// Om Namah Shivaya
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)
template<typename T>
void amin(T &a, T b) {
a = min(a, b);
}
template<typename T>
void amax(T &a, T b) {
a = max(a, b);
}
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
/*
refs:
https://codeforces.com/blog/entry/68269?#comment-527139
edi
choose l <= a < b < c <= r s.t:
b-a <= c-b
arr[a]+arr[b]+arr[c] is max
let's say we fix (a,b)
when can we consider (a,b) as an option
if there is a guy in between with a higher val, then we dont have to consider pair (a,b)
if there exists k s.t a < k < b with arr[k] >= arr[a] or arr[k] >= arr[b], then we can do this:
arr[k] >= arr[a]: set a = k
arr[k] >= arr[b]: set b = k
by doing this, we reduce the jump distance from a to b
so we have more options for c
in short, consider pair (a,b) only if there is nobody in between with a higher or equal value
how many such pairs are there?
iterate over b from 1 to n and find all valid a
maintain all valid a values when increasing b
once some a becomes bad (some >= guy appears in between), he never becomes good for bigger b
we want to find all a for whom nobody >= them has appeared in [a,b] so far
when moving to a new b, we can make pairs (a,b) for all active a
then we will remove all a values for which arr[a] <= arr[b]
(yet to complete explanation)
*/
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
template<typename T>
struct segtree {
// https://codeforces.com/blog/entry/18051
/*=======================================================*/
struct data {
ll mx1, mx2, res;
};
data neutral = {-inf2, -inf2, -inf2};
data merge(data &left, data &right) {
data curr;
curr.mx1 = max(left.mx1,right.mx1);
curr.mx2 = max(left.mx2,right.mx2);
curr.res = max({left.res, right.res, left.mx2 + right.mx1});
return curr;
}
void create(int i, T v) {
tr[i].mx1 = v;
}
void modify(int i, T v) {
amax(tr[i].mx2, v);
tr[i].res = tr[i].mx1 + tr[i].mx2;
}
/*=======================================================*/
int n;
vector<data> tr;
segtree() {
}
segtree(int siz) {
init(siz);
}
void init(int siz) {
n = siz;
tr.assign(2 * n, neutral);
}
void build(vector<T> &a, int siz) {
rep(i, siz) create(i + n, a[i]);
rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
}
void pupd(int i, T v) {
modify(i + n, v);
for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
}
data query(int l, int r) {
data resl = neutral, resr = neutral;
for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
if (l & 1) resl = merge(resl, tr[l++]);
if (!(r & 1)) resr = merge(tr[r--], resr);
}
return merge(resl, resr);
}
};
void solve(int test_case)
{
ll n; cin >> n;
vector<ll> arr(n+5);
rep1(i,n) cin >> arr[i];
vector<ll> good;
vector<pll> pairs;
rep1(b,n){
while(!good.empty() and arr[b] >= arr[good.back()]){
pairs.pb({good.back(), b});
good.pop_back();
}
if(!good.empty()){
pairs.pb({good.back(), b});
}
good.pb(b);
}
vector<ll> enter[n+5];
for(auto [a,b] : pairs){
enter[a].pb(b);
}
ll q; cin >> q;
vector<pll> queries[n+5];
rep1(i,q){
ll l,r; cin >> l >> r;
queries[l].pb({r, i});
}
vector<ll> ans(q+5);
segtree<ll> st(n+5);
st.build(arr,n+1);
rev(l,n,1){
trav(b,enter[l]){
ll dis = b - l;
ll c = b + dis;
st.pupd(c, arr[l] + arr[b]);
}
for(auto [r, id] : queries[l]){
ans[id] = st.query(l,r).res;
}
}
rep1(i,q) cout << ans[i] << endl;
}
int main()
{
fastio;
int t = 1;
// cin >> t;
rep1(i, t) {
solve(i);
}
return 0;
}
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
1 ms |
212 KB |
Output is correct |
| 2 |
Runtime error |
1 ms |
468 KB |
Execution killed with signal 6 |
| 3 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
1 ms |
212 KB |
Output is correct |
| 2 |
Runtime error |
1 ms |
468 KB |
Execution killed with signal 6 |
| 3 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
97 ms |
34336 KB |
Output is correct |
| 2 |
Correct |
57 ms |
30052 KB |
Output is correct |
| 3 |
Correct |
61 ms |
31544 KB |
Output is correct |
| 4 |
Correct |
91 ms |
34428 KB |
Output is correct |
| 5 |
Correct |
97 ms |
34424 KB |
Output is correct |
| 6 |
Incorrect |
91 ms |
34356 KB |
Output isn't correct |
| 7 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
1 ms |
212 KB |
Output is correct |
| 2 |
Runtime error |
1 ms |
468 KB |
Execution killed with signal 6 |
| 3 |
Halted |
0 ms |
0 KB |
- |
