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.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = Integer.parseInt(scan.next());
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(scan.next());
        }
        scan.close();

        // 配列
        long[] even = new long[(n + 1) / 2];
        long[] odd = new long[n / 2];
        for (int i = 0; i < even.length; i++) {
            even[i] = a[2 * i];
        }
        for (int i = 0; i < odd.length; i++) {
            odd[i] = a[2 * i + 1];
        }
        // 累積和
        long[] evenSum = new long[(n + 1) / 2 + 1];
        long[] oddSum = new long[n / 2 + 1];
        for (int i = 1; i < evenSum.length; i++) {
            evenSum[i] = evenSum[i - 1] + even[i - 1];
        }
        for (int i = 1; i < oddSum.length; i++) {
            oddSum[i] = oddSum[i - 1] + odd[i - 1];
        }

        long sum = Long.MIN_VALUE;
        if (n % 2 == 0) {
            sum = Math.max(evenSum[(n + 1) / 2], oddSum[n / 2]);
        } else {
            // 偶数のみ
            for (int i = 0; i < even.length; i++) {
                long value = evenSum[n / 2 + 1] - even[i];
                if (sum < value) {
                    sum = value;
                }
            }

            // 奇数あり,偶数から
            for (int i = 0; i < even.length; i++) {
                for (int j = i; j < even.length; j++) {
                    long value = evenSum[i] - evenSum[0] + oddSum[j] - oddSum[i] + evenSum[even.length] - evenSum[j + 1];
                    if (sum < value) {
                        sum = value;
                    }
                }
            }

            // 奇数あり,奇数から
            for (int i = 0; i < odd.length; i++) {
                for (int j = i; j < odd.length; j++) {
                    long value = oddSum[i] - oddSum[0] + evenSum[j] - evenSum[i] + oddSum[odd.length] - oddSum[j + 1];
                    if (sum < value) {
                        sum = value;
                    }
                }
            }
        }

        System.out.println(sum);
    }

}