May 19, 2023 what

//I don't know why, but my comment got sent 2 times, don't pay attention

hoping to solve atleast 3 problems

it's ironic that all the testers are not actually eligible for div 3...

Hoping to become Expert in this round

why is there no cyan, green or gray testers?

i hope to be blue this time

I have a really good feeling about that this would be my last rated Div. 3 contest and I would have become a blue expert after this 888ᵗʰ round of Codeforces!! May the Force be with us

I have exam (TT — TT)

Hopefully, we will see a problem connected with number $$$8$$$, lol.

I think most of Div.3's are made by Vladosiya

best part of the comment section is I see anime pfpS and discover new interesting animes

I think this will be a good one

I hope this round will be the great round of Div.3!

Good luck

Out of competition, but I'll try my best to do it ^_^

It is so good to have a contest every few days, hopefully i can cross 1500 today

I don't know if this is the right place to say this, but I've noticed for a while now that div3 round announcements say that you can only qualify as a trusted div3 participant if you "do not have a point of 1900 or higher in the rating." Shouldn't that number be 1600 instead?

Experts to lower rankings in Div.3 be like:

After playing div2,I feel that div3 and div4 are still suitable for me

Tends to Pupil:)

hoping to solve at least 4 problems

In first line of blog "You will be offered 6-8 problems" .Later "You will be given 7 problems and 2 hours and 15 minutes to solve them ".So, 7 problems will be there ?.

It's fun that the round will be rated for me.

hope to be specialist in this contest

I am at the lowest confidence in my life.

Good Luck everyone

I will solve atleast 5 problems.

so she knows what awaits her. emmmmmmhahahahaha~

Thank you Vladosiya very much. This is the best Div.3 contest I've ever written.

Why I an rated ??????

Readforces

any hints to G?

Yes/NoForces

D: Let difference d[0]=a[0], d[i]=a[i]-a[i-1] when i>0. If the missing prefix sum is n*(n+1)/2, then {d[i]} will be a permutation of [1, n] missing one element. Therefore, for this case we need to check in {d[i]} are pairwise distinct and inside range [1, n]. Otherwise, we will have a[n-1]=n*(n+1)/2, and n-2 elements of {d[i]} will be inside [1, n] and distinct, the last element will be the sum of 2 missing elements. So for this case we need to check if {d[i]} contains n-2 different numbers inside [1, n].

E: First for potions with infinite supply we can just set their cost to 0. Then, if potion i can be obtained mixing {j[1], j[2], ...}, we add a directed edge (j[t] --> i) for each j[t]. Since no potion can be obtained by mixing itself, these edges will from a DAG (directed acyclic graph), and we can run dp on the topological order of the DAG: Let dp[i]=sum(cost[j]) where edge (j --> i) exists, and cost[i]=min(cost[i], dp[i]).

F: If a[i], a[j] and x have only 1 bit, then if a[i]==0, a[j]==0, we can set x=1 then (a[i]^x)&(a[j]^x)=1, if a[i]==1, a[j]==1, we can set x=0 then (a[i]^x)&(a[j]^x)=1, if a[i]!=a[j], no matter how we set x, (a[i]^x)&(a[j]^x)=0. So we need to find (i, j) such that a[i]==a[j]. For general cases, we need to minimize (a[i]^a[j]) so that sum(t: 0<=t<k, the t-th bits of a[i] and a[j] are same)(1<<t) = (1<<k)-1-(a[i]^a[j]) is maximized. This can be processed using a trie, or sort a[i] and find the minimal a[i]^a[i+1].

G: For a single query (a, b, e), the maximum height of nodes on the path cannot exceed h[a]+e, which means, we can pass an edge (u, v) if max(h[u], h[v])<=h[a]+e. Therefore, we can sort queries by h[a]+e and sort edges by max(h[u], h[v]), then solving queries using dsu.

Bad statement for most problems :(

Yes-NoForces!!

i feel language was bit difficult.

Learning tries was after all worth it saving me today. Problem F felt really good to AC :)

It was hard for me to concentrate on reading those problems, I liked it tho

D was awful, imho

I think my F is hackable

Submitted just after the contest got over.

Overall looks like a decent div3 contest, I have enjoyed solving A-E.

• A — might be a little bit too much text for a div3A problem, but nothing too extreme
• B and C — both are fine problems imo
• D — some case work, nothing too dramatic, but I still have no idea how to prove my solution (also randomly got 50 additional penalty because I trolled myself again)
• E — a problem about my favorite topic, I can see how this statement can confuse some people, but I myself probably couldn't phrase the condition of a graph being a DAG better

How to solve E ?

I thought that G was about finding a path, so that $$$max - min \le e$$$. Btw, would there be a solution if this was asked?

Lets Go Pupil here I come..........

rank 834 can add 39 to become expert?

Nice problems but bad statments for most of them , thanks anyway!

What the hell

Can someone share their solution to B, C and D.

problem E is an easy version of one of the provincial team selection test for NOI of Ahhui, China 2014(AHOI2014) problem. See in here:https://www.luogu.com.cn/problem/P4042

a

D was frustrating.

can some one explain this sample case for problem E

4 2

1 1 5 4

2 4

3 2 4 3

0

2 2 4

1 2

how is the answer 0 0 0 0 , here

In problem A, is there any specific reason why Vlad cannot talk to the person on the same steps of the escalator??

Hello, could someone please help me figure out why my code is giving a runtime error on E? I don't usually use python https://mirror.codeforces.com/contest/1851/submission/215620727

Lol I didn't even understand the statement of problem E. Eventually, i just did wat-task-told-me-to-do + my guess and intuition and AC-ed :D

if problem E, the statement "Moreover, no potion can be obtained from itself through one or more mixing processes." is removed, I guess we need a scc to solve it?

Time to switch to hackforces

Problem statement is very hard

Thank you for having blank lines after the answer after every test case in Problem G. It made understanding the sample answer easier.

Hope to become newbie after this contest

I couldn't debug G in time.Loved all the problems.

My thiscode gives correct output on my compiler and also on online compilers but somehow this is giving wrong answer on the Codeforces judge , what's the reason , please look into it.

I have doubt in problem statement of problem E.It is not mentioned whether cycle will exists or not.

In F there is no need to know Trie , my solution is just greedy + binary search

Can anyone please explain the statement of problem e i still can't get it

i took this contest while in an airport lol

can someone help me understand why am I getting runtime error for this code (problem C round 888 div3):

#include<bits/stdc++.h>
using namespace std;
#define int long long int
int32_t main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,k,w,flag1=0,flag2=0;
        cin>>n>>k;
        int l=k;
        vector<int> v;
        for(int i=0;i<n;i++)
        {
            int m;
            cin>>m;
            v.push_back(m);
        }
        int cnt=0;
        if(k==n && n==1)
        {
            cout<<"Yes"<<endl;
        }
        else if(v[0]==v[n-1])
        {
            for(int i=0;i<n;i++)
            {
                if(v[i]==v[0])
                {
                    cnt++;
                }
            }
            if(cnt>=k)
            {
                cout<<"Yes"<<endl;
            }
            else
            {
                cout<<"No"<<endl;
            }
        }
        else
        {
            for(int i=n-1;i>=0;i--)
            {
                if(v[i]==v[n-1])
                {
                    k--;
                }
                if(k==0)
                {
                    w=i;
                    flag1=1;
                    break;
                }
            }
            for(int i=0;i<w;i++)
            {
                if(v[0]==v[i])
                {
                    l--;
                }
                if(l==0)
                {
                    flag2=1;
                    break;
                }
            }
            if(flag1==1 && flag2==1)
            {
                cout<<"Yes"<<endl;
            }
            else
            {
                cout<<"No"<<endl;
            }
        }
        
    }
}

Can E be solved using Dijkstra?

Why am I not in the standings?

F is similar to 1721D.

waiting for editorial...

in D testcase 10:
prefix sum is: [13 21 36 42]
could not we assume initial array like this? [13, 8, 15, 6, 2]

[13, 8, 15, 6, 2] -prefix sum-> [13, 21, 36, 42, 44] and assume he lost [44]

what i am doing wrong?

I was able to solve one problem, but still am unrated. Any reason for that?? (This was my first ever contest. Sorry if I'm being dumb)