The contest time is unusual. But, I hope, the 'Dashboard' will be usual insha-Allah.

As a tester, I tested the round yesterday

Thank you, and I wish everyone success!

Please notice the unusual time!

Comeback Round

As a tester, do try all the problems.

Score Distribution is really good. Excited for tomorrow.

If FynjyBath = tester, round = good

Hoping to reach pupil ;)

scoring distribution is asking me to solve abcd. I hope I can full fill its question

Clashing with Shanghai CPC, skipping this one

good luck, hoping for +delta

Hopping for +delta

OMG Mikhango!!!!

Potential SpeedForces on the way!

sounds like CodeForces Round 879 Div. 2

https://codeforces.me/contest/1834

bashkort orz . Just wanted to tell you that your pfp is my favourite pfp I have seen on codeforces. It looks cool af.

Woowww score distribution so interesting

Good start time for Chinese.And I hope I can become Expert in this contest!!!

Pre-Eid Day Contest , Participating

GLHF, hope I become expert today

Getting good vibes for this round

WOW, this time the problems are more interesting but easier (I think) than normal div.2. This is my first time to AK in div.2 haha.

C would be enjoyable if we didn't have to print the permutation.

darn I was an hour late! still fun contest tho

div 2.5?

Why did you not let the much harder segment tree solution pass in E? I mean TLE

submitted D with just 2 min left

Can someone explain why my code for D fails?

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

void Solve() {
  ll n, c;
  cin >> n >> c;
  vector<ll> a(n);
  for (ll i = 0; i < n; i++) {
    cin >> a[i];
  }
  a[0] += c;
  vector<ll> mx_suf(n);
  for (ll i = n - 2; i >= 0; i--) {
    mx_suf[i] = max(mx_suf[i + 1], a[i + 1]);
  }
  if (*max_element(a.begin(), a.end()) == a[0]) {
    cout << "0 ";
  } else {
    cout << "1 ";
  }
  ll pre = a[0], mx = a[0];
  for (ll i = 1; i < n; i++) {
    ll ans = 0, cur = a[i];
    bool added = false;
    if (mx >= a[i]) {
      ans += i;
      cur += pre;
      added = true;
    }
    if (mx_suf[i] > cur) {
      if (!added) {
        ans += i + 1;
      } else {
        ans++;
      }
    }
    pre += a[i];
    mx = max(mx, a[i]);
    cout << ans << ' ';
  }
  cout << '\n';
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tt;
  cin >> tt;
  for (int i = 0; i < tt; i++) {
    Solve();
  }
  return 0;
}

D felt little easier than C.

5 seconds away from submitting D... I was too drowned on E+F and that wasted my time... :(

UPD: That D actually AC'd. I have myself to blame, I guess.

For E, I resolved the problem into the following Given a list of segments, for each query l, r find number of segments that fall into [l,r]. I'm not sure how to build a segtree for this (the merging part). Any hints?

Imho E was quite easy, easier than D and probably even C, I am surprised there are not that many AC.

D and E are both easier than C.

why it's not working 266036235 problem D

I know this will TL anyway but, quite shock that even brute force solution got WA on pretest 2 am I understand something wrong completely?

does someone know why I got idleness limit exceeded on C. I'm so sad right now :(

https://codeforces.me/contest/1978/submission/266042353

How to solve F?

Whats in test 2 of F? Any counterexamples? (My solution is kind of scanline on dividers).

Thanks for a great round! I hope my C won't FST... I don't understand what is wrong with my solution for F, btw, like no idea. So sad couldn't find mistake during the round.

long longforces

I guess many people, including me, got confused that the word "number" was used instead of index.

Just Found out that B can be done without Binary search.. Maybe I think too much LOL. 265999363

how to proof C that we have to take the two pointer and if 2*diff<=k then we reduce it else either p1++ or p2--. Anyone?

Such a fool contest! For prob. A to prob E, there's nothing harder than suffix sum! How can this Contest be Div.2 ???????

why does the idea of swapping $$$i th$$$ and $$$i+k/2$$$ th elements not work for task C?

void init() {
    int n, k;
    cin >> n >> k;
    if ((k & 1) or (k > (n * (n + 1)) / 2)) // higher bound
        cout << "NO" << endl;
    else {
        vector<int> res;
        res.reserve(n);
        for (int i = 1; i <= n; i++) res.pb(i);
        int i = 0, j = min(n - 1, k / 2);
        int cnt = 0;
        while (k > 0 and i < n and i < j) {
            swap(res[i], res[j]);
            k -= 2 * (j - i);
            i++;
            j = min(n - cnt - 1, i + (k / 2));
            cnt++;
        }
        if (k != 0)
            cout << "NO" << endl;
        else {
            cout << "YES" << endl;
            for (auto x : res) cout << x << " ";
            cout << endl;
        }
    }
}

oops!

It was not possible to make a better contest !! Hats Off to vaaven

Can someone please share a typical case for D?

Can someone tell me the TC where this code gets runtime error[submission:266046159]. Approach:We can always create any K greedily if its even and is less than the max value achievable.Then we know that if we swap two values,then twice of their difference contributes to K,so i have halved the value of K to x.Then using binary search on set to get maximum possible value which we can attain to decrease x, we swap both of them and decrease x.We repeat until x is 0.

System test when?

I am so ded rn,i forgot that iterator doesn't copy it points to original location,which returned ruined my C solution.

solved my first div 2 problem in contest (just A but still happy about it)

on problem A and D why don't you just say index instead of number ... what a problem statement

Will this solution work for E?

https://www.ideone.com/PApbrg

I was very slow in contest, I assumed that E would be harder and started it very late.

LOL it turned out to be easier

I solved B using calculus first derivation. I know this is not intended solution. But I would love to share here because it was interesting.

Noticed that the answer should be

$$$a(n - k) + bk - \frac{k(k + 1)}{2} + k$$$

you can use gaussian for this computation.

You can modify this to make $$$f(k) = -\frac{1}{2}k^{2} + (b - a + \frac{1}{2}) k + an$$$

You can look for the maximum $$$f(k)$$$ using first derivation, $$$k = b - a + \frac{1}{2}$$$. Don't forget to check the answer with $$$k = 0$$$ and $$$k = min(n, b)$$$ as the boundaries.

Statements were a bit unlcear, in my opinion.

• A: The statement says "highest number", but I can't find where 'number' is defined. There are many other places where 'number' is used, but some of them are referring to 'number of pages'. There is no place where it says 'number' actually means $$$i$$$. It is not obvious that '$$$1$$$-st book' means that the 'number' of the book is $$$1$$$. It could be more clearly stated, or could just be replaced with another word.

• B: The story of the legend is misleading. Like, Bob organized the promotion to "attract customers", but he's trying to sell them even more expensive than they currently are? Why would customers be attracted to such promotion?

I just realized my solution for F had a very terrible sieve, that it didn't really puncture composite numbers — so instead of listing all prime divisors for integers in range $$$[1, 10^6]$$$, it lists all divisors instead (this will affect the DSU process later on since the number of edges are increased drastically).

Solution: 266020666

Can it be uphacked? Thanks in advance.