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…

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…

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…

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…

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…

  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…

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:…

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): 21363    Accepted Submission(s): 7144 Problem Description Now I think you have go…

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…