It's div3 for AIs of course. They think the contest is so easy with a human to help them submit.

F Too simple.

G Just randomly guess some sufficient/necessary conditions.

AI shitted the contest. Bad round.

Don't get the point of this round being put under the "Hello 2026" tag.

A,B ok.

C: A good problem, but it looks like a copy-paste of generals.io and the smallest time to capture a specific set of tiles is exactly generals.io's game-opening strategy (wait until you have a number of army and then make the other tiles yours).

D1: ok.

D2: I know the authors probably put D2 because contestants might guess the lower bound of the answer in D1. But it made the problem too difficult for this position. (I can say for sure that Not only E but also F<D1).

E: ok problem, but it's a bit boring if you've seen this formula before(?). So it's too easy a problem.

F: just open the square and it's brainless.

So it's clear that E and F can't be put in that place. I'd say E 1800, F 2000? if F is put in E position then obviously many more will pass it.

G/H/I1/I2: I didn't think carefully, but tourist didn't solve any of them. Too hard. They together with E,F made the big gap. Seen from the ranklist, the round failed to separate CM~IGM contestants with the number of problems they solved — only the penalty.

So it has bad differentiation, and its A~F is not good either.

I'd say bad round.

wtf G BM??? anyone has easier solution

wtf H brute searching is correct

so shocked, so what's the intended solution

IOI2028 Winner

wow 6 prob arc!

Just type 10/6/2025.

So is the contest going to be unrated or not, want my positive delta :(

Nailong — Too lazy to take contests or see the problems, lie on the sofa, watch cartoons, and be a couch potato.

I thought it wrote "(Div. 2)" on the title of the contest.

Why have I found five Div. 4 problems in the "tasks" session?

There must be something wrong.

Congrats, piggy!

1c very weak pretest... i hacked >3 people

Use n=12900 and a[1]=110.

Good D! However my MO classmate (doesn't know how to write code; just solves it on the paper) solves ABD, and solves E after given some hints.

OI<<<MO :D

A wasted me half a round ... Nice problem but it's too alien-like.

Make the round 2h45min, and rainboy can AK.

BTW, congrats to rainboy!

UPD: ok I can see now. thx

Why can't I see others' code?

Too much case work in C.

Okay, I'm cyan, so I'm in.

L_Wave

Yes — I solved A, then thought about B

then solved D then thought about B

then I knew how to solve C... but at the same time I knew how to solve B

It's too hard, but at the same time a fresh-style task.

I noticed that and passed lol. However the sample of the 2x2 grid full of 1s is misleading. I first thought that the real method is just a corner case!

Very good problems, I love them! Especially B&C.

Congrats to programpiggy for finally reaching M!

Okay, I'm on my way to the bar.

Candidate Master L_Wave

The never submit strategy isn't good. You can try to hack others with the samples. In this way you will gain a big rank, about 10000. In this way you will get to expert very fast!

$$$\forall n\in\mathbb{N}_+,\sum_{k\mid n}a_k=m^n$$$: For every positive integer $$$n$$$, the sum of $$$a_k$$$ where $$$k$$$ is a divisor of $$$n$$$ is $$$m^n$$$.

Speedrun Yes. BCDE are all easier than ARC184_B.

But I optimized my algorithm (not my constant) and got AC. Your code is easy to optimize.

I got a TLE on this :(

It seems like a speedrun. Anyway, gl&hf!

E2 => Knapsack

E3 => Use Minkowski to optimize that knapsack

1776J — Italian data centers

G has an originate problem (and I submitted my code for that problem changing N from 105 to 305), but my code got WA*3. It turned out that, the eps for precision error should be $$$10^{-14}$$$ or something less, which is quite annoying.

Lost my first blood. btw congrats to maspy

I think atcoder should have more testers like codeforces, and they should cover most of the rating levels, in this way the difficulty will be more balanced.

I guess the two testers (one IGM and one LGM) solve at least ABCD so they don't realize that B,C,D are almost the same level (and what's more, I think D<B).

The $$$900-1000-1000$$$ seems a bit tough, but anyway hope I can solve 3.

So can we become rank Tourist (or Champion, whatever) in Christmas?

It's Chinese editorial. English editorial will be published soon (probably since the translations are done).

Yes! I debugged using that case. Maybe authors want to make A a bit friendly (?)

And even for a period of time jiangly is the only aker!

zhoukangyang will win IOI24!

Kevin114514 is in team China? Heard it from a coach of the team.

Why downvotes? Of course they are not the same person or Mike would have already banned one of them.

Orz wmh.

It makes me think of Teacher Ka (Karry5307) in HNOI.

Hope it's a good round with strong pretests!

zhoukangyang will win the world tour!

I thought of tetr.io when I first saw the idea. In the game website, the users can only enter the league (i.e. official game) when they are level 10 or higher. This annoys me because getting to level 10 can take a week or more time.

The same as above, setting "unrated 20" contests for beginners isn't a good idea. They would become bored and quit cf.

OMG, I don't know how to explain it in English... I have a Chinese blog (OI-Wiki). Maybe you can look at the gif (which scans parallel to the x-axis) or translate it using ChatGPT.

Click me

Discrete the rectangles, put them into segment trees, and brute force to get the answer. It's an $$$\Theta(n\log n)$$$ approach.

Forall x, iterate the multiple of x. The time conplexity is approx. $$$O(A\log A)$$$, where $$$A=\max{a}$$$.

Well, I swapped $$$(i,j)$$$ such as $$$a_j$$$ is the biggest and $$$a_i$$$ is the smallest.

I think B<<<<D. B is easy to construct the answering pairs when you guessed the number of operations.

Ohhhh, I got my 3rd rk145 in arc! So unbelievable!

It's one of the jokes between me, him and a few other friends like mhb2010,ballpoint_pen.

zhoukangyang will win the semifinal!

He should achieve his rank in reverse order lol. First getting LGM, then IGM, and then ..., and finally Newbie.

But he added a '\n' before a { in D, but he didn't do so in F. Also, the type of main()'s return value is different.

D:

signed main()
{

F: int32_t main() {

Consider $$$A=-4,2,3,3$$$, and $$$R=4$$$.

A well known saying in China: "Instead of beating the problem, I only need to beat the author." The person who said it got accepted in C with $$$n=75$$$, and his algorithm's easy to hack.

I'm in Hangzhou now, where it's held. Good luck!

Finally a div.1 standing with tourist rk1 and jiangly rk2! BTW, congrats to Kevin!

Maybe you can look at the following cases:

One of the ways of n=6,m=1,(a,b)=[(1,3)]:
ooxooo
oooxoo
ooooxo
ooooox
xooooo
oxoooo
---
One of the ways of n=6,m=2,(a,b)=[(1,3),(5,4)]:
ooxooy
yooxoo
oyooxo
ooyoox
xooyoo
oxooyo

Where x and y means $$$1$$$ and o means $$$0$$$. You can divide them into $$$n$$$ sets that in each set, the elements has a same $$$(b_i-a_i)\bmod n$$$.

A was kinda tricky if you don't get the point :( I spent 30+ min on it :(

I won two TON coins in CodeTON Round 8, but now I don't know how I can receive them. I have a TON account

and also I updated my wallet before April 16 UTC.

What can I do beside just waiting?

Hey guys, I have a way to deal with division and modular in $$$\Theta(f(n)\log n)$$$ time, where $$$f(n)$$$ is the complexity of multiplying two $$$n$$$-length integer (which means, the complexity is $$$\Theta(n\log^2n)$$$ altogether). The code was written a long time ago, so I can't remember how did I implement it :(

	friend pair<Unsigned, Unsigned> div_bf(Unsigned a, const Unsigned& b) {
		if (a < b) return {(Unsigned)0, (Unsigned)a};
		// cout << '$' << b.size() << endl << flush;
		assert((bool)b);
		Unsigned c(0);
		c.init_size(a.size() - b.size() + 1);
		for (int i = c.size(); i; i--) {
			int lo = 0, hi = Z - 1, mi, res = 0;
			while (lo <= hi) {
				mi = (lo + hi) >> 1;
				if (((b * mi) << (i - 1)) <= a) {
					res = mi;
					lo = mi + 1;
				} else
					hi = mi - 1;
			}
			c[i] = res;
			a -= ((b * res) << (i - 1));
		}
		c.flat();
		return {c, a};
	}
	friend Unsigned num_inv(const Unsigned& a) {
		// cout << a << ' ' << a.size() << endl << flush;
		if (a.size() <= 20) {
			// cout << "FINISH RECURSION\n";
			// cout << a.size() << endl << flush;
			Unsigned b;
			b.init_size(a.size() << 1 | 1);
			b.back() = 1;
			return div_bf(b, a).ff;
			// cout << "EIOFJEOIFJOWE\n" << flush;
		}
		Unsigned b;
		int m = a.size(), k = (m + 5) >> 1;
		b.init_size(k);
		for (int i = b.size(), j = a.size(); i; i--, j--) b[i] = a[j];
		b.flat();
		Unsigned c = num_inv(b), d = a * c * c;
		d >>= (k * 2);
		b = c + c;
		b <<= m - k;
		Unsigned res = b - d - 1;
		d.init_size(m << 1 | 1);
		d.back() = 1;
		c = d - res * a;
		if (c >= a) ++res;
		return res;
	}
	friend Unsigned div_newton(const Unsigned& a, const Unsigned& b) {
		if (a.size() < b.size()) return Unsigned();
		Unsigned x = a, y = b;
		int n = x.size(), m = y.size();
		if (n > m * 2) {
			x <<= n - m * 2;
			y <<= n - m * 2;
			m = n - m;
			n = m << 1;
		}
		Unsigned z = num_inv(y);
		y = x * z;
		y >>= m * 2;
		x = b * y;
		// cout << x.size() << "#\n" << flush;
		z = a - x;
		if (z >= b) ++y;
		return y;
	}

UPD: note that the << and >> operators are "shifting", or "multiplying $$$10^x$$$" in my code, instead of multiplying $$$2^x$$$!

Yes, maybe B and E harder as ARC173B and this contest's E?

The code of C is actually very short.

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = (a), i##ABRACADABRA = (b); i <= i##ABRACADABRA; i++)
#define drep(i, a, b) for (int i = (a), i##ABRACADABRA = (b); i >= i##ABRACADABRA; i--)
using namespace std;
using ll = long long;

int n;
ll L[500010],R[500010],res[500010];

int main() {
  scanf("%d",&n);
  rep(i,1,n)scanf("%lld%lld",&L[i],&R[i]),res[i]=L[i];
  rep(i,2,n)if (res[i]<res[i-1])res[i]=min({res[i-1],R[i]});
  drep(i,n-1,1)if (res[i]<res[i+1])res[i]=min({res[i+1],R[i]});
  rep(i,2,n)if (res[i]>res[i-1])res[i]=max({res[i-1],L[i]});
  drep(i,n-1,1)if (res[i]>res[i+1])res[i]=max({res[i+1],L[i]});
  rep(i,1,n)printf("%lld ",res[i]);
  return 0;
}

Finally broke the streak of rk145 and got rk149... Anyway, good contest.

I totally agree as there are $$$201$$$ participants who solved $$$5$$$ problems.

He means problem A.

Yes, it's tourist ranking first, finally! Hope tourist can get back to the rating leader again!

No, it only needs std::set and a little bit observation.

Damn I got WA*2,AC*142 on E... ): ): ):

cc_hash_table, of course.

Also I once heard that the gp_hash_table in pbds contains bugs. Not sure if it's true or not, I decided not to use it any more.

Well, when I first saw this blog I took it as "A person who shared the same room in the previous contest is green" :).

BTW, congrats!

  scanf("%d%d%d",&n,&l,&d);
  w[0]=u[0]=1;
  ld sum=1,res=0;
  rep(i,1,n){
    w[i]=sum/d;
    if (i<l)sum+=w[i];
    if (0<=i-d&&i-d<l)sum-=w[i-d];
    // cout<<i<<' '<<w[i]<<' '<<sum<<'\n';
  }
  sum=1;
  su[0]=1;
  rep(i,1,n){
    u[i]=sum/d;
    sum+=u[i];
    if (0<=i-d)sum-=u[i-d];
    su[i]=su[i-1]+u[i];
    // cout<<i<<' '<<u[i]<<' '<<sum<<'\n';
  }
  rep(i,0,l-1)w[i]=0;
  drep(i,n,1)w[i-1]+=w[i];
  sum=0;
  drep(i,n,0){
    t[i]=(1-w[i])/d;
    st[i]=st[i+1]+t[i];
    ut[i]=ut[i+1]+t[i]*su[i-d-1<0?n+1:i-d-1];
    res=max(res,su[i-1]*(st[i]-st[i+d])-ut[i]+ut[i+d]);
    // cout<<i<<' '<<dp[i]<<' '<<w[i]<<' '<<u[i]<<' '<<t[i]<<'\n';
  }
  printf("%.12Lf",res);

He is my classmate so I'm sure he is Chinese (:

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Of course no lol. adamant isn't very good at Chinese.

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Really? I didn't think of the gf solution. orz qjm

Ahhh I've just noticed that. Updated. orz zifan

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Yes it's the reference trick.

Or you can use iterators: for (auto it=mp.begin();it!=mp.end();it++) sort(it->second.begin(),it->second.end());

Well, I usually use it when I have to copy a part of the code into another one, in this way it is annoying to delete all of the declaration parts.

Such as:

{
  int a = 5;
  printf("%d\n",a);
}
{
  // this part is copied from the part above
  int a = 10; // You don't need to delete this 'int'.
  printf("%d\n",a);
}

Very dramatic that 74 solved F in the last 10 minutes. Maybe he is (re-)learning to setting problems.

I didn't wrote this algo but I read a code when hacking. I think this algo is right, because it's a quadratic function which has only $$$\le 2$$$ max value. But I prefer the $$$\Theta(\sum c_i)$$$ solution better :).

Happy Chinese New Year!

Good Bye, Lunar 2023! Hello, Lunar 2024 — the year of dragon!

sszcdjr will win IOI2027!

mhb2010 will win IOI2028!

Well, I've just found that Mike is also a Barcelona fan!

It is unaccessible again.

Shame on the DDOS-ers!

Okay, "The USACO website is currently down due to being hit by a massive denial of service attack. Please check back later. We appreciate your understanding during this challenging and frustrating situation.".

And I didn't save the problem statements.