First**

How to turn off the sound on contest page?

Cake walk A-E.

Give a thumbs up to ABC this time! The question is very concise.

I think D is much harder then E.

Why D and E giving TLE and RE tried using difference array and BFS technique. BFS tried to optimize using DP but failed, why man ?

D was really educational on how to use Difference Array Dynamically with prefix sums which something i never encountered

Only $$$8$$$ contestants solved all the problems without any penalties. The symbol $$$\color{red}{(x)}$$$ is almost everywhere!

How to do E? welp

E is too easy for E and the difficulty gap between E and F is not reflected in the point value at all. F is a great task btw.

problem E is the same as this one:

https://mirror.codeforces.com/problemset/problem/372/A

Did anybody else get WA on 2 tests in F? How did you fix it

Liked the contest. Thanks !

Can anyone HELP debug this solution of F?

Why it is giving wrong answer

I am just finding the pair element for the element present at the beginning and remove the first element and the paired element , i am going to do this until s.is exhausted() or i can not find paired element for my current element which indicates that if there is no elemnet for me then all the lements after me is bigger than me then for them also there is no elment in the set.

// Source: https://usaco.guide/general/io

#include <bits/stdc++.h>
using namespace std;
#define ll long long 

void solve(){
	int n;cin>>n;
//	int arr[n];
	multiset<int>s;
	for(int i=0;i<n;i++){
		int val;cin>>val;
		s.insert(val);
	}
     int ans=0;
     while(s.size()){
		
		
		// check if 2*val is there or not 
		auto it = lower_bound(s.begin(),s.end(),*s.begin()*2);

		if(it!=s.end()){
			 
			//cout<<"removing "<<*beginIt<<" and "<<*it<<'\n';
			s.erase(s.begin());
			s.erase(it);
			
			ans+=1;}

		else{
             // no 2*val is ofund
			 break;
		}

		 
	}
	cout<<ans;

}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	solve();
	// int a, b, c; cin >> a >> b >> c;
	// cout << "The sum of these three numbers is " << a + b + c << "\n";
}

Now i edited it to make it as follows:-

Traversing from the last of the set and finding the paired element for it .

If found remove last element and paired one

IF not found remove last element Do it until I have elements in the set

// Source: https://usaco.guide/general/io

#include <bits/stdc++.h>
using namespace std;
#define ll long long 

void solve(){
	int n;cin>>n;
//	int arr[n];
	multiset<int>s;
	for(int i=0;i<n;i++){
		int val;cin>>val;
		s.insert(val);
	}
     int ans=0;
     while(s.size()){
		
		
		// check if 2*val is there or not 
		auto it =s.end();
		--it;
		int val = *it;
		auto it2 = lower_bound(s.begin(),s.end(),val/2);

		if(it2!=s.end()){
			 if(*it2==(val/2)){
				 s.erase(it);
				 s.erase(it2);
				 ans+=1;
			 }else{
				   //check if 
				   if(it2==s.begin()){
                     // all are bigger than half of it
					 s.erase(it);
				   }else{
					   --it2; // as lower bound points to at leas tgreate rthan val/2
					    s.erase(it);
				 s.erase(it2);
				 ans+=1;

				   }
			 }
		}
        else{
             // no 2*val is ofund
			 break;
		}

		 
	}
	cout<<(ans);

}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	solve();
	// int a, b, c; cin >> a >> b >> c;
	// cout << "The sum of these three numbers is " << a + b + c << "\n";
}

Any help would be grateful

Fkd up on D clutched on E . D was so annoying

I solved problem G 10 minutes after the contest ended

Why E and G have (basically) same statements but different constraints? I solved G first, then got RE on E because of keeping the size of arrays $$$2\times10^5$$$...

Problem F and G are good

In D first make every element give to its right and take it from left, some values will be negative now then just iterate and every negative value a tells to reduce 1 from last a elements. ex 2 9 1 2 0 4 6 7 1 5 make it -7 2 -4 -1 -1 5 9 12 8 14 then first make last 7 elements reduce by 1-> 0 2 -4 -2 -2 4 8 11 7 13 now for 2 do nothing now for -4 make last 4 elements reduce by 1 and same until last index it gives correct answer but i am not getting how to implement this in linear time, try to help

can anyone explain in G why are we checking for the max value of j- i; such that a[j]>=2*a[i] and j>i? i came across one explanation that said that L+K−1+(max(j-i) for all i<L+mid and i>=L) <=R, but why is it so?

How to solve G?

I think problem F is really a brain storm, and thank you atcoder team for providing such a nice problem!

I solved it as follows. I set E = 22, and at first have lvr = [1]. Then, for each given interval [l, r], we first check which points from [l — E, l — 1] can be reached from points belonging to lvr, and this can be done by some simple mathematical computing. We denote these points as xvr. Next, we can use BFS to check which points from [r + 1, r + E] can be reached from xvr. Then, we update lvr according to these points, and move on to the next given interval.

The key idea is that for two given intervals, we only care about some of their neighboring points, at most 22(maybe 20), and for other points, we should deal with them by mathematical computing quickly.

Will the english editorials be released for this contest?