Submission #983361

#	Time	Username	Problem	Language	Result	Execution time	Memory
983361		c2zi6	Sequence (APIO23_sequence)	C++17	28 / 100	848 ms	17492 KiB

sequence

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "sequence.h"

namespace TEST2 {
	int n;
	VI a;
	int solve(VI A) {a = A, n = a.size();
		int ans = 0;
		replr(l, 0, n-1) {
			indset<PII> st;
			int UID = 0;
			replr(r, l, n-1) {
				st.insert({a[r], UID++});
				int med1 = st.find_by_order((st.size()-1)/2)->ff;
				int med2 = st.find_by_order(st.size()/2)->ff;
				setmax(ans, int(st.order_of_key(*st.upper_bound({med1, 2e9})) - st.order_of_key(*st.lower_bound({med1, -2e9}))));
				setmax(ans, int(st.order_of_key(*st.upper_bound({med2, 2e9})) - st.order_of_key(*st.lower_bound({med2, -2e9}))));
			}
		}
		return ans;
	}
};

namespace TEST3 {
	int n;
	VI a;
	int solve(VI A) {a = A, n = a.size();
		VI cnt, poqr, mec;
		cnt = poqr = mec = VI(n+2);
		for (int x : a) cnt[x]++;
		replr(i, 1, n) poqr[i] = poqr[i-1] + cnt[i-1];
		reprl(i, n, 1) mec[i] = mec[i+1] + cnt[i+1];
		
		int ans = 0;
		replr(i, 1, n) {
			int a = poqr[i], b = cnt[i], c = mec[i];
			if (a >= c - b) {
				setmax(ans, b);
				/* cout << "AY SIK KRNA EXNI MEDIAN " << i << endl; */
			}
		}
		return ans;
	}
};

int sequence(int n, VI a) {
	if (n <= 2e3) return TEST2::solve(a);
	return TEST3::solve(a);
	return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 17 / 17
• Subtask 3: 0 / 7
• Subtask 4: 0 / 12
• Subtask 5: 0 / 13
• Subtask 6: 0 / 22
• Subtask 7: 0 / 18