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.

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).

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.

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

Ignatius.L
分析：这是一道动态规划题，刚开始做的时候想了好久，第一次接触感觉真的好难，提交了好多次，，，终于过了

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int test;
    int n;
    int a[100050];
    cin>>test;
    int k=1;
    while(test--)
    {
        cin>>n;
        for(int i=0;i<n;i++)
            cin>>a[i];
        int max=a[0],now=a[0];
        int p1=0,p2=0;
       int x=0;
        for(int i=1;i<n;i++)
        {

            if(now+a[i]<a[i])   
            {
                x=i;
                now=a[i];
            }
            else
                now=now+a[i];
            if(max<now)
            {
                p1=x;
                max=now;
                p2=i;
            }
        }
        printf("Case %d:\n",k++);
        printf("%d %d %d\n",max,p1+1,p2+1);
        if(test)
            printf("\n");
    }
    return 0;
}