hdu1003最大连续子序列

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>
using namespace std;

int main()
{
    int n,t,i,j,k,first,end,sum,x,max;
    cin>>t;
    for(j=1;j<=t;j++)
    {
        cin>>n;
        sum=0;
        max=-1001;
        first=end=k=1;
        for(i=1;i<=n;i++)
        {
            cin>>x;
            sum+=x;
            if(sum>max)
            {
                max=sum;
                first=k;
                end=i;
            }
            if(sum<0)
            {
                sum=0;
                k=i+1;
            }
        }
         if(j!=1)
         printf("
");
         printf("Case %d:
",j);
         printf("%d %d %d
",max,first,end);

    }
    return 0;
}