.h> #include <ctype.h> #include <math.h> #incl…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17336    Accepted Submission(s): 5701 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

注:摘的老师写的 最大m子段和问题 以-1 4 -2 3 -2 3为例最大子段和是:6最大2子段和是:4+(3-2+3)=8所以,最大子段和和最大m子段和不一样,不能用比如先求一个最大子段和,从序列中去掉已求子段,再求下一个最大子段和的方法,这种方法有点贪心的味道,但是不行.所以,还得用动态规划. 1.基本思路:  首先,定义数组num[n],dp[m][n].   num[n]用来存储n个整数组成的序列. dp[m][n]代表n个整数序列求m个子段和. 按照分阶段的思想,我们首先考虑最后一项,…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44371    Accepted Submission(s): 16084 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1024 Problem 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 mor…

转载请注明出处,谢谢:http://www.cnblogs.com/KirisameMarisa/p/4302208.html   ---by 墨染之樱花 dp是竞赛中常见的问题,也是我的弱项orz,更要多加练习.看到邝巨巨的dp专题练习第一道是Max Sum Plus Plus,所以我顺便把之前做过的hdu1003 Max Sum拿出来又做了一遍 HDU 1003 Max Sum 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003 题目描述:…

虽然这道题看起来和 HDU 1024  Max Sum Plus Plus 看起来很像,可是感觉这道题比1024要简单一些 前面WA了几次,因为我开始把dp[22][maxn]写成dp[maxn][22]了,Orz 看来数组越界不一定会导致程序崩溃,也有可能返回一个错误的结果 dp[i][j]表示前j个数构成前i段所得到的最大值 状态转移方程: dp[i][j] = max{dp[i][j-1],  dp[i-1][j-len[i]] + sum[j] - sum[j-len[i]]} 分别对应…