1003

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

这里就是要考虑全为负数的情况

#include <iostream>
using namespace std;
int main()
{
	int n,i=0,m,j;
	int* arry;
	int start,end,sum,maxs,q;
	cin>>n;
	while (i<n)
	{
		cin>>m;
		arry=new int[m];
		j=0;
		while (j<m)
		{
			cin>>arry[j];
			++j;
		}
		int flag=0;
		j=0;
		maxs=arry[0];
		start=0;
		while(j<m)
		{
			if (arry[j]>0)
				flag=1;
			else if (arry[j]>maxs)
			{
				maxs=arry[j];
				start=j;
			}
			j++;
		}
		if (!flag)
		{
			cout<<"Case "<<i+1<<":"<<endl;
			cout<<maxs<<" "<<start+1<<" "<<start+1<<endl;
		}
		else
		{
			q=maxs=sum=start=end=j=0;
			while (j<m)
			{
				int test=sum+arry[j];
				if (sum+arry[j]<0)
				{
					q=j+1;
					sum=0;
				}
				else if (sum+arry[j]<maxs)
					sum+=arry[j];
				else
				{
					start=q;
					end=j;
					maxs=sum+arry[j];
					sum=maxs;
				}
				++j;
			}
			cout<<"Case "<<i+1<<":"<<endl;
			cout<<maxs<<" "<<start+1<<" "<<end+1<<endl;
		}
		++i;
		if (i<n)
			cout<<endl;
		delete []arry;
	}
	return 0;
}