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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)


import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {
    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();
        final List<Long> list = IntStream.range(0, n)
                .mapToObj(i -> scanner.nextLong())
                .collect(Collectors.toList());

        if (n % 2 == 0) {
            long sumOfEvens = 0;
            long sumOfOdds = 0;
            for (int i = 0; i < list.size(); i++) {
                if (i % 2 == 0) {
                    sumOfEvens += list.get(i);
                } else {
                    sumOfOdds += list.get(i);
                }
            }
            System.out.println(max(sumOfEvens, sumOfOdds));
            return;
        } else if (n == 3) {
            System.out.println(list.stream().max(Long::compareTo).orElseThrow());
            return;
        }

        final long[][] dpTable = new long[n][3];
        dpTable[0][0] = list.get(0);
        dpTable[1][1] = list.get(1);
        dpTable[2][2] = list.get(2);
        for (int i = 2; i < n; i++) {
            final long tmp = list.get(i);
            dpTable[i][0] = dpTable[i - 2][0] + tmp;
            if (i >= 3) {
                dpTable[i][1] = max(dpTable[i - 2][1], dpTable[i - 3][0]) + tmp;
            }
            if (i >= 4) {
                dpTable[i][2] = max(dpTable[i - 2][2], dpTable[i - 3][1], dpTable[i - 4][0]) + tmp;
            }
        }

        System.out.println(max(dpTable[n - 1][2], dpTable[n - 2][1], dpTable[n - 3][0]));
    }

    private static long max(long... numbers) {
        return Arrays.stream(numbers).max().orElseThrow();
    }
}