Max Sum Plus Plus (动态规划) HDU1024】的更多相关文章

题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=1024 (http://www.fjutacm.com/Problem.jsp?pid=1375) 题意:长度为n的序列里,m段不相关区间的最大和 思路:我们先要确定一个东西,就是状态,这里我用dp[i][j]表示前j个数在取a[j]情况下分i段的最大和: 那么我们为了找规律,可以先来一发Excel,就以样例为例子: 然后我们可以发现其实红圈里的8是状态dp[2][6](i=2, j=6),那么我们可…
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…
Max Sum Plus Plus Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22262    Accepted Submission(s): 7484   Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. T…
Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37418    Accepted Submission(s): 13363 Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem.…
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…
最近想学DP,锻炼思维,记录一下自己踩到的坑,来写一波详细的结题报告,持续更新. 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the…
题目链接:https://cn.vjudge.net/problem/HDU-1024 题意 给n, m和一个序列,找m个不重叠子串,使这几个子串内元素和的和最大. n<=1e6 例:1 3 1 2 3 答:6 (唯一的子串1 2 3) 思路 先顺便记录一下动态规划的一般解题思路: 原问题->子问题->状态->转移->边界 再顺便记录一下最大值最小化这类问题套路解法: 二分 贪心 不能二分的问题,贪心八九不离十. 一般是AB和BA这两个元素的顺序,不影响前后变化时,直接算目标…
http://acm.hdu.edu.cn/showproblem.php?pid=1003 / 给组测试数据 1 7 2 3 -4 -5 6 7 8 一个关键问题 : 什么时候将开始位置重新赋值 即当连续序列和小于等于零时 / include int main() { int t; scanf("%d",&t); int k=1; while(t--) { int n; scanf("%d",&n); int m,max1=-0x3f3f3f3f,…
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=1003 给出一个包含n个数字的序列{a1,a2,..,ai,..,an},-1000<=ai<=1000 求最大连续子段和及其起始位置和终止位置,很基础的动态规划(DP)问题,看完DP第一次做的DP题目 DP真的是一种很优美的算法,或者说思想,但是比较难理解,我对DP的理解还很浅薄 # include <stdio.h> # define INF 1000000000 int main() { i…