InputThe 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).
OutputFor 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 Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6
題意：找出最大的連續字串和；

#include<iostream>
#include<map>
using namespace std;
int main()
{
    int t,mark=0,cnt=0;
    cin >> t;
    while (t--)
    {
        if (cnt) cout << endl;
        cnt = 1;
        mark
 
++;
        int temp = 1, frist, end;
        int n,ko,sum=0,max=-100000;
        cin >> n;
        for (int i = 0; i < n; i++)
        {
            cin >> ko;
            sum += ko;
            if (sum > max)
            {
                max = sum; frist = temp; end = i + 1;
            }
            
 
if (sum < 0)
            {
                sum = 0; temp = i + 2;
            }
        }
        cout << "Case " << mark << ":" <<endl<< max << " " << frist << " " << end << endl;
    }
    return 0;
}

註意：因為要輸出下標，必須靈活應用temp這個中間值，剛開始因為一直沒仔細考慮下標的情況，導致負數情況下不能出正確答案；

通過temp的加入後，可以在大於max的條件滿足下更改最後的下標，小於零的情況下可以更改temp，到累計大於max時，直接改frist；

（經典DP題）

Max Sum （dp）