Is this new judging system an update to the judge machine’s configuration, or does it simply add support for more programming languages? If they’ve actually changed the judge machine, maybe programs could run a bit faster? (I have a bit of a TLE phobia…)

No C++ 17 or 20 is wild

language brainfuck is among availible languages, wtf

Why 30min later?

Strange start time.

Did you update the contest start time?

Is the contest delayed by 30min? Technical issues of the new judge?

upd:according to contest page it is delayed and new judges won't be used

It seems to be slower and slower for us to visit AtCoder... Even submitting a problem takes a few minutes :(

Is it rated?

Hope it will be high quality.

why did E<<<D???

The comparision of tuple in E gives TLE under correct time complexity.

Don't know why E was even included in this contest. D >>> E imo.

E is such a classic problem.Bad E.

(The favs only)

This is too bad.How the difficulty gap between D,E and F,G so big？

And E is too classic that it is a shit here.

Though I spent too much time on D,I think it's a good problem here.

q=int(input())
res=[0]*q
stk1=[]
stk2=[]
for _ in range(q):
    query=input()
    if len(query)==1:
        if len(stk1)==1:
            stk1.pop()
            stk2.pop()
        else:
            prev=res[len(stk1)-2]
            cur=res[len(stk1)-1]
            stk1.pop()
            if cur>prev:
                stk2.pop()
            else:
                stk2.append(stk1[-1])
        if len(stk2)==0:
            print("Yes")
        else:
            print("No")
    else:
        stk1.append(query[-1])
        if query[-1]=="(":
            stk2.append("(")
        else:
            if stk2 and stk2[-1]=="(":
                stk2.pop()
            elif len(stk2)==0:
                stk2.append(")")
        res[len(stk1)-1]=len(stk2)
        if len(stk2)==0:
            print("Yes")
        else:
            print("No") 

What is wrong with my code for c

i think E > D :(

E solution is that take the diameter of the tree and do the bfs statistics to the dist of these two nodes.

vector<vector<int> > g;

vector<int> bfs(int s, int n)
{
    vector<int> dist(n+1, -1);
    queue<int> q;
    dist[s]=0;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front(); q.pop();
        for(int v : g[u])
        {
            if(dist[v]==-1)
            {
                dist[v] = dist[u] + 1;
                q.push(v);
            }
        }
    }
    return dist;
}


int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n; cin >> n;
	g.resize(n+1);
	for(int i=0; i<n-1; i++)
	{
		int u,v; cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	vector<int> d=bfs(1,n);
	int a=1;
	for(int i=1; i<=n; i++)
		if(d[i]>d[a] || (d[i]==d[a] && i>a))a=i;
	vector<int> da=bfs(a,n);
	int b=a;
	for(int i=1; i<=n; i++)
		if(da[i]>da[b] || (da[i]==da[b] && i>b))b=i;
	int maxn=max(a,b);
	vector<int> db=bfs(b,n);
	for(int i=1; i<=n; i++)
	{
		if(da[i]>db[i])cout << a << endl;
		else if(da[i]<db[i])cout << b << endl;
		else cout << maxn << endl;
	}	

	return 0;
}

But question D seems to be able to be written using binary search. Please give me some advice.

Wow! Coder ME competed in Atcoder Beginner Contest 427&428 and gained -97 rating points! Share it!

Rubbish Round

swap(D,E) plz

I think D>>>E.

another speedcoder?? btw is F an ad-hoc? I simply didn't understand how the official editorial works

Could the new judging system be opened for testing before the next competition starts? I don't want to face a completely new and unfamiliar system when I get to the ABC contest.

By the way, long live chez scheme!

My Java code strangely TLE and has an O(n) complexity. I tested it locally, timing the read from n to the output, and it only took 1300+ms. I can't figure out why it times out. I've never encountered such a Java timeout before. Does this mean Java won't be able to use DFS for 5e5 data in the future? Here's the submission link. I think it's due to the OJ system changes. Could atcoder officials please check it out?

https://atcoder.jp/contests/abc428/submissions/70268852

swap(D,E) please

D is terrible for me and I got rating-9 at last

E is too simple for this place.

How to approach problem C can anyone please help

my approach for D (This could help to understand better)

1 <= x <= d
F(c , c + x) => to_string(c) + to_string((c + x))
             => fixed prefix + varying suffix

and this suffix varies from : [c + 1 , c + d]

Initial thoughts calculate : find L , R:
    L = F(c + minSuffix)
    R = F(c + maxSuffix)

    ans = count(0....R) - count(0....L-1)

But this wont work ... why ?

say, c = 4 and d = 80
     suffix varies : [c + 1 , c + d]
                   : [5 , 84]

    L = "4" + "5" = 45
    R = "4" + "84" = 484

    when we use count(0....N), it consides all numbers in [0....N]
    which includes perfect squares with prefix other then 'c' => (81 , 100 , ...)


Solution!! : Break the suffix range into blocks

something like this,
[1 , 9] [10 , 99] , [100 , 999] , ....

Once again say, c = 4 and d = 80

L = "4" + "1" = 41
R = "4" + "9" = 49

L = "4" + "10" = 410
R = "4" + "99" = 499 -> greater than (maxR:484), so bound to maximum
               = 484

ans = [count(0...49) - count(0...41-1)] + [count(0...484) - count(0...410-1)]

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

ll count(ll n){
    return sqrtl(n); 
}

int main(){
    int t ; cin >> t;

    while(t--){
        ll c, d ; cin >> c >> d;
        ll ans = 0;

        ll L = stoll(to_string(c) + to_string(c + 1));
        ll R = stoll(to_string(c) + to_string(c + d)); 

        for(ll x = 1 ; ; x *= 10){
            ll currL = max(L , stoll(to_string(c) + to_string(x)));
            ll currR = min(R , stoll(to_string(c) + to_string(10LL * x - 1)));

            ll LR = max(0ll , count(currR) - count(currL - 1));
            ans += LR;

            if(currR >= R) break;
        }

        cout << ans << endl;
    }

    return 0;
}

Could someone call me what's wrong with my code in problem F QAQ. I found that when migrating operations 1 and 2, there are actually intervals smaller than v that may move, so I tried using a segtree to solve this problem. At the same time, the line segment tree can also be used to split operation 3. But I successfully passed the sample and 10 points, but still had 6 WA points QAQ. Here is my code, because I am not very familiar with it. The lazy markers are relw_ay and rel_num, and lazy_ay and lazy_num correspond to whether the interval is offset to the left or right and the corresponding offset position. (I am a Chinese, please forgive me for not reading correctly. qwq)

#include <iostream>
#include <cstdio>
#include <cstring> 
#include <algorithm>

using namespace std;
const int N = 200010;
typedef pair<int, int> PII;

int n, q;
int w[N]; 

struct STree
{
	int l, r;
	int num;
	int rel_way = 0, rel_num = -1; 
	int lazy_way = -1, lazy_num = 0; 
} stree[N << 2];

void pushdown(int plc)
{
	int way = stree[plc].rel_way, lz = stree[plc].rel_num;
	if (way == 0 || (stree[plc].l == stree[plc].r))
	{
		stree[plc].rel_way = 0;
		stree[plc].rel_num = -1;
		return ;
	}
	stree[plc << 1].rel_way = way;
	stree[plc << 1].rel_num = lz;
	stree[plc << 1].lazy_way = way;
	stree[plc << 1].lazy_num = lz;
	
	stree[plc << 1 | 1].rel_way = way;
	stree[plc << 1 | 1].rel_num = lz;
	stree[plc << 1 | 1].lazy_way = way;
	stree[plc << 1 | 1].lazy_num = lz;
	
	stree[plc].rel_way = 0;
	stree[plc].rel_num = -1;
}

void init(int l, int r, int plc)
{
	stree[plc].l = l, stree[plc].r = r;
	if (l == r)
	{
		stree[plc].num = w[l];
		return ;
	} 
	int mid = (l + r) >> 1;
	init(l, mid, plc << 1);
	init(mid + 1, r, plc << 1 | 1);
	stree[plc].num = max(stree[plc << 1].num, stree[plc << 1 | 1].num);
}

PII query(int pos, int plc)
{
	if (stree[plc].r < pos || stree[plc].l > pos) return {-1, -1};
	if (stree[plc].l == pos && stree[plc].r == pos)
	{
		if (stree[plc].lazy_way == -1)
		{
			return {stree[plc].lazy_num, stree[plc].lazy_num + stree[plc].num};
		}
		else
		{
			return {stree[plc].lazy_num - stree[plc].num, stree[plc].lazy_num};
		}
	}
	pushdown(plc);
	PII res = {0, 0};
	res = max(res, query(pos, plc << 1));
	res = max(res, query(pos, plc << 1 | 1));
	return res;
}

void update(int l, int r, int lz_way, int lz_num, int plc)
{
	if (stree[plc].r < l || stree[plc].l > r) return ;
	if (stree[plc].l >= l && stree[plc].r <= r)
	{
		stree[plc].rel_way = lz_way;
		stree[plc].rel_num = lz_num;
		stree[plc].lazy_way = lz_way;
		stree[plc].lazy_num = lz_num;
		return ;
	}
	pushdown(plc);
	int mid = (stree[plc].l + stree[plc].r) >> 1;
	if (mid >= l) update(l, r, lz_way, lz_num, plc << 1);
	if (mid < r) update(l, r, lz_way, lz_num, plc << 1 | 1);
}

int query_insert(int pos, int plc)
{
	if (stree[plc].l == stree[plc].r) return stree[plc].l;
	pushdown(plc);
	int lz_way = stree[plc << 1].lazy_way, lz_num = stree[plc << 1].lazy_num, nm = stree[plc << 1].num;
	if (lz_way == -1)
	{
		if (lz_num <= pos && lz_num + nm > pos)
		{
			return query_insert(pos, plc << 1);
		}
	}
	if (lz_way == 1)
	{
		if (lz_num > pos && lz_num - nm <= pos)
		{
			return query_insert(pos, plc << 1);
		}
	}
	return query_insert(pos, plc << 1 | 1);
}

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i ++ )
	{
		scanf("%d", &w[i]);
	}
	init(1, n, 1);
	
	scanf("%d", &q);
	for (int i = 0; i < q; i ++ )
	{
		int op, v;
		scanf("%d%d", &op, &v);
		if (op == 1)
		{
			if (v == 1) continue;
			PII mov = query(v, 1);
			update(1, v - 1, -1, mov.first, 1);
		}
		else if (op == 2)
		{
			if (v == 1) continue;
			PII mov = query(v, 1);
			update(1, v - 1, 1, mov.second, 1);
		}
		else
		{
			printf("%d\n", n - query_insert(v, 1) + 1);
		}
	}
	
	return 0;
}

I didn't even have time to solve E as I spent a lot of time of D! E is too classic!

My issue

Why is my own code for D wrong? In the editorial, it says that we can use floor(sqrt(x)) instead of the function f(x) to compute $$$\lfloor \sqrt{k} \rfloor$$$. But when I use floor(sqrt(x)), the judge result is this. However, when I finish my code according to the editorial, my judge result is this. Could anyone explain the reason please?