Max Sum（最大子序和）

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 <stdio.h> 
using namespace std;
int main()
{
    int T,c=1;
    scanf("%d",&T);
    while(T--)
    {
        int N,i,sum=0,max=-1001,t=1,start,end,n;
        cin>>N;
        for(i=1;i<=N;i++)
        {
            scanf("%d",&n);
            sum=sum+n;
            if(sum>max)
            {
                max=sum;
                start=t;
                end=i;
            }
            if(sum<0)
            {
                sum=0;t=i+1;
            }
        }
        cout<<"Case "<<c++<<":"<<endl;
        cout<<max<<" "<<start<<" "<<end<<endl;
        if(T!=0)
            cout<<endl;
    }
32}