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 <iostream>
#ifdef MBP
#include "stdc++.h"
#else
#include <bits/stdc++.h>
#endif

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define REP(i, n) for (long long int i = 0, i##_len = (n); i < i##_len; ++i)
#define REP1(i, n) for (long long int i = 1, i##_len = (n); i <= i##_len; ++i)
#define REP_R(i, n) for (long long int i = (n - 1); i >= 0; --i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((long long int)(x).size())
#define BIT(n) (1LL << (n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

//#define DEBUG

#ifdef DEBUG
#define OUT(x) std::cout << (x) << std::endl
#define OUT2(x, y) std::cout << (x) << "," << (y) << std::endl
#define OUT3(x, y, z) std::cout << (x) << "," << (y) << "," << (z) << std::endl
#define OUTSP(x) std::cout << (x) << " ";
#define OUTLF std::cout << std::endl
#define OUTPA(p) OUT2(p.first, p.second)
#else
#define OUT(x) NULL
#define OUT2(x, y) NULL
#define OUT3(x, y, z) NULL
#define OUTPA(p) NULL
#define OUTSP(p) NULL
#define OUTLF NULL
#endif

#define INF INT_MAX
#define INFLL LLONG_MAX
#define NINF INT_MIN
#define PB push_back
#define MOD 1000000007LL

#define LSB(x) __builtin_ctzll(x)
#define MSB(x) 64LL - __builtin_clzll(x) - 1LL
#define MSBINT(x) 32LL - __builtin_clz(x)
#define BITCNT(x) __builtin_popcountll(x)

using namespace std;
using uif = uint_fast64_t;
using ll = long long int;
using vec = vector<ll>;
using mat = vector<vector<ll>>;
using pa = pair<ll, ll>;
using tTree = __gnu_pbds::tree<ll, __gnu_pbds::null_type, less<ll>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;

template <class T>
inline void input(T &arr)
{
  ll n = arr.size();
  REP(i, n)
  cin >> arr[i];
}
template <class T>
inline void input2(T &arr1, T &arr2)
{
  ll n = arr1.size();
  REP(i, n)
  cin >> arr1[i] >> arr2[i];
}
template <class T>
inline void inputPA(T &arr)
{
  ll n = arr.size();
  REP(i, n)
  cin >> arr[i].first >> arr[i].second;
}
template <class T>
inline void outArr(T &arr)
{
  for (auto x : arr)
  {
    OUTSP(x);
  }
  OUTLF;
}

template <class T>
using PQ = priority_queue<T>;
template <class T>
using PQG = priority_queue<T, vector<T>, greater<T>>;

const ll dx[4] = {1, 0, -1, 0};
const ll dy[4] = {0, 1, 0, -1};
const ll dx8[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy8[8] = {0, 1, 1, 1, 0, -1, -1, -1};

template <class T>
bool chmax(T &, const T &);
template <class T>
bool chmin(T &, const T &);

ll qp(ll a, ll b);
ll qp(ll a, ll b, ll mo);

ll mod_pow(ll, ll);
ll mod_add(ll, ll);
ll mod_mul(ll, ll);
ll mod_div(ll, ll);
void make_fact(ll);
void make_invfact(ll);
void init_fact();

ll combination(ll, ll);

#define FACSIZE 200002LL

ll invfac[FACSIZE];
ll fac[FACSIZE];

/*
i/2 - 1
to
(i+1)/2

i%2==0 i/2-1 to i/2 (i==4, 1,2)

i%2==1 i/2 to (i+1)/2 (i==7, 3,4)

1->i/2

//*/

int main(void)
{
  ll n;
  cin >> n;
  vec A(n);
  input(A);
  ll ans = -INFLL;
  mat DP(3, vec(200020, -INFLL));
  DP[0][0] = 0;

  for (ll i = 1; i <= n; i++)
  {
    ll Ax = A[i - 1];

    if (i == 1)
    {
      DP[2][i] = A[i - 1];
      DP[1][i] = 0;
    }
    else if (i == 2)
    {
      DP[1][i] = max(A[i - 1], A[i - 2]);
      DP[0][i] = 0;
    }
    else
    {
      if (i % 2 == 0)
      {
        DP[1][i] = max(DP[1][i - 2] + Ax, DP[2][i - 1]);
        DP[0][i] = max(DP[0][i - 2] + Ax, DP[1][i - 1]);
      }
      else
      {
        DP[2][i] = DP[2][i - 2] + Ax;
        DP[1][i] = max(DP[1][i - 2] + Ax, DP[1][i - 1]);
        DP[0][i] = max(DP[0][i - 2] + Ax, DP[0][i - 1]);
      }
    }
  }
  if (n % 2 == 0)
  {
    chmax(ans, DP[1][n]);
    chmax(ans, DP[2][n - 1]);
  }
  else
  {
    chmax(ans, DP[1][n]);
    chmax(ans, DP[1][n - 1]);
    chmax(ans, DP[2][n - 2]);
  }

  cout << ans << endl;

  return 0;
}

template <class T>
bool chmax(T &a, const T &b)
{
  if (a < b)
  {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
  if (b < a)
  {
    a = b;
    return 1;
  }
  return 0;
}

ll qp(ll a, ll b)
{
  ll ans = 1LL;
  do
  {
    if (b & 1LL)
      ans = 1LL * mod_mul(ans, a) % MOD;
    a = 1LL * mod_mul(a, a) % MOD;
  } while (b >>= 1LL);
  return ans;
}
ll qp(ll a, ll b, ll mo)
{
  ll ans = 1LL;
  do
  {
    if (b & 1LL)
      ans = 1LL * (ans % mo) * (a % mo);
    a = 1LL * (a % mo) * (a % mo);
  } while (b >>= 1LL);
  return ans;
}

ll mod_pow(ll x, ll r)
{
  if (r == 0)
    return 1;
  else if (r == 1)
    return x % MOD;
  else if (r % 2 == 0)
  {
    ll t = mod_pow(x, r / 2) % MOD;
    return t * t % MOD;
  }
  else
  {
    ll t = mod_pow(x, r / 2) % MOD;
    ll k = t * t % MOD;
    return k * x % MOD;
  }
}
ll mod_add(ll a, ll b)
{
  return (a + b) % MOD;
}

ll mod_mul(ll a, ll b)
{
  return (a * b) % MOD;
}

ll mod_div(ll a, ll b)
{
  return mod_mul(a, mod_pow(b, MOD - 2));
}

void make_fact(ll n)
{
  fac[0] = 1;
  REP(i, n)
  {
    fac[i + 1] = mod_mul(fac[i], i + 1);
  }
}

void make_invfact(ll n)
{
  invfac[n] = mod_pow(fac[n], MOD - 2);
  REP_R(i, n)
  {
    invfac[i] = mod_mul(invfac[i + 1], i + 1);
  }
}

void init_fact()
{
  make_fact(FACSIZE - 1);
  make_invfact(FACSIZE - 1);
}

ll combination(ll n, ll c)
{

  return mod_mul(fac[n], mod_mul(invfac[c], invfac[n - c]));
}
///////////////fact
/*
  init_fact();
  cout << fac[5] << endl;
  cout << invfac[5] << endl;
  cout << combination(8, 4) << endl;
  cout << combination(31, 8) << endl;
  //*/
///////////////////