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

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;

}

巴特西

HDU-1858-Max Partial Value I,有坑点，不难；

Max Partial Value I

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