hdu1003 Max Sum---最大子段和+记录开始结束点

【hdu1003 Max Sum---最大子段和+记录开始结束点】的更多相关文章

hdu1003 Max Sum---最大子段和+记录开始结束点

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1003 题目大意: 求最大子段和,并且输出最大子段和的起始位置和终止位置. 思路: 根据最大子段和基本方法,直接在线处理(如果对最大子段和不熟悉,点这里) 需要增加两个变量,start和finish,记录最大子段和的起点和终点.如果thissum > maxsum,更新这三个值.如果thissum < 0,设置thissum = 0,并且thisstart = i + 1,因为thissum<…

解题报告：hdu1003 Max Sum - 最大连续区间和 - 计算开头和结尾

2017-09-06 21:32:22 writer:pprp 可以作为一个模板 /* @theme: hdu1003 Max Sum @writer:pprp @end:21:26 @declare:连续区间最大和 @data:2017/9/6 */ #include <bits/stdc++.h> using namespace std; int main() { //freopen("in.txt","r",stdin); int cas; cin…

hdu1003 Max Sum（经典dp ）

A - 最大子段和 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u 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 max sum in…

hdu1003 Max Sum【最大连续子序列之和】

题目链接:https://vjudge.net/problem/HDU-1003 题目大意:给出一段序列,求出最大连续子序列之和,以及给出这段子序列的起点和终点. 解题思路:最长连续子序列之和问题其实有很多种求解方式,这里是用时间复杂度为O(n)的动态规划来求解. 思路很清晰,用dp数组来表示前i项的最大连续子序列之和,如果dp[i-1]>=0的话,则dp[i]加上dp[i-1]能够使dp[i]增大:若dp[i-1]<0的话,则重新以dp[i]为起点,起点更新. #include <cs…

HDU-1003 Max Sum(动态规划，最长字段和问题)

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 193355 Accepted Submission(s): 45045 Problem Description Given a sequence a[1],a[2],a[3]--a[n], your job is to calculate the max sum of a su…

杭电60题--part 1 HDU1003 Max Sum（DP 动态规划）

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…

HDU1003 Max Sum(求最大字段和)

事实上这连续发表的三篇是一模一样的思路,我就厚颜无耻的再发一篇吧! 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003 ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:…

HDU--1003 Max Sum（最大连续子序列和）

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 max sum in this sequence is 6 + (-1) + 5 + 4 = 14. Input The first line of the input contai…

[ACM_动态规划] hdu1003 Max Sum [最大连续子串和]

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 max sum in this sequence is 6 + (-1) + 5 + 4 = 14. Input The first line of the input contai…

ACM学习历程—HDU1003 Max Sum（dp && 最大子序列和）

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 max sum in this sequence is 6 + (-1) + 5 + 4 = 14. Input The first line of the…

最大子序列和——HDU-1003 Max Sum

题目大意:给定序列个数n及n个数,求该序列的最大连续子序列的和,要求输出最大连续子序列的和以及子序列的首位位置解题思路:经典DP,可以定义dp[i]表示以a[i]为结尾的子序列的和的最大值,因而最大连续子序列及为dp数组中的最大值. 状态转移方程:dp[1] = a[1]; //以a[1]为结尾的子序列只有a[1]: i >= 2时, dp[i] = max( dp[i-1]+a[i], a[i] ); dp[i-1]+a[i] > a[i]时,即dp[i-1](以a[i-1]为结尾…

hdu1003 Max Sum(最大子串)

https://vjudge.net/problem/HDU-1003 注意考虑如果全为负的情况,特判. 还有输出格式,最后一个输出不用再空行. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #define lson l, m, rt<<1 #define…

HDU1003 Max Sum

解题思路:最大连续和,此题多了记录的下标,具体见代码. #include<cstdio> #include<algorithm> using namespace std; #define inf 0x3f3f3f3f int main() { , s, e, a; scanf("%d", &t); int tmp = t; while(t--) { scanf("%d", &n); s = e = k = ; sum = ,…

hdu1003 1024 Max Sum&Max Sum Plus Plus【基础dp】

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:…

HDOJ-1003 Max Sum（最大连续子段动态规划）

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HDU 1024 Max Sum Plus Plus【动态规划求最大M子段和详解】

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…

（最大m子段和） Max Sum Plus Plus （Hdu 1024）

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…

HDU 1003：Max Sum（DP，连续子段和）

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 292444 Accepted Submission(s): 69379 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

HDU 1024：Max Sum Plus Plus（DP，最大m子段和）

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…

hdu 1024 Max Sum Plus Plus (子段和最大问题)

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…

最大m段子段和 Day9 - E - 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 consecutive number sequence S 1, S 2, S 3,…

[ACM] hdu 1003 Max Sum（最大子段和模型）

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 135262 Accepted Submission(s): 31311 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

Max Sum—hdu1003（简单DP）标签： dp 2016-05-05 20:51 92人阅读评论(0)

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 207989 Accepted Submission(s): 48681 Problem Description Given a sequence a[1],a[2],a[3]--a[n], your job is to calculate the max sum of a su…