Day 31 C. Alternating Subsequence

Problem
Recall that the sequence b is a a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For example, if a=[1,2,1,3,1,2,1], then possible subsequences are: [1,1,1,1], [3] and [1,2,1,3,1,2,1], but not [3,2,3] and [1,1,1,1,2].

You are given a sequence a consisting of n positive and negative elements (there is no zeros in the sequence).

Your task is to choose maximum by size (length) alternating subsequence of the given sequence (i.e. the sign of each next element is the opposite from the sign of the current element, like positive-negative-positive and so on or negative-positive-negative and so on). Among all such subsequences, you have to choose one which has the maximum sum of elements.

In other words, if the maximum length of alternating subsequence is k then your task is to find the maximum sum of elements of some alternating subsequence of length k.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1≤n≤2⋅10^5) — the number of elements in a. The second line of the test case contains n integers a1,a2,…,an (−10^9 ≤ ai ≤ 10^9,ai≠0), where ai is the i-th element of a.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5 (∑n≤2⋅10^5).

Output
For each test case, print the answer — the maximum sum of the maximum by size (length) alternating subsequence of a.

Example
input
4
5
1 2 3 -1 -2
4
-1 -2 -1 -3
10
-2 8 3 8 -4 -15 5 -2 -3 1
6
1 -1000000000 1 -1000000000 1 -1000000000
output
2
-1
6
-2999999997

题目大致意思为：给定一串n个含有负数和正数的数组 然后要分成新的子数组 要求每个数组里元素正负相间 求每个子数组的最大总和

#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<stdio.h>
#include<limits.h>
#include<cmath>
#include<set>
#include<map>
#define ll long long
using namespace std;
int a[200005];
//思路很明确 找每个正数数组或是负数数组的最大值 然后加到一起 
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		memset(a, 0, 200005);
		int n;
		ll sum = 0;
		cin >> n;
		for (int i = 0; i < n; i++)
		{
			cin >> a[i];
		}
		int p = 0;
		long long ans = 0;
		while (p < n)
		{
			int mx = a[p];
			while (p + 1 < n && (ll)a[p] * a[p + 1] > 0)
			{
				p++;
				mx = max(mx, a[p]);
			}
			ans += mx;
			p++;
		}
		cout << ans << endl;
	}
	return 0;
}