标签：acm dp
Max Sum
Time Limit: 2000ms Memory limit: 32768K 有疑问？点这里^_^
题目描述
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.
输入
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
输出
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position
of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
示例输入
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
示例输出
Case 1:
14 1 4
Case 2:
7 1 6
提示
hdoj1003 注意：本题后台测试数据量比较大，请使用C语言scanf输入。避免超时！！！！
又是一道很经典的dp，题意是让求最大连续字段和 不难写出状态转移方程 dp[i]=max(dp[i-1]+i,i);
HDU--Max sum---DP练习,布布扣,bubuko.com
HDU--Max sum---DP练习
标签：acm dp
原文地址：http://blog.csdn.net/qq_16255321/article/details/37325513
