Submission #936652

#	Time	Username	Problem	Language	Result	Execution time	Memory
936652		JuanJL	Sequence (APIO23_sequence)	C++17	28 / 100	605 ms	93504 KiB

sequence

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp> 

#define fst first 
#define snd second
#define pb push_back
#define forn(i,a,b) for(int i = a; i < b; i++)
#define ALL(x) x.begin(),x.end()
#define SZ(x) (int)x.size()
#define mset(a,v) memset((a),(v),sizeof(a))
#define FIN ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define dbg(v) cout<<"Line("<<__LINE__<<"): "<<#v<<" = "<<v<<'\n';
#define pi pair<int,int>
#define pll pair<ll,ll>
typedef long long ll;
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_multiset;

ll n; 
vector<ll> a;


int sequence(int N, std::vector<int> A) {
	n=N;
	a.clear();
	a.resize(n);
	forn(i,0,n) a[i]=A[i];
	ll res = 0;
	
	if(n<=2*pow(10,3)){
		//Subtasks 1 , 2
		forn(i,0,n){
			indexed_multiset im;
			ll ocurr[n+1];
			mset(ocurr,0);
			forn(j,i,n){
				im.insert(a[j]);
				ocurr[a[j]]++;
				res = max(res,ocurr[*im.find_by_order((SZ(im)-1)/2)]);
				if(SZ(im)%2==0) res = max(res,ocurr[*im.find_by_order((SZ(im)-1)/2+1)]);
			}
		}
	}else{
		vector<vector<ll>> p;
		set<ll> vals;
		map<ll,ll> cI;
		forn(i,0,n) vals.insert(a[i]);
		ll ind = 0;
		for(auto i:vals) cI[i]=ind,ind++;
		p.resize(SZ(cI));
		forn(i,0,n)	p[cI[a[i]]].pb(i);
		/*forn(i,0,n){
			ll igu = SZ(p[cI[a[i]]]);
			ll men = p[cI[a[i]]][0]+(n-1)-p[cI[a[i]]][igu-1];
			ll may = (p[cI[a[i]]][igu-1]-p[cI[a[i]]][0])-igu;
			if(may<=men+igu){
				res=max(res,igu);
			}else{
				ll preRes = 0;
				forn(j,0,SZ(p[cI[a[i]]])){
					if(abs(p[cI[a[i]]][j]-p[cI[a[i]]][j-1])==1) preRes++;
					else break;
				}
				res=max(res,max(preRes,SZ(p[cI[a[i]]])-preRes));
			} 
		}*/
		ll men = 0;
		ll may = 0;
		ll igu = 0;
		forn(i,0,SZ(p)){
			men=0; may = 0; igu = 0;
			forn(j,0,SZ(p[i])){
				if(abs(p[i][j]-p[i][j-1])==1) igu++;
				else break;
			}
			res=max(res,igu);
			res=max(res,SZ(p[i])-(igu+1));
			may=(p[i][igu+1]-p[i][igu])-1;
			if(may<=men+igu) res = max(res,(ll)SZ(p[i]));
		}
	}
	return res;
}

• 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