So the intention of $$$N \le 2.5 \times 10^6$$$ in G is not to hack $$$O(N \log^2 N)$$$ but to hack $$$O(N \log N)$$$ with a large constant factor???

Oh, fixed.

You can check the innermost spoiler, or: https://uoj.ac/problem/953

"I often recall the past" is the English translation of the first sentence.

See (some of) them here: https://mirror.codeforces.com/blog/entry/151401

But sorry...I forgot to shoot that.

The reason is that y1 is defined as a function in the <cmath>library. Directly using y1 sometimes causes Compilation Error if the coder uses <bits/stdc++.h>. It is heard that someone once failed a whole problem (in an OI contest) due to this.

But seemingly this does not happen in Codeforces?

Well, I can just say I hardly use variable names consisting of more than 2 chars! However, I use "1 letter + 1 digit" more often. Below is an example (guess what it does):

#define y1 Y1 // you know why
bool ok(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4)
{
	if ((x2 - x1) * (y4 - y3) == (x4 - x3) * (y2 - y1))
	{
		if ((x2 - x1) * (y3 - y1) != (x3 - x1) * (y2 - y1)) return true;
		if (x1 == x2 && max(y1, y2) <= min(y3, y4)) return true;
		if (x1 == x2 && max(y3, y4) <= min(y1, y2)) return true;
		if (x1 != x2 && max(x1, x2) <= min(x3, x4)) return true;
		if (x1 != x2 && max(x3, x4) <= min(x1, x2)) return true;
		return false;
	}
	int a1 = (x3 - x1) * (y4 - y3) - (y3 - y1) * (x4 - x3);
	int a2 = (x2 - x1) * (y4 - y3) - (y2 - y1) * (x4 - x3);
	int b1 = (x3 - x1) * (y2 - y1) - (y3 - y1) * (x2 - x1);
	int b2 = (x2 - x1) * (y4 - y3) - (y2 - y1) * (x4 - x3);
	if (a1 * a2 < 0 || abs(a1) > abs(a2)) return true;
	if (b1 * b2 < 0 || abs(b1) > abs(b2)) return true;
	if (a1 == a2 * 0 && b1 == b2 * 0) return true;
	if (a1 == a2 * 1 && b1 == b2 * 0) return true;
	if (a1 == a2 * 0 && b1 == b2 * 1) return true;
	if (a1 == a2 * 1 && b1 == b2 * 1) return true;
	return false;
}

I have to say such variable names give this code..."neatness"?

...but it should be pointed out that such behavior is slightly dishonorable (except in AHCs).

According to an experiment, (clock() - timecode) > x * CLOCKS_PER_SEC would regard 1 real second as around 1.05 ~ 1.1 seconds (whose accurate value is $$$\dfrac{2^{20}}{10^6}$$$ or $$$\dfrac{2^{30}}{10^9}$$$, I guess), so you might pass even if you set it to around 6.3 ~ 6.6 seconds.

This feature is really common among different judgers.

So when is a round considered popped from the queue? After the round ended or whatever?

The answer might be clear.

#include<bits/stdc++.h>
using namespace std;
int read()
{
	int x=0,f=1; char c=getchar();
	while(!isdigit(c)) f=c=='-'?0:1,c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return f?x:-x;
}
const int N=100010;
int n,a[N],num[N],ans[N];
int head[N],cnt;
struct edge
{
	int nxt,to;
}e[N];
void add(int u,int v)
{
	e[++cnt]={head[u],v};
	head[u]=cnt;
}
int root[N],tot,siz[N<<5],ls[N<<5],rs[N<<5];
void insert(int &rt,int l,int r,int pos)
{
	if(!rt) rt=++tot;
	siz[rt]++;
	if(l==r) return;
	int mid=(l+r)>>1;
	if(pos<=mid) insert(ls[rt],l,mid,pos);
	else insert(rs[rt],mid+1,r,pos);
}
int query(int rt,int l,int r,int L,int R)
{
	if(!rt) return 0;
	if(L<=l&&r<=R) return siz[rt];
	int mid=(l+r)>>1,res=0;
	if(L<=mid) res+=query(ls[rt],l,mid,L,R);
	if(R>mid) res+=query(rs[rt],mid+1,r,L,R);
	return res;
}
int merge(int p,int q,int l,int r)
{
	if(!p||!q) return p|q;
	siz[p]+=siz[q];
	int mid=(l+r)>>1;
	ls[p]=merge(ls[p],ls[q],l,mid);
	rs[p]=merge(rs[p],rs[q],mid+1,r);
	return p;
}
void dfs(int u)
{
	for(int i=head[u];i;i=e[i].nxt)
	{
		int v=e[i].to; dfs(v);
		root[u]=merge(root[u],root[v],1,n);
	}
	ans[u]=query(root[u],1,n,a[u],n);
	insert(root[u],1,n,a[u]);
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++) a[i]=num[i]=read();
	sort(num+1,num+n+1);
	for(int i=1;i<=n;i++) a[i]=lower_bound(num+1,num+n+1,a[i])-num-1;
	for(int i=2;i<=n;i++) add(read(),i);
	dfs(1);
	for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
	return 0;
}

And as we know, in where zhoujiarun0216's avatar comes from (Luogu), a user will be banned if he cheat for 2/2 times.

zhoujiarun0216

Note: we will publish the Chinese editorials at https://hydro.ac/d/FAOI/contest/66b2e4be8aad713736f45885 one millisecond (yeah I'm serious!) after the contest ends.

As a result, we would welcome _istil and his great translations!

As a staff who verified D using 90 minutes, welcome to the contest!

(I won't tell you that I was told "f**k you and your translations" yesterday and my translations are replaced.)

Had 2 * x and enum (y mod w)? [w = 2 ^ k / w > x]

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MikeMirzayanov As you have seen, the 'jokers' open two accounts to the max Rating from dropping. However, Rating dropping is a necessary part of CP:

So this do be a good lesson.

Some people might say "I'm just disguising so that my friends would't discover that I'm participating Codeforces contests". I failed to understand why they do that at a risk of losing their accounts while such actions are obviously of no benefit. It is sharing and communicating instead of disguising that should be couraged.

Some other people might say "I'm just examining myself by endeavoring to reach Master in 3/6 participations (orz as I can't)". Yet why don't you try VP? Don't you value Codeforces' VP system while someone in Luogu is still appealing for it (or even part of it)? Again, is this worth risking losing your account? Definitely not!

As a result, I suggst them correcting these mistakes instead of revenging by their unfair and irrational downvotes.

Yeah.

Hey! Why giving mike downvotes? Has he done anything wrong? Let's upvote!

In my opinion the score of Div. 1 E should be 1250 + 2250 instead of 2000 + 1500...

Given sequence $$$a_n$$$ and suppose $$$b_{a_i}=i$$$. In each query given $$$x,y$$$, print $$$x$$$ if $$$b_x \gt b_y$$$, or $$$y$$$ if not.

I observed that, but what's next?

imo: Brute force (enuming) with Segment Tree.

F<<<E!!!

omg this is the first time that I see a video in the contest description!

So can I say there will be more Ad-hoc problems (I mean the problems that don't have the same difficulty to everyone) on Codeforces and Atcoder because of their "Russian-ness"?

Yes: 220553759

Not all codes with while(1) are unable to terminate...

In my opinion F<<D... So why is a BST written in the statement? Just to make it scary and match our past impressions of the difficulty of problem F?

btw why is C (in F) up to 1e9, not 998244352?

This?

I solved F just because I saw this before (The problem called Mowing the Lawn in USACO 2011 Open Gold Group)...

Why is F given 1200 score...

I don't know its sex actrually though it's probably male, so here I regard it as a common rabbit. :)

No. I can't use PS, the meme is made by PinkieRabbit itself.

It's a good choice to:

1. Shot a share page
2. Drag it into Photoshop (PS)
3. Change the delta

Number of Tasks: 8

The point values will be 100-200-300-400-450-550-575.

No, you get no penalties while I get 18. :(

Yes.

Any, DFS doesn't either.

https://atcoder.jp/contests/abc318/submissions/45166197

...or DFS, here is the code.

https://atcoder.jp/contests/abc318/submissions/45166197

What was mine... (OMG my code was even faster!)

Yes.

4 3
2 1 3
2 4
1 4
3 4

Why not DFS with a tiny cut?

I get 14 penalties in D...

CMs (or people with higher ranks) can uphack in the first week after contest.

This is easy. Just find the 'Settings' sidebar, then uncheck 'Show tags for unsolved problems'.

Thanks a lot because I haven't noticed this before...

Thanks for sharing!

I found an excluded one with your help.

The memory limit of E may be a little small because...

A poor guy is hacked!

Therefore ABC311 should be also skipped because of the earthquake in 2011.

.

.

.

.

Oh it works again!

Omg the queue is down...

Why am I still in queue...

Woohoo it starts!

220542661 I hope I won't TL...

220542661

How? Do you mean down/sumto/diffto arrays? I think it should be from the lazytags in Segment Trees..

.

I think it's a greedy & brute force problem. :)

No.

Just do greedy?

Me too. I'm surprised that there are no DP problems from A to E...

Obibously, themselves'.

In my mind E is always a DP problem. Will it...?

My code for B:

...
bool h[400012];
int de[400012];
bool alr[400012];
int cnt=0;
bool dfs(int x,int d,int y)
{
	cnt++;
	if(!y) alr[x]=true;
	if(de[x]!=-1&&de[x]%2!=d%2&&y==1) return true;
	if(y>=2) return false;
	if(de[x]!=-1) return false;
	de[x]=d;
	int ii=fdlks.first[x];
	while(ii!=-1)
	{
		if(dfs(fdlks.sds[ii].to,d+1,y+(h[x]==h[fdlks.sds[ii].to]))) return true;
		ii=fdlks.sds[ii].next_side;
	}
	de[x]=-1;
	return false;
}
...

This one got AC, but get $$$cnt = 1.26 \cdot 10^8$$$ in Test 40, which is obivously not $$$O(N)$$$. Is it $$$O(N^2)$$$?

(Whole code: Submission 43433400)

exactly 3 in fact...but still too few.

Hacks:

A: WA: 665; RT: 1; TL: 1; ML: 1.

B: TL: 2.

C: TL: 85; WA: 2; ML: 1. [An uphack to C (TL) has been counted.]

D: TL: 14; WA: 3. [An uphack to D (WA) has been counted. ]

E: TL: 2.

F: (None.)

Total: WA: 670; TL: 104; ML: 2; RT: 1.

Count: 777

Could anybody explain why there are so many hacks?

Sorry, I used another similar plan to solve it. It has been corrected now.

So for 5x5 this is ok:

6 7 8 9 10

16 17 18 19 20

1 2 3 4 5

...

Just think about:

n+1 n+2 ...

3n+1 3n+2 ...

...

1 2 3 4 ...

2n+1 2n+2 ...

...

So C<<A for me...

It's a great pity that I divided B into 8 situations and got +10...I was shocked when I tried hacking.

You have just took part in three contests and I think it's too early to ask this question.

It seems that Problem C can be solved with $$$O(n^2)$$$ time and $$$O(n \cdot \log n)$$$ memory.

Good problem!

So why not make the value of $$$n$$$ larger?

It's from a game that developed by me. It has several levels. A lucky player passed Level 3 (the meaning of "L3") by getting 35966 points (the meaning of "035966") in May 2020. Sadly, the game has been closed in July 2021.

.