Submission #1293553

#TimeUsernameProblemLanguageResultExecution timeMemory
1293553harryleeeGap (APIO16_gap)C++20
100 / 100
40 ms3236 KiB
#include "gap.h"
#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;
typedef tree<long long, null_type, less<long long>, rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_set;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_multiset;
#define ll long long
#define ld long double
#define iloop(m, h) for (auto i = m; i != h; i += (m < h ? 1 : -1))
#define jloop(m, h) for (auto j = m; j != h; j += (m < h ? 1 : -1))
#define kloop(m, h) for (auto k = m; k != h; k += (m < h ? 1 : -1))
#define lloop(m, h) for (auto l = m; l != h; l += (m < h ? 1 : -1))
#define iloop_(m, h, g) for (auto i = m; i < h; i += g)
#define jloop_(m, h, g) for (auto j = m; j < h; j += g)
#define kloop_(m, h, g) for (auto k = m; k < h; k += g)
#define lloop_(m, h, g) for (auto l = m; l < h; l += g)
#define getchar_unlocked _getchar_nolock // comment before submission
#define pll pair<ll, ll>
#define plll pair<ll, pll>
#define pllll pair<pll, pll>
#define vll vector<ll>
#define qll queue<ll>
#define dll deque<ll>
#define pqll priority_queue<ll>
#define gll greater<ll>
#define INF 1000000000000000
#define MOD1 1000000007
#define MOD2 998244353
#define MOD3 1000000009
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
long long findGap(int T, int N) {
	if (T == 1) {
		ll a[N];
		iloop(0, (N+1)/2) {
			if (!i) MinMax(0, 1000000000000000000LL, &a[0], &a[N-1]);
			else MinMax(a[i-1]+ 1, a[N-i] - 1
			, &a[i], &a[N-i-1]);
		}
		ll cans = 0;
		iloop(0, N-1) cans = max(cans, a[i+1] - a[i]);
		return cans;
	}
	ll x, y, t1, t2, cans = 0;
	MinMax(0, 1000000000000000000LL, &x, &y);
	ll prv = x;
	iloop(0, N-1) {
		MinMax(x + (((long double)(y-1-x))*(long double)i/(long double)(N-1)) + 1, x + (((long double)(y-1-x))*(long double)(i+1)/(long double)(N-1)), &t1, &t2);
		if (t1 != -1) {cans = max(cans, t1 - prv), prv = t2;}
	}
	return max(cans, y-prv);
}
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