J - Max Sum

J - Max Sum

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status

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>
#define  MAX 100005


using namespace std;


int a[MAX];


int main()
{
	int T,N,t=0,sum[MAX],s[MAX],ans,tt;
	cin>>T;
	tt=T;
	while(tt--)
	{
		cin>>N;
		for(int i=0;i<N;i++)
			cin>>a[i];
		sum[0]=a[0];
		s[0]=0;
		ans=0;
		for(int i=1;i<N;i++)
		{
			if(sum[i-1]>=0)
			{
				sum[i]=sum[i-1]+a[i];
				s[i]=s[i-1];
			}
			
			else
			{
				sum[i]=a[i];
				s[i]=i;
			}
			 if(sum[ans]<sum[i])
			 	ans=i;
		}
		
		printf("Case %d:\n%d %d %d\n",++t,sum[ans],s[ans]+1,ans+1);
		if(t!=T)
			cout<<endl;
		
	}
	return 0;
}