#636 (Div. 3)C. Alternating Subsequence

题目分析

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≤104) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1≤n≤2⋅105) — the number of elements in a. The second line of the test case contains n integers a1,a2,…,an (−109≤ai≤109,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⋅105 (∑n≤2⋅105).

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

Note

In the first test case of the example, one of the possible answers is [1,2,3–,−1–––,−2].
In the second test case of the example, one of the possible answers is [−1,−2,−1–––,−3].
In the third test case of the example, one of the possible answers is [−2–––,8,3,8–,−4–––,−15,5–,−2–––,−3,1–].
In the fourth test case of the example, one of the possible answers is [1–,−1000000000,1,-1000000000,1,−1000000000].

题目大意

找到最长的+ - + - + - + - + -（+ -交替即可）的序列，求最长序列的最大值。

这道题本来做出来了，结果到最后的时候被Hacked了。。。。。

题目分析

这道题好像没什么好说的，就是找出每一段全正或全负的子序列最大值作为最长序列的数即可，可以用双指针算法。

ac代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
using namespace std;
const int N=2e5+5;
int a[N];
bool check(int x,int y)
{
	if((x>0&&y>0)||(x<0&&y<0)) return true;
	else return false;
}
int main()
{
    int t;
	cin>>t;
	while(t--)
	{
		int n;
		cin>>n;
		for(int i=0;i<n;i++)
		cin>>a[i];
		
		int i=0,j=0;
		LL sum=0;     //结果可能爆int
		while(i<n)
		{
			j=i;
			int num=a[j];
			while(j+1<n&&check(a[j],a[j+1]))//判断a[j]和a[j+1]是否同号
			{//注意要判断j是否越界，我就是因为没加这个判断被Hacked了
				num=max(num,a[j+1]);   //找出全正或全负子学列的最大值
				j++;
			}
			sum+=num;  //加上每段的最大值
			i=j+1;
		}
		cout<<sum<<endl;
	}
    return 0;
}