why is kenkoooo not working these days?

Recent ABC contests are pretty good. Wish another fascinating ABC tonight!

The special effect when clicking the left mouse button is quite interesting.

Hello AtCoder,

Please stop blocking kenkoooo thanks. Or please add pages like My Submissions or Friends' Submissions on AtCoder. Thank you!

F is too standard.

what happened to kenkooo? It has been freezing for over a week.

D is too hard for me

F

why was tle in c ? this code is n log n , is not ?

#include <bits/stdc++.h>

using namespace std;
#define FAST_IO ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);

typedef long long ll;

void solve(){
	FAST_IO
	ll n; cin >> n;
	vector<ll> a(n);
	multiset<ll> ms;
	for(ll i  = 0; i < n; i++){
		cin >> a[i];
		ms.insert(a[i]);
	}
	
	bool c = true;

	vector<pair<ll,ll>> aux;
	for(ll i = 0; i < n; i++){
		if(ms.count(a[i]) == 1){
			aux.push_back({a[i],i+1});
			c = false;
		}
	}

	if(c){
		cout << -1 << endl;
		return;
	}

	ll ans = -1;
	ll comp = -1;

	for(ll i = 0; i < aux.size(); i++){
		if(aux[i].first > comp){
			comp = aux[i].first;
			ans = aux[i].second;
		}
	}

	cout << ans << endl;



	return;
}

int main(){
	solve();
}

What a shit contest it is!

problem F is so shit!

For problem C,

#include <bits/stdc++.h>

using namespace std;

int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int inp, flag=0;
  vector <pair<int, int>> vp;
  for(int i=0; i<n; i++){
    cin >> inp;
    vp.push_back(make_pair(inp, i+1));
  }
  if(n == 1){
    cout << "1\n";
  } else {
    sort(vp.begin(), vp.end());
    for(int i=n-1; i>0; i--){
      if(vp[i-1].first != vp[i].first){
	cout << vp[i].second << "\n";
	flag =1;
	break;
      } else {
	while(vp[i-1].first == vp[i].first || i == 1){
	  i--;
	}
      }
    }
    if(flag==0){
      cout << "-1\n";
    }
  }
}

I can't figure out how this is failing?

Absolutely terrible contest.

F has no "takahash" substring.(Takahashi, same pronunciation with 它卡哈希 in Chinese which means the problem can't solve by hashing.) So it can be solved easily by hashing :) LOL

More, G is too hard for rated participant :(

problem G is well known in India

Was it just me who found the statement and samples for C contradicting (somehow or I am just dumb). (I eventually guessed it from samples).

What was the answer to problem G

Louissun used AI in the competition.

Problem G is a harder version of Canada MO 2019/5.

I used Manacher algorithm to solve problem F, and I found it in the Japanese version of editorial. In fact, I just learned this algorithm from Codeforces Round 524 (Div. 2) E. Sonya and Matrix Beauty， maybe several months ago.

This really feels good, and makes me believe that everyone would always benefit from your practice sooner or later :D

For the number of connected components where the number of vertices in both parts are both oddeditorial has oeinstead of oo

E is disturbing for such a strict output format.

Who can help me with problem E?

#include<bits/stdc++.h>
using namespace std;
int n,cnt1,cnt2,A[105],B[105],ft[105];
vector<int> zi[105];
set<pair<int,int> > s;
void dfs(int u,int fa,int s){
	ft[u]=fa;
	if(s)
		A[++cnt1]=u;
	else
		B[++cnt2]=u;
	for(int i=0;i<zi[u].size();i++){
		int v=zi[u][i];
		if(v==fa)
			continue;
		dfs(v,u,s^1);
	}
}
void solve(){
	while(1){
		set<pair<int,int> >::iterator it=s.begin();
		cout<<(*it).first<<" "<<(*it).second<<endl;
		s.erase(it);
		int u,v;
		cin>>u>>v;
		pair<int,int> p1={u,v},p2={v,u};
		if(u==-1)
			break; 
		set<pair<int,int> >::iterator it1,it2;
		it1=s.find(p1),it2=s.find(p2);
		if(it1==s.end())
			s.erase(it2);
		else
			s.erase(it1);
	}
}
int main(){
	cin>>n;
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		zi[u].push_back(v);
		zi[v].push_back(u);
	}
	dfs(1,0,1);
	int sum=cnt1*cnt2-n+1;
	for(int i=1;i<=cnt1;i++)
		for(int j=1;j<=cnt2;j++){
			int u=A[i],v=B[j];
			if(ft[u]==v||ft[v]==u)
				continue;
			s.insert(make_pair(u,v));
		}
	if(sum%2){
		cout<<"First"<<endl;
		solve();
	}
	else{
		cout<<"Second"<<endl;
		int u,v;
		cin>>u>>v;
		if(u==-1)
			return 0;
		pair<int,int> p1={u,v},p2={v,u};
		set<pair<int,int> >::iterator it1,it2;
		it1=s.find(p1),it2=s.find(p2);
		if(it1==s.end())
			s.erase(it2);
		else
			s.erase(it1);
		solve(); 
	}
	return 0;
}

Thanks!