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 Rust (1.42.0) (judge verdict: Time Limit Exceeded)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char) 
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn read_vec<T: FromStr>(n: usize) -> Vec<T> {
    return (0..n).map(|_| read()).collect();
}

fn max<T: std::cmp::PartialOrd>(a: T, b: T) -> T {
    if a > b {
        return a;
    } else {
        return b;
    }
}

// const I64_MIN: i64 = -9223372036854775808;
const I64_MIN: i64 = -922337203685477580;

fn main() {
    let n: usize = read();
    let ai: Vec<i64> = read_vec(n);

    let size = n / 2;

    // dp[i番目まで考える][jつすでに利用している][i番目を利用しているflag]
    let mut dp = vec![vec![vec![I64_MIN; 2]; size+1]; n+1];

    for i in 0..n {
        for j in 0..size {
            // k = 0
            dp[i+1][j+1][0] = max(dp[i+1][j+1][0], dp[i][j+1][0]);
            dp[i+1][j+1][0] = max(dp[i+1][j+1][0], dp[i][j+1][1]);

            // k = 1
            if j == 0 {
                dp[i+1][j+1][1] = max(dp[i+1][1][1], ai[i]);  // 1つ前からの遷移
            } else {
                dp[i+1][j+1][1] = max(dp[i+1][j+1][1], dp[i][j][0] + ai[i]);  // 1つ前からの遷移
            }
            for m in 1..4 {  // 2つ前からの遷移
                if i >= m {
                    if j == 0 {
                        dp[i+1][j+1][1] = max(dp[i+1][1][1], ai[i]);
                    } else {
                        dp[i+1][j+1][1] = max(dp[i+1][j+1][1], dp[i-m][j][0] + ai[i]);
                        dp[i+1][j+1][1] = max(dp[i+1][j+1][1], dp[i-m][j][1] + ai[i]);
                    }
                }
            }
            // println!("{} {}", dp[i+1][j+1][0], dp[i+1][j+1][1]);
        }
        // println!("----");
        // println!("{} {}", dp[i+1][size][0], dp[i+1][size][1]);
    }

    let ans = max(dp[n][size][0], dp[n][size][1]);
    println!("{}", ans);
}