| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 775557 | OrazB | Gap (APIO16_gap) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#include "gap.h"
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
//Dijkstra->set
//set.find_by_order(x) x-position value
//set.order_of_key(x) number of strictly less elements don't need *set.??
#define N 100005
#define wr cout << "Continue debugging\n";
#define all(x) (x).begin(), (x).end()
#define ll long long int
#define pii pair <int, int>
#define pb push_back
#define ff first
#define ss second
const ll inf = 1e18;
ll a[N];
vector<ll> vec;
// void rec(ll l, ll r){
// if (l > r) return;
// ll mn, mx;
// MinMax(l, r, &mn, &mx);
// if (mn == -1) return;
// vec.pb(mn);
// if (mn != mx)
// vec.pb(mx);
// ll md = (mx+mn)/2;
// rec(mn+1, md);
// rec(md+1, mx-1);
// }
ll findGap(int t, int n)
{
// if (t == 1){
// int l = 1, r = n;
// ll mn = 0, mx = 1e18, ans = 0;
// while(l <= r){
// ll s = mn, t = mx;
// MinMax(s, t, &mn, &mx);
// a[l] = mn;
// a[r] = mx;
// if (l > 1) ans = max(ans, a[l]-a[l-1]);
// if (r < n) ans = max(ans, a[r+1]-a[r]);
// l++;
// r--;
// mn++;
// mx--;
// }
// if (n == 2) return a[2]-a[1];
// return ans;
// }
ll ans = 0;
ll mn, mx;
MinMax(0, inf, &mn, &mx);
ll l = mn, r = mx;
while (l < r){
if (l+1 == r){
ans = max(ans, 1);
break;
}
ll md = (l+r)/2;
ll mn, mx, mn1, mx1;
MinMax(l, md, &mn, &mx);
mn1 = mn; mx1 = mx;
MinMax(md+1, r, &mn, &mx);
if (mn1 == -1) l = mn, r = mx;
else if (mn == -1) l = mn1, r = mx1;
else{
ans = max(ans, mn-mx1);
if (mx1-mn1 > mx-mn) l = mn1, r = mx1;
else l = mn, r = mx;
}
}
// rec(0, inf);
// sort(all(vec));
// for (int i = 1; i < (int)vec.size(); i++) ans = max(ans, vec[i]-vec[i-1]);
return ans;
}