Submission #775558

#	Time	Username	Problem	Language	Result	Execution time	Memory
775558		OrazB	Gap (APIO16_gap)	C++17	0 / 100	22 ms	1152 KiB

gap

#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, 1ll);
			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;
}

• Subtask 1: 0 / 30
• Subtask 2: 0 / 70