HDU-1858-Max Partial Value I,有坑点，不难；

Max Partial Value I

Time Limit: 1000/5000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)

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

记得以前做过一样的题，而且还水过了，但在这里WA了一整天，整个人都不好了，如果要求最大和的话分分钟水过，但又要求把下标求出，，，”
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) “，千万要注意这点，是要在最大和相同的前提下把最靠近左边的输出，如 0 1，输出应该是 1 1 2；而不是1 2 2；

解决了这个问题还要注意用long long ,不过目测不会超int啊，奇怪；

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;
const int N=1000000+10;
int a[N];
int main()
{
    int t,n,i,j,k;
    long long maxsum,x;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        x=0;
        maxsum=-10000000;
        j=k=1;
        int j1=1,k1=1;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(x>=0)
            {
                 x+=a[i];
                   k1=i;
            }
            else
            {
                x=a[i];
                j1=k1=i;
            }
           if(x>maxsum)
           {
              maxsum=x;
              j=j1;
              k=k1;//核心也就这么一点，连动规都不算；
           }
        }
        printf("%I64d %d %d\n",maxsum,j,k);
    }
    return 0;
}