Problem:
Score : 600 points

Problem Statement

Given is an integer sequence A_1, ..., A_N of length N.

We will choose exactly \left\lfloor \frac{N}{2} \right\rfloor elements from this sequence so that no two adjacent elements are chosen.

Find the maximum possible sum of the chosen elements.

Here \lfloor x \rfloor denotes the greatest integer not greater than x.

Constraints

2 \leq N \leq 2\times 10^5

|A_i|\leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the maximum possible sum of the chosen elements.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

12

Choosing 2, 4, and 6 makes the sum 12, which is the maximum possible value.

Sample Input 2

5
-1000 -100 -10 0 10

Sample Output 2

0

Choosing -10 and 10 makes the sum 0, which is the maximum possible value.

Sample Input 3

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

5000000000

Watch out for overflow.

Sample Input 4

27
18 -28 18 28 -45 90 -45 23 -53 60 28 -74 -71 35 -26 -62 49 -77 57 24 -70 -93 69 -99 59 57 -49

Sample Output 4

295

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)


#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 5;
ll dp[N][2],arr[N];
void solve()
{
	int n;
	cin >> n;

	for (int i = 1; i < n + 1; i++)
		cin >> arr[i];
	memset(dp, 0, sizeof(dp));
	dp[2][1] = arr[2];
	dp[2][0] = arr[1];
	for (int i = 3; i < n + 1; i++)
	{
		if (i & 1)
		{
			ll temp = max({dp[i-3][1],dp[i-3][0],dp[i-2][1],dp[i-2][0]});
			//i为奇数 i要选x个  x=i/2
			//i-1 位置已经维护好 x个  那么只能从 选了x-1的 地方进行转移，很明显i-2   i-3  是维护好了x-1个
			dp[i][1] = arr[i] + temp;

			dp[i][0]=max({dp[i-1][1],dp[i-1][0],arr[i-2]+dp[i-3][0]});
			//该位置不选，从x继承dp[i-1]
			//从 x-1 的一个位置：i-3不选  选i-2  转移
		}
		else
		{
			dp[i][1]=arr[i]+max({dp[i-2][1],dp[i-2][0],dp[i-1][0]});
			///偶数位置 i  选当前位置 
			///1、要选 i/2个 dp[i-2]的位置已经维护好了（i/2-1）个
			///2、前一个不选
			dp[i][0] = arr[i-1]+dp[i-2][0];
			///i不选，
			///1、一定选i-1  维护前i个必须选i/2个的特性
		}
	}
	cout << max(dp[n][1], dp[n][0]) << endl;
	return;
}

int main()
{
	solve();
	return 0;
}
/*
没有两个相邻的被选上
8
1 2 3 4 -1 -2 -3 -4
*/