1989번: 부분배열 고르기 2 ↗

/*
[1989: 부분배열 고르기 2](https://www.acmicpc.net/problem/1989)

Tier: Platinum 5 
Category: data_structures, segtree, prefix_sum, divide_and_conquer, stack
*/


#include <bits/stdc++.h>

using namespace std;

#define for1(s, e) for(int i = s; i < e; i++)
#define for1j(s, e) for(int j = s; j < e; j++)
#define forEach(k) for(auto i : k)
#define forEachj(k) for(auto j : k)
#define sz(vct) vct.size()
#define all(vct) vct.begin(), vct.end()
#define sortv(vct) sort(vct.begin(), vct.end())
#define uniq(vct) sort(all(vct));vct.erase(unique(all(vct)), vct.end())
#define fi first
#define se second
#define INF (1ll << 60ll)

typedef unsigned long long ull;
typedef long long ll;
typedef ll llint;
typedef unsigned int uint;
typedef unsigned long long int ull;
typedef ull ullint;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<double, int> pdi;
typedef pair<string, string> pss;

typedef vector<int> iv1;
typedef vector<iv1> iv2;
typedef vector<ll> llv1;
typedef vector<llv1> llv2;

typedef vector<pii> piiv1;
typedef vector<piiv1> piiv2;
typedef vector<pll> pllv1;
typedef vector<pllv1> pllv2;
typedef vector<pdd> pddv1;
typedef vector<pddv1> pddv2;

const double EPS = 1e-8;
const double PI = acos(-1);

template<typename T>
T sq(T x) { return x * x; }

int sign(ll x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(int x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(double x) { return abs(x) < EPS ? 0 : x < 0 ? -1 : 1; }

struct Node {
  ll sm;
  ll mn;
};

struct SegTree {
  ll n;
  llv1 ar;
  vector<Node> tree;

  SegTree(ll n) {
    this->n = n;
    ar.resize(n + 1);
    tree.resize(4 * n + 5);
    ar[0] = INF;
  }

  void init() {
    init(1, n, 1);
  }

  Node init(ll left, ll right, ll idx) {
    if(left == right) return tree[idx] = { ar[left], left };

    ll mid = (left + right) / 2;

    Node l_ret = init(left, mid, idx * 2);
    Node r_ret = init(mid + 1, right, idx * 2 + 1);

    return tree[idx] = {
      l_ret.sm + r_ret.sm,
      (ar[l_ret.mn] <= ar[r_ret.mn] ? l_ret.mn : r_ret.mn),
    };
  }

  Node query(ll left, ll right, ll idx, ll q_start, ll q_end) {
    if(right < q_start || q_end < left) return { 0, 0 };

    if(q_start <= left && right <= q_end) {
      return tree[idx];
    }

    ll mid = (left + right) / 2;
    Node l_ret = query(left, mid, idx * 2, q_start, q_end);
    Node r_ret = query(mid + 1, right, idx * 2 + 1, q_start, q_end);

    return {
      l_ret.sm + r_ret.sm,
      (ar[l_ret.mn] <= ar[r_ret.mn] ? l_ret.mn : r_ret.mn)
    };
  }

  Node query(ll q_start, ll q_end) {
    return query(1, n, 1, q_start, q_end);
  }

  pair<ll, pair<ll, ll>> maxArea(ll q_start, ll q_end) {
    if(q_start > q_end) return { 0, { q_start, q_end }};
    if(q_start == q_end) {
      return { ar[q_start] * ar[q_start], { q_start, q_end } };
    }

    Node query_ret = query(q_start, q_end);
    ll min_idx = query_ret.mn;
    ll min_value = ar[min_idx];
    ll sm_value = query_ret.sm;

    pair<ll, pair<ll, ll>> ret = { 
      sm_value * min_value,
      { q_start, q_end }
    };

    pair<ll, pair<ll, ll>> left_ret = maxArea(q_start, min_idx - 1);
    pair<ll, pair<ll, ll>> right_ret = maxArea(min_idx + 1, q_end);

    if (left_ret.first > ret.first) {
      ret = left_ret;
    }

    if (right_ret.first > ret.first) {
      ret = right_ret;
    }
    
    return ret;
  }
};

void solve() {
  ll N;
  cin >> N;
  SegTree tree(N);

  for1(1, N + 1) {
    cin >> tree.ar[i];
  }

  tree.init();

  pair<ll, pair<ll, ll>> ans = tree.maxArea(1, N);

  cout << ans.first << "\n";
  cout << ans.second.first << " " << ans.second.second;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);cout.tie(NULL);
  int tc = 1; // cin >> tc;
  while(tc--) solve();
  
}