【经典DP类型】 最大连续子序列和

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

34 32 -39 -30 -288 1 2 3 -10 1 -1 5 110 14 -12 -8 -13 3 5 42 -24 -32 -12 

 

Sample Output

32 1 16 1 350 5 7 

Hint

 Huge input and output,scanf and printf are recommended. 

大致题意：找出子串中和最大的，并且标注区间， 套路题；

AC代码：

#include <iostream>using namespace std;int main(){    int n,m,i,l ,r ,y;    long long max,p1,p2,now,temp;    cin>>n;    while(n--)    {        cin>>m>>temp;        now=max=temp;        r=l=p1=p2=1;        for(i=2; i<=m; i++)        {            cin>>temp;            if(temp>now+temp)            {                now=temp;                p1=i;            }            else            {                now+=temp;            }            if(now>max)            {                l=p1;                r=i;                max=now;            }        }        cout<<max<<" "<<l<<" "<<r<<endl;    }    return 0;}