As a tester, it's the first time I tested an actual Codeforces round!

As a tester,the problemset is incredible!Hope you enjoy it :)

As a tester, it’s my first round that I tested, good luck to everyone!

As a tester, it is very good round! Good luck!

Lets goooo , its time i become pupil

I loved testing the round, problems are of high quality. Good luck!

5-6 or 7-8 tasks

Score Distribution??

Score Distribution?? Edit : Very Balanced Score Distribution according to toughness..

Hoping to stay cyan

As a tester, I hope you enjoy the round as much as I did.

As not tester, idk

As a tester, the problems are very good. I encourage you do participate in the round!

2 hours and 7 tasks :O

How to reach Expert?

How to reach expert ?

53 points to Master. Hope to be a master tonight!

53 points to master. Hope to be a master tonight!

As a tester, good luck to everyone! It's the first time i tested a round.

Score distribution?

Let's go =)

Hopefully I move to green this time! Excited for this round! Btw when can I become a problem setter / tester?

Hopefully I move to green this time! Excited for this round! Btw which rating should I be to become a problem setter / tester?

What will be the score distribution?

Best of luck, guys. It's time for me to become a specialist....

hope tonight will lucky night and the line graph will reverse!!

waiting for that tree based div 2 D since forever

Knowing Anton, I will be very surprised to see a segment tree in any of his problems

Lost 50 pt in previous div2. Hope to recover

are the round writers siblings?

All the best guys! Hoping to reach 1300 today.

Good luck and We hope good problems' statement to solve that :D

ace5 will the score distribution be announced?

hope to become CM in this contest. Now only 72 rating.

damn it's going to be speedforces according to the score distribution

hope reaches pupil in this contest.

As a participant, I hope I can get non negative delta again.

I will lose pupil at this round :(

hope I can do well

Hope not to drop down to pupil, spec is fine

GLHF

they write to read about interactive problems but dont give any, hope to see one today (ofc i wont be able to solve it)

good luck!

Is D so easy that 1000 people are able to solve and I am not even able to figure out from where to start? :(

Is D that easy? 1000+ solves and I am not able to figure out from where to begin? :(

I have no idea how to start thinking about D!

Binary search? NO

Two Pointers? NO

DP? NO

Gready? NO

Graphs? NO

Range queries? NO

Math? NO

Thanks a lot for this round! Problems are really interesting, especially E with it's idea of solution(just asking questions as in quicksort?).

Is D really this easy? or am I just dumb? 1k+ already solved. It's a recursion problem I know that but somehow not able to form the code ughhhhhhhh :D

Ternary search in D?

Guys why my this code is wrong for codeforces question 3 xor distance one

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

#define ll long long

void solve()
{
    ll a, b, r;
    cin >> a >> b >> r;
    if (r == 0)
    {
        cout << abs(a - b) << endl;
        return;
    }

    vector<int> a1(64, 0), b1(64, 0);
    int greater = 0;
    for (int i = 0; i <=63; i++)
    {
        if ((a & (1LL << i)))
        {
            a1[i]++;
            greater = i;
        }
        if ((b & (1LL << i)))
        {
            b1[i]++;
            greater = i;
        }
    }

    ll min1 = LLONG_MAX;
    for (int i = 0; i <=greater+1; i++)
    {
        for (int j = 0; j <=greater+1; j++)
        {
            ll k = (1LL << i) + (1LL << j);
            ll o = (1LL << i);
            ll p = (1LL << j);
           

            if (k > r)
                break;
            min1 = min(min1, abs((a ^ k) - (b ^ k)));
            min1 = min(min1, abs((a ^ o) - (b ^ o)));
            min1 = min(min1, abs((a ^ p) - (b ^ p)));
            
        }
    }
    if (min1 == LLONG_MAX)
        min1 = 1;
    cout << min1 << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    ll t;
    cin >> t;
    while (t--)
        solve();

    return 0;
}

Do we use divide and conquer on E?

how C?

Oh god, I'm so lucky. I freakin stared at monitor looking for overflow errors in D, but couldn't find anything, since locally everything works ok. Decided to go ooga-booga mode and wrote #define int long long, and it freakin passed. After it passed I realized the error. No -Wconversion moment, shame on me

This is a cool contest with a cool problemset ☝☝☝☝

How C?

is C brute force?

from topbit(r) to 0, if we can flip current bit then copy it and compute abs(a^r — b^r) to determine bit should be flipped or not.

had 10 WA pretest2 still no clue

Can someone please explain, how to solve D?

how to manage the states, within the constraints

Man, I was struggling with C :( TLE or MLE was troubling me? Can anyone give the solution?

Spent 1h+ on C because wrote 1<<i instead of 1ll<<i =(

The problems are really interesting!

B is similar to ARC149B, and I think they probably have somthing in common.

Today i solved 3 problems, and enjoy this contest, Thankyou

Can someone please explain, how to solve D?

how to manage the states, within the constraints

Oh bad I spent 1:10 and almost 100 lines of code C Why are solutions from my room so short...

Was the intended solution for F sth like: sort the sons for each node by greatest depth of node in its subtree, then do dfs in that order and maybe get the profit of jumping from each node on the trampoline when you traverse the tree in dfs order, and take the best profits? I did that without having an actual proof or something and got WA on test 27, but maybe I was missing something.

Please stop create monkey-problems like B

Can someone please tell me what's wrong in this submission for C? Spent over an hour but I couldn't find anything :( Submission

Thanks to preparing the contest! I enjoeyed A,B,C, and G That's how I solved G

import sys
from collections import deque
input=sys.stdin.readline
def iinput():return int(input())
def linput():return list(map(int,input().split()))
mod=998244353

n=iinput()
def solve(n):
    if n==3 or n==5:return False
    if n==2:return [1,1]
    if n%2:
        L=[-1,-1,2,-2,-3,4,1]
        for i in range((n-7)//2):
            L.append(-L[-1])
            L.append(L[-3]-L[-2])
        return L
    else:
        L=[1,2,-2,-1]
        for i in range((n-4)//2):
            L.append(-L[-1])
            L.append(L[-3]-L[-2])
    return L
L=solve(n)
if L==False:print('NO')
else:
    print('YES')
    for i in L:print(i,end=' ')

I'm just curious. To solve G, I have worked out the cases when n is even. But when n is odd, I have completely no idea.

I wonder if we randomly choose numbers from {-3,-2,-1,1,2,3}, is is possible to "random" out a possible permutation?

can someone suggest how to approach problems like B. Tried proving many greedy ideas but failed

did 1<<i instead of 1ll<<i :(

did 1<<i instead of 1ll << i :(

can someone suggest how to approach problems like B. Tried proving many greedy ideas but failed

hopefully i will reach Specialist

i hate bitmask problems a little bit

My solution for problem D

We will binary search on the answer. Let's call the currrent value we are deal with is X Let's call dp[i] : Consider position 1 to i, if we take element i, what will be the sum of blocked elements So : - dp[i] = dp[j] + a[i] - sum from j + 1 to i — 1 must be <= X. We can find the range of j by binary search and get the minimum value by segment tree simutanously. - The function check on binary search return true if there is exist an indice i such that : sum(i + 1, n) <= X && dp[i] <= x Hope it make sense

why is my solution wrong:

problem C made life hard

My solution for problem D

We will binary search on the answer. Let's call the currrent value we are dealing with is X Let's call dp[i] : Consider position 1 to i, if we take element i, what will be the sum of blocked elements. So :

dp[i] = min(dp[j] + a[i]) and sum from j + 1 to i — 1 must be <= X. We can find the range of j by binary search and get the minimum value by segment tree simutanously. The function check on binary search return true if there is exist an indice i such that : sum(i + 1, n) <= X && dp[i] <= x Hope it make sense

hopefully i will reach Specialist

C is stolen from codechef just expanded the constraints https://www.codechef.com/problems/XORDIF

Did someone prove that in B we are getting the best number of inversions by just sorting one of the permutations? That is not that intuitive for me...