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 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 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

//前1个数的DP值>=0，就加；DP值<0,就重新开始，自己本身。

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstring>

using namespace std;

int data[],dp[];

int main()

{

    int c,num=,n;

    cin>>c;

    while(c--)

    {

        num++;

        int maxx=-,a=,b=;

        memset(dp,,sizeof(dp));

        cin>>n;

        for(int i=;i<n;i++)

        scanf("%d",&data[i]);

        dp[]=data[];

        for(int i=;i<n;i++)

        {

            if(dp[i-]>=)

            {

                dp[i]=dp[i-]+data[i];

            }

            else

            {

                dp[i]=data[i];

                a=i;

            }

            if(dp[i]>maxx)

            {

                maxx=dp[i];

                b=i;

            }

        }

        printf("Case %d:\n",num);

        printf("%d %d %d\n",maxx,a+,b+);

    }

    return ;

}