　　　　　　　　　　　　Max Sum

　　　　　　　　　　Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　　Total Submission(s): 294096 Accepted Submission(s): 69830

Problem 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

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

int num[100086];

int main()

{

    int T;

    scanf("%d",&T);

    int cases=0;

    while(T--){

        cases++;

        int n;

        scanf("%d",&n);

        for(int i=1;i<=n;i++){

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

        }

        int l=1,r=1;

        int x,y;

        int maxx=-999999999,sum=0;

        for(int i=1;i<=n;i++){

            sum+=num[i];

            if(sum>maxx){

                x=l;y=i;

                maxx=sum;

            }

            if(sum<0){sum=0;l=i+1;}

        }

        if(cases!=1){printf("\n");}

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

        printf("%d %d %d\n",maxx,x,y);

    }

}

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