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>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using dijkstra = priority_queue<T, vector<T>, greater<T>>;

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define fbo find_by_order  // k th index
#define ook order_of_key   // strictly smaller than k
#define PI acos(-1.0)
#define inf 0x3f3f3f3f
#define max_ull 18446744073709551615
#define max_ll 9223372036854775807
#define lead_zero(x) __builtin_clzll(x)
#define trail_zero(x) __builtin_ctzll(x)
#define total_1s(x) __builtin_popcountll(x)
#define first_1(x) __builtin_ffsll(x)
#define log2_(x) __builtin_clz(1) - __builtin_clz(x)
#define isLeap(x) ((x % 400 == 0) || (x % 100 ? x % 4 == 0 : false))
#define QUERY           \
    int test;           \
    scanf("%d", &test); \
    for (int _T = 1; _T <= test; _T++)
#define FAST ios_base::sync_with_stdio(0), cin.tie(0)
#define all(v) v.begin(), v.end()
#define reunique(v) v.resize(std::unique(all(v)) - v.begin())
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pul pair<ull, ull>
#define ff first
#define ss second
#define MERGE(v1, v2, v) merge(all(v1), all(v2), back_inserter(v))
#define MP make_pair
#define EB emplace_back
#define read freopen("input.txt", "r", stdin)
#define write freopen("output.txt", "w", stdout)
#define in_range(v, r, l) upper_bound(all(v), r) - lower_bound(all(v), l)
#define LCM(a, b) (a / __gcd(a, b)) * b;

ll rdn(int y, int m, int d) {
    /* Rata Die day one is 0001-01-01 */
    if (m < 3)
        y--, m += 12;
    return 365 * y + y / 4 - y / 100 + y / 400 + (153 * m - 457) / 5 + d - 306;
}
/* Direction arrays */
/*int dx[] = {1,-1,0,0}, dy[] = {0,0,1,-1};                             */  // 4Direction
/*int dx[] = {1,-1,0,0,1,1,-1,-1}, dy[] = {0,0,1,-1,1,-1,1,-1};         */  // 8Direction
/* int dx[] = {1,-1,1,-1,2,2,-2,-2} , dy[] = {2,2,-2,-2,1,-1,1,-1};     */  // KnightDirection
/* int dx[] = {2,-2,1,1,-1,-1} , dy[] = {0,0,1,-1,1,-1};                */  // HexagonalDirection
/* int day[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; */
const int N = 2e5 + 9;
ll ara[N];
ll dp[N][3];
int visited[N][3];
int n;
int cas;
int st = 1;
int jumpc = 0;
ll recur(int pos, int skipped) {
    if (pos > n)
        return 0;
    if (st == 1 && skipped == 0 && pos == n)
        return 0;
    ll& ans = dp[pos][skipped];
    int& visi = visited[pos][skipped];
    if (visi == cas)
        return ans;
    visi = cas;
    ans = -1e17;
    ans = max(ans, ara[pos] + recur(pos + 2, skipped));
    if (skipped < jumpc) {
        ans = max(ans, recur(pos + 1, skipped + 1));
    }
    return ans;
}
int main() {
    scanf("%d", &n);
    ll sum1 = 0, sum2 = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &ara[i]);
    }

    cas++;
    if(n%2)jumpc = 2;
    else jumpc = 1;
    ll ans = recur(1, 0);
    cas++;
    st++;
    jumpc--;
    ans = max(ans, recur(2, 0));
    cout << ans << "\n";
}