hdu 1858 Max Partial Value I

Problem Description
HenryFour has a number of stones which have different values from -4444 to 4444. He puts N stones in a line and wants to find the max partial value of these N stones.

Assume the values of the N stones in line are: v1, v2, v3, v4, ..., vN. The partial vaule of stones from Lth stone to Rth stone (1 ≤ L ≤ R ≤ N) is the sum of all the stones between them. i.e. PartialV(L, R) = v[L] + v[L+1] + .... + v[R] (1 ≤ L ≤ R ≤ N)

Since the number of stones (N) is very very large, it is quite difficult for HenryFour to find the max partial value. So could you develop a programme to find out the answer for him?


Input
There are several test cases in the input data. The first line contains a positive integer T (1 ≤ T ≤ 14), specifying the number ot test cases. Then there are T lines. Each of these T lines contains a positive number N followed by N integers which indicate the values of the N stones in line.
1 ≤ N ≤ 1,000,000
-4444 ≤ v[i] ≤ 4444


Output
Your program is to write to standard output. For each test case, print one line with three numbers seperated by one blank: P L R. P is the max partial value of the N stones in line. L and R indicate the position of the partial stones. If there are several Ls and Rs that have the same value PartialV(Li, Ri) = P, please output the minimum pair. For pair (Li, Ri) and (Lj, Rj), we define (Li, Ri) < (Lj, Rj) if and only if: Li < Lj or (Li == Lj and Ri < Rj)


Sample Input
3
4 32 -39 -30 -28
8 1 2 3 -10 1 -1 5 1
10 14 -12 -8 -13 3 5 42 -24 -32 -12


Sample Output
32 1 1
6 1 3

50 5 7

让求连续子序列的最大和，并输出相应的起始位置和结束位置；

代码：

#include <iostream>
using namespace std; 

int main()
{  int n,m,i,j,k,y;
   long long max,p1,p2,now,temp;
     cin>>n;
     while(n--)
     {
     	cin>>m>>temp;
     	now=max=temp;
     	y=k=p1=p2=1;
     	for(i=2;i<=m;i++)
     	{
     		cin>>temp;
     		if(temp>now+temp)
     		{
     			now=temp;k=i;//记录起始位置 
     		}
     		else 
            {
			   now+=temp;y=i;//记录结束位置 
            }
            if(now>max)
            { p1=k;p2=y;max=now;//最大值放到max中，p1为起始位，p2位结束位置； 
            }
           
     	}
     	cout<<max<<" "<<p1<<" "<<p2<<endl;
     }
     return 0;
}