Given a sequence of K integers { N1, N2, …, N**K }. A continuous subsequence is defined to be { N**i, N**i+1, …, N**j } where 1≤ijK. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

## Input Specification

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

## Output Specification

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

## Sample Input

``````10
-10 1 2 3 4 -5 -23 3 7 -21``````

## Sample Output

``10 1 4``

## 实现思想

temp = temp + v[i]，当temp比sum大，就更新sum的值、leftindex和rightindex的值；当temp < 0，那么后面不管来什么值，都应该舍弃temp < 0前面的内容，因为负数对于总和只可能拉低总和，不可能增加总和，还不如舍弃～

## 代码实现

``````#include <iostream>
#include <vector>

using namespace std;

int main() {
int n;
cin >> n;
vector<int> v(n);
int sum = -1, leftIndex = 0, rightIndex = n - 1, temp = 0, tempIndex = 0;
for(int i = 0; i < n; i++) {
cin >> v[i];
temp = temp + v[i];
if(temp < 0) {
temp = 0;
tempIndex = i + 1;
} else if(temp > sum) {
sum = temp;
leftIndex = tempIndex;
rightIndex = i;
}
}
if(sum < 0) sum = 0;
cout << sum << " " << v[leftIndex] << " " << v[rightIndex];
return 0;
}``````