Max Sum（最大子序和）

Description

Given a sequence a[1],a[2],a[3]......a
, 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}