interesting score distribution

culver0412 orz

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.

Hopefully I solve a,b and c .

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

If I solve a,b,c in first hour I give my team eidia else Mohamed Hassan daeeeeef

good luck, hoping for +delta

Hopping for +delta

OMG Mikhango!!!!

Potential SpeedForces on the way!

sounds like CodeForces Round 879 Div. 2

https://mirror.codeforces.com/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

my rating is below 1000. should i participate in this?

I am on the verge of becoming a specialist. Lets hope for the best !!!!

Let's do it..

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

excited for this ,lets goooooo

Pre-Eid Day Contest , Participating

GLHF, hope I become expert today

Getting good vibes for this round

Meow.

Lol, we thought too hard, the problems were pretty good and balanced in my opinion

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.

good logic used today

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

mathforces strikes again

darn I was an hour late! still fun contest tho

It was super gang bang

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://mirror.codeforces.com/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;
        }
    }
}

would have been great for me if i had not wasted time in B

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.

Nice overall round with +ve delta forces. I think this round was a great opportunity to gain some positive delta bcz the problems were a bit easier than a normal div 2 round. Kudos to the tester who gave us the hint to read all the problems bcz I think there was a good chance an average newbie coder could have also solved D if he was a bit sincere on the clock(unlike me).

why its not working???? PROBLEM D

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; 


#define getar(ar,n) for(ll i=0;i<n;++i) cin>>ar[i]
#define show(n) cout<<n<<'\n'
#define all(v) v.begin(), v.end()
#define br cout<<"\n"
#define pb push_back
#define nl '\n'
#define yes cout<<"YES\n"
#define no cout<<"NO\n"
#define ret return
#define ll long long
#define ld long double
#define sza(x) ((int)x.size())

const int mxN = 1e5 + 5;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
const ld EPS = 1e-9;

void solve(ll tc) {
    ll n,k;cin>>n>>k;
    ll ar[n];getar(ar,n);
    ar[0]+=k;
    if(n==1){
    	cout<<0<<nl;
    	return;
    }
   	ll mx=0,sum=0;
   	vector<ll> rghtMx(n,0),rghtMxFrq(n,1);
   	rghtMx[n-1]=ar[n-1];
   	for(ll i=n-2;i>=0;--i){
   		if(rghtMx[i+1]==ar[i]) rghtMx[i]=rghtMx[i+1],rghtMxFrq[i]+=rghtMxFrq[i+1];
   		else if(rghtMx[i+1]<ar[i]) rghtMx[i]=ar[i];
   		else rghtMx[i]=rghtMx[i+1],rghtMxFrq[i]=rghtMxFrq[i+1];
   	}
   	for(ll i=0;i<n;++i){
   		if(i==n-1){
   			if(mx>=ar[i]) cout<<n-1;
   			else cout<<0;
   		}else if(i==0){
   			if(rghtMx[i+1]<=ar[i]) cout<<0;
   			else cout<<1;
   		}else{
   			ll tot=ar[i],ans=0;
   			if(mx>=ar[i]){
   				tot+=sum;
   				ans=i;
   			}
   			if(rghtMx[i+1]>tot){
   				++ans;
   			}
   			cout<<ans;
   		}
   		cout<<" ";
   		mx=max(mx,ar[i]);
   		sum+=ar[i];
   	}
   //	br;
   	//for(auto it:rghtMx) cout<<it<<" ";
   	br;
}

int main() {
    
    #ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
    freopen("output.out","w",stdout);
    #endif

    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    ll tc;//=1;
    cin>>tc;
    for (ll t = 1; t <= tc; t++) {
        solve(t);
    }
    return 0;
}

System test when?

testing please we wanna submit unsolved

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

Can anyone give the test case for problem C where this submission is giving wrong answer 266027942

UPD: Nvm got it

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

Ayyo... Why am I getting Idlesness error in C??? 266040778

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.

Great Contest. Absolutely Loved it. This was my first time solving Div2 B & C

why does this solution to E give wrong answer? Can you suggest a countercase?

The wording threw me off so bad for A and D that it took more time to solve A than it took to solve D. v_v

Can anybody point out the error in my code of c or give me a test case where my submission is failing?

Submission id is 266045808

Best contest for me.

Can someone suggest me how to plan to solve the correct questions?? Since many of you are saying that D and E were both easier than C. But I solve in order and presume that E>>D>>C in difficulty so no point in reading it. I solver A and B under 30 mins, figured the logic of C easily but it took me 2 hours to implement it and AC'd after the contest was over. I am still a newbie (hopefully will become Pupil now) so can you guys please suggest me how to plan to solve the right questions??

i was doing Binary seach on the problem B for optimal k which gives me the max profit but getting wromg answer i dont know which case is not being handled pair<ll,ll> possible(ll n,ll a,ll b,ll mid,ll initial) {

ll val=mid*b;
    // cout<<mid<<endl;

    ll val2=((mid-1)*mid)/2;
    ll sell=val-val2;
    ll sell2=(n-mid)*a;

    ll profit=sell+sell2;

    if(profit>initial)
    {
        return{1,profit};

    }
    else{
        return {0,initial};

    }

}

void solve() { ll n,a,b; cin>>n>>a>>b;

ll low=0; ll high=n; ll initial=n*a;

while(low<=high) { ll mid=(low+(high-low)/2); pair<ll,ll>temp =possible(n,a,b,mid,initial);

if(temp.first==1)
        {
            initial=temp.second;
            low=mid+1;

        }

        else{
            high=mid-1;

        }

} ll val=n*b; // cout<<mid<<endl;

ll val2=((n-1)*n)/2;
    ll sell=val-val2;
   if(sell>initial)
   {
    cout<<sell<<endl;
    return ;

   }

cout<<initial<<endl; return; can check my code..

Bro... 2 rating points away from pupil...

round is relatively div2

I can't believe how 6000 coders solved C.I would like account verification to be more difficult. Because it's unreal to watch how many guys who write for the first time pass ABC.

CDE should've been in reverse order, C took too much of my time and couldn't solve D due to a small implementation error. Upsolved E pretty quickly.

Editorial is not published in contest materials

why did my rating drop down by 4 even after solving 3 problems(ABC) today?Can someone explain how does this rating change works? Is it coz I got ranked lower than my rank in previous div 2 round?

Before Eid-day, I am back to CYAN!!!