Max Sum

一、题目

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 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).
Output
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.
Sample
Inputcopy Outputcopy
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Case 1:
14 1 4

Case 2:
7 1 6

二、分析

计算连续数列的最大和，使用贪心算法。
如果sum小于0了，下一段sum重新开始，更新左端点

#include<iostream>
using namespace std;
int main()
{int T;cin>>T;for(int k=1;k<=T;k++){int n;cin>>n;int a[n+2];for(int i=1;i<=n;i++) cin>>a[i];int dp[n+2]={0};int st=1;int ansl=-1,ansr=-1,mx=-1e9,sum=0;//mx往开到极小for(int i=1;i<=n;i++){sum+=a[i];if(sum>mx){mx=sum;ansl=st;ansr=i;}if(sum<0){sum=0;st=i+1;}}cout << "Case " << k << ":" << endl;cout << mx << " " << ansl << " " << ansr <<endl<<endl;}
}