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.*;

public class Main {
    void run() {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] aArr = new int[n];

        for (int i = 0; i < n; i++) {
            aArr[i] = sc.nextInt();
        }

        long[][][] dp = new long[n + 1][3][2];
        for (long[][] x : dp) {
            for (long[] x2 : x) {
                Arrays.fill(x2, Integer.MIN_VALUE);
            }
        }
        // x o x o x o x
        // o x x x o x o
        dp[0][0][0] = 0;


        // dp[i][j][k] ... i番目までの整数列で、ひとつおきじゃないやつj個、(直前の数を使っていなければk = 0)の最大値
//        debug(dp);
        for (int i = 0; i < n; i++) {
            dp[i + 1][0][1] = dp[i][0][0] + hoge(dp[i][0][0], aArr[i]);
            dp[i + 1][1][1] = dp[i][1][0] + hoge(dp[i][1][0], aArr[i]);
            dp[i + 1][2][1] = dp[i][2][0] + hoge(dp[i][2][0], aArr[i]);
            dp[i + 1][0][0] = dp[i][0][1];
            dp[i + 1][1][0] = Math.max(dp[i][0][0], dp[i][1][1]);
            dp[i + 1][2][0] = Math.max(dp[i][1][0], dp[i][2][1]);
        }

        List<Long> list;
        if (n % 2 == 0) {
            // o x o x o x
            // x o x o x o
            // o x x o x o
            list = Arrays.asList(dp[n][0][0], dp[n][1][0],
                    dp[n][0][1], dp[n][1][1]);
        } else {
            // x o x o x o x
            // o x x x o x o
            // x x o x o x o
            list = Arrays.asList(dp[n][0][0], dp[n][1][0], dp[n][2][0],
                    dp[n][0][1], dp[n][1][1], dp[n][2][1]);

        }
        System.out.println(Collections.max(list));
    }

    long hoge(long i, long j) {
        return i == Integer.MIN_VALUE ? 0 : j;
    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}