RobinFromTheHood's blog

Codeforces Round 974 (Div. 3)
By RobinFromTheHood, history, 2 years ago, In English

Hello Codeforces!

Greetings from Nottingham, England! We are delighted to invite you to Legend of Robin Hood, a round inspired by our local folklore.

Codeforces Round 974 (Div. 3) will start on Sep/21/2024 17:45 (Moscow time).

You will be offered 6-8 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have a rating of 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ICPC). Thus, solutions will be judged on preliminary tests during the round, and after the round, it will be a 12-hour phase of open hacks. After open hacks all accepted solutions will be rejudged on successful hacks.

You will be given 6-8 problems and 2 hours and 15 minutes to solve them.

Note that the penalty for the wrong submission in this round is 10 minutes.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a trusted participant of the third division, you must:

  • take part in at least five rated rounds (and solve at least one problem in each of them)
  • do not have a point of 1900 or higher in the rating.

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.

Problems have been prepared by ChairmanFMao, Filikec and RobinFromTheHood.

We would like to thank:

  1. Vladosiya for brilliant coordination, and for improving all problems.

  2. 18o3, cry, ikrpprppp, pavlekn for orange testing.

  3. employed, raztun, Proof_by_QED for purple testing.

  4. Alenochka, FBI, macaquedev, Non-origination, JuicyGrape, umezo for blue testing.

  5. gbula, Pa_sha, 1165MOHITSINGHAL, NotHosam for Div. 3 testing :)

  6. MikeMirzayanov for Polygon and Codeforces platforms.

  7. You for participating in the round!

Good luck!

UPDATE: Editorial is out!

Image by Zmarlen with our thanks!

  • Vote: I like it
  • +248
  • Vote: I do not like it

Comments Comments (136)
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2 years ago, hide # |
 
Vote: I like it +59 Vote: I do not like it

As a setter, if you cheat I'll be hiding under your bed

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2 years ago, hide # |
 
Vote: I like it +43 Vote: I do not like it

as a tes, give me contribution

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2 years ago, hide # |
 
Vote: I like it +3 Vote: I do not like it

As a Div.3 contestant, RobinFromTheHood orz.

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

But is it AI proofed?

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2 years ago, hide # |
← Rev. 2  
Vote: I like it -13 Vote: I do not like it

edited

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    2 years ago, hide # ^ |
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    Vote: I like it +41 Vote: I do not like it

    I think u guys should already stop with this gpt's discussion. It has nothing to do with the round, as usage of the AI for solving is not allowed in codeforces.

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2 years ago, hide # |
 
Vote: I like it +12 Vote: I do not like it

"You" are in black and red. so funny

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2 years ago, hide # |
 
Vote: I like it +10 Vote: I do not like it

after such a bad contest in today's div2 hoping for some good delta positive in this div3 tomorrow

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

When did Robin last wear glasses?

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2 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

The 3rd trial to be a pupil from Div.3 I hope it works this time :D

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    2 years ago, hide # ^ |
     
    Vote: I like it +1 Vote: I do not like it

    you'll be able to do it bro, only 20 more rating needed. Just solve 4 problems and you should be able to get at least 20 from it

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2 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

I have love for coding but I get frustrated when I can't solve any problem... What is the solution?plz help me.

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    2 years ago, hide # ^ |
     
    Vote: I like it +1 Vote: I do not like it

    leave the problem, Do some other problem, or do something else like go play with your dog or go make some coffee.

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

+1500 rate ? lezgo

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2 years ago, hide # |
← Rev. 2  
Vote: I like it +4 Vote: I do not like it

WOWSER! Cant wait for this round! Im gonna shoot an arrow straight through probem A in the first 2 minutes!!

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2 years ago, hide # |
 
Vote: I like it +6 Vote: I do not like it

the robinhood gang in the picture is awesome. xd.

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2 years ago, hide # |
 
Vote: I like it +6 Vote: I do not like it

This contest will get rating from specialists and will give it to newbies?

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2 years ago, hide # |
 
Vote: I like it -8 Vote: I do not like it

As ChatGPT I can confirm I cannot solve the problems.

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2 years ago, hide # |
← Rev. 2  
Vote: I like it +18 Vote: I do not like it
this can't be just me
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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

I turned blue in the last div2 so now this is unrated T_T

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    2 years ago, hide # ^ |
     
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    Congrats bro. But your graph is just unbeleivable. Do you have a second ID or did olympiads before?

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      2 years ago, hide # ^ |
       
      Vote: I like it +2 Vote: I do not like it

      Thank you so much ^^ I've actually won a silver medal in the Italian Olympiad in Informatics and I'm currently training to be potentially chosen for IOI :3

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2 years ago, hide # |
 
Vote: I like it +3 Vote: I do not like it

I am so excited for the contest!

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2 years ago, hide # |
 
Vote: I like it +2 Vote: I do not like it

I'm trusted, why I'm "out of competition"?

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Can't wait for this round!

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2 years ago, hide # |
 
Vote: I like it +11 Vote: I do not like it

one refresh costed me 10 mins :(

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2 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

delayforces?

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    2 years ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    Yeah. But why the 10 min delay

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      2 years ago, hide # ^ |
       
      Vote: I like it +1 Vote: I do not like it

      It happens often a while last year, I remember the main reason is ddos attack (I mean today's round even has less people than yesterday's div2, so it can't be overcrowded)

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2 years ago, hide # |
 
Vote: I like it +4 Vote: I do not like it

10 minute penalty for entering the contest.

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2 years ago, hide # |
 
Vote: I like it +4 Vote: I do not like it

delay forces?

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

why is it pending brof

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2 years ago, hide # |
← Rev. 2  
Vote: I like it +6 Vote: I do not like it

Lol rk1 solved till F in 12 mins and now it shows User is disabled
Live vac ban

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2 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Types of headache meme was supposed to be about Div2 D

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2 years ago, hide # |
 
Vote: I like it +12 Vote: I do not like it
  1. Not sure if it was just me being bad today, but I fell into a dilemma of absolutely hating G and feeling it should be something of my usual favorite and specialty a.k.a. simulation implementation. It still stings that the whole difficulties of G were based on the implementation of the simulation alone, the statements told you all you needed to do upfront...
  2. Is there anyone having a clean $$$\mathcal{O}(n \sqrt{n})$$$ solution for H? At first I TLE'd when doing $$$\mathcal{O}(n \sqrt{n} \log{n})$$$, and the reduction of the $$$\log$$$ is pretty tedious...
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    2 years ago, hide # ^ |
    ← Rev. 3  
    Vote: I like it +28 Vote: I do not like it

    nvm

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      2 years ago, hide # ^ |
      ← Rev. 2  
      Vote: I like it +13 Vote: I do not like it

      Oh yeah, neat trick! Thanks!

      UPD: Why editing? I think the trick is actually reasonable, it's just not trivial on sight.

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        2 years ago, hide # ^ |
         
        Vote: I like it 0 Vote: I do not like it

        Hey if you've solved H using Mo's can you please see where this code fails? I do not know where it's going wrong 282342460

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          2 years ago, hide # ^ |
           
          Vote: I like it 0 Vote: I do not like it

          At a glance, probably the sweeping order was wrong. Usually, I tend to check for moving the L/R pointers to increase the range (L to left, R to right) first before moving to decrease the range (L to right, R to left).

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            2 years ago, hide # ^ |
             
            Vote: I like it 0 Vote: I do not like it

            uhm, my way of moving the pointers worked for queries on distinct values in a range. the only change i did here was pertaining to frequency change, that's all

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        2 years ago, hide # ^ |
         
        Vote: I like it 0 Vote: I do not like it

        you were asking about sqrt solution, so I thought I wrote an off topic

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          2 years ago, hide # ^ |
          ← Rev. 3  
          Vote: I like it +1 Vote: I do not like it

          Maybe that'd apply to other, but I absolutely won't mind an offtopic that is resourceful to everyone. So yeah, I have no issue with it at all, keep it!

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    2 years ago, hide # ^ |
    ← Rev. 2  
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    You are not the only one in your opinion of Problem G. I also thought that it had straight forward idea for stimulation with the implementation difficulty as the deciding factor for the contestants.

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    2 years ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    For $$$O(n \sqrt n)$$$ in H, you can use Mo's while keeping track of frequencies. At the end you just need to check if there exist any odd frequency element or not. My Submission

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      2 years ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      I looked at your code and realized I have always implemented Mo's algorithm inefficiently, i.e. excessive case-handling when moving from a block to another. That's on me I guess.

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

just please someone tell me why in problem D last case the boy can't start at day 1 i read the statement 789 time and still have no clue

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    2 years ago, hide # ^ |
     
    Vote: I like it +1 Vote: I do not like it

    if the brother visits on day 3, he will overlap with 3 jobs. if he visits on day 1 he will only overlap with 2 jobs

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    2 years ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    The boy is the one to maximize visits. You gets three overlaps at $$$3$$$ ($$$2$$$, $$$3$$$ and $$$4$$$), while only two ($$$3$$$ and $$$4$$$) at $$$1$$$.

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      2 years ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      sorry I meant 2 if the the boy starts at day 2 he will intersect with (1,3) (2,3) (4,8)

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        2 years ago, hide # ^ |
         
        Vote: I like it +1 Vote: I do not like it

        Starting at $$$2$$$ only covers the day range $$$[2, 3]$$$. It doesn't touch $$$[4, 8]$$$ at all.

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Auto comment: topic has been updated by RobinFromTheHood (previous revision, new revision, compare).

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    2 years ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    i think because it recalculates the number of overlaps for every possible starting day of the visits.

    you will have to use prefix sums to avoid TLE

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2 years ago, hide # |
 
Vote: I like it -15 Vote: I do not like it

can someone please tell me what am I doing wrong in Problem C 



#include <bits/stdc++.h>
using namespace std;
 
 
bool check(int mid, vector<float> &arr) {
    float avg = 0;
    int sum = 0;
    for(int i = 0; i < arr.size(); i++) {
        sum += arr[i];
    }
    sum += mid;
    avg = static_cast<float>(sum) / (2 * arr.size());
    int pop = 0;
    for(int i = 0; i < arr.size(); i++) {
        if(arr[i] < avg) {
            pop++;
        }
    }
    return pop > arr.size()/ 2? true: false;
}

void solve(){
    int n;
    cin>>n;
    vector<float> arr(n, 0);
    for(int i = 0; i < n;i++) {
        cin>>arr[i];
    }
    if(n <= 2) {
        cout<<"-1"<<'\n';
        return;
    }
    sort(arr.begin(), arr.end());
    int left = 0, right = 1e6;
    int ans = -1;
    while(left <= right) {
        int mid = left + (right - left)/2;
        if(check(mid, arr)) {
            ans = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    cout<<ans<<'\n';
}
 
 
int main() {
    ios_base :: sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
 
#ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
#endif 
    int t;
    cin>>t; 
    while(t--) {
        solve();   
    }
    return 0;
}
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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

I suck at questions like D. Can anyone give any guidance about how to approach questions involving ranges.

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    2 years ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    I solved with prefix sum range or whatever thats called. 282320011 Now i am also not that good at range problems cuz G is also a range type of problem and i couldnt do it

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    2 years ago, hide # ^ |
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    I used sort and heap to solve these 282360541. Overall the algo will also work with bigger l,r up to 10**9...

    But I guess it's overkill because I don't notice the limit is small and can use other method to solve

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      2 years ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      heap is just priority queue right ? if that is the case can you just briefly talk about your method thanks.

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        2 years ago, hide # ^ |
         
        Vote: I like it 0 Vote: I do not like it

        yes you can say heap and priority queue are interchange-able.

        Let's say I brute force of all time point from 1 to n-d+1 and calculate how many job intervals it collides.

        First step is sort the job intervals [l,r] So the intervals I pushed in the heap at time point i, will no longer needed to check them again at time point i+1. So I can apply sliding pointer technique.

        Second step is the pushed data in the heap is only r for each interval, because the job intersect with time point [i, i+d-1] is the heap size, and to maintain the heap you need to check the smallest r that r<i will need to be pushed out (they're no longer intersect with the current time point).

        Bonus: if the time point limit is 10**9 and intervals limit also 10**9, then I won't for loop from 1 to n-d+1. I will check i = left interval.

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Can someone please tell me why does my solution for D(282356603) get TL? I thought the complexity was O(n+k)

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2 years ago, hide # |
 
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What do we need to do to solve H?

I think we just need to check whether all elements from l to r are the same. If they are, we then check whether the length is even or odd

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2 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Casino H

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2 years ago, hide # |
← Rev. 3  
Vote: I like it +3 Vote: I do not like it

how come when I view "common standings" I have some rank (e.g. 1500), but when I view "friend standings" I have a worse rank (e.g. 2000) ? I made sure to have "show unofficial" disabled both times.

If I look back at previous contests, it would appear that the one in "friends standings" is the actual rank.

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2 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

what is the upper bound on x in problem C? my intial submission uses 1e12 and I fear hacks so I resubmitted with 1e18. can someone tell if 1e12 works, then why does it work?

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    2 years ago, hide # ^ |
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    I think $$$4 \cdot 10^{11}$$$ is one upper bound. The problem can be solved straight-up from maths.

    In short, we need a person with $$$x$$$ golds at maximum to feel dissatisfied (so that there are at least strictly more than half people having no more than $$$x$$$ each). We need to satisfy $$$x \lt \frac{s}{2n}$$$, or $$$s \gt 2nx$$$, with $$$s$$$ being the total amount of golds from everyone. Clearly $$$2nx \le 2 \cdot 2 \cdot 10^5 \cdot 10^6$$$, or $$$2nx \le 4 \cdot 10^{11}$$$.

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Not good not bad contest D destroyed me today need to upsolve it

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2 years ago, hide # |
 
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Is C intended to be binary search ? cuz i thought it is easier with maths 282287861

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2 years ago, hide # |
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The statement for problem C is probably the most try-hardest to be confusing statement I have ever seen. Why "half of the average"? On top of that, real number, really?

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    2 years ago, hide # ^ |
     
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    since everything is "strictly less" or "strictly more", you can simply round up.

    for example, strictly half of the population = half of the population rounded up.

    one's wealth is strictly less than half of the average wealth = average must be strictly greater than double that person's wealth. that person's wealth is always an integer so you can treat average as an integer too

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2 years ago, hide # |
 
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how did you solved Problem C with binary search? anyone?

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    2 years ago, hide # ^ |
     
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    basically you want to test whether amount of gold makes robin hood appear. to do this, you make your guess (mid) and then find the average wealth: ceil(total + mid)/n.

    now you simply compare this to the median wealth: a[n/2].

    if the average is greater than 2*median, this pot will summon robinhood: r = mid.

    if the average is less than or equal to 2*median, this pot will not summon robinhood: l = mid+1.

    return l after binary search terminates, it is the smallest pot that will summon robinhood.

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    2 years ago, hide # ^ |
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    (Edit: I apologise for misreading the question. I mistakenly wrote the solution for D instead)

    Let's say the guest's stay starts at day $$$i$$$. Then, the guest leaves at day $$$i+d$$$ (day $$$i+d$$$ is excluded). Which jobs are excluded? Only those which either stop before day $$$i$$$ starts, or start after or on day $$$i+d$$$. Note that there is no intersection between these 2 sets of jobs. Thus you can sort $$$l$$$ and $$$r$$$ separately and solve for each day using binary search.

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2 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Read the problem B wrong cost me 30 minutes. After that screw up implementation at problem D cost another 30 minutes. Then all I have is ~ 15 minutes for E, so I choose to stare in the ceiling to recollect my dumb mistakes...

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    2 years ago, hide # ^ |
     
    Vote: I like it +2 Vote: I do not like it

    Why not trying? E is (somewhat) basic Dijkstra though, it might still be tough for a green but not impossible.

    Moral of the story: just fight to the very last seconds. You never know how much power you got if you resigned to fate.

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      2 years ago, hide # ^ |
       
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      Yep thanks for the advice. Probably I need to train in VC to strengthen my mental till the end of contest time limit...

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      2 years ago, hide # ^ |
       
      Vote: I like it +1 Vote: I do not like it

      Finally, It took me about an hour to AC the problem. But maybe I can do it faster next time.

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2 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

omg double dijkstra on E ??? Niceee.

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2 years ago, hide # |
 
Vote: I like it +5 Vote: I do not like it

is it only me or i feel like H was easier than E in terms of the idea (I didn’t solve either of them :I)

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2 years ago, hide # |
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Can someone kindly help me in finding the mistake with my solution for question E: https://codeforces.me/contest/2014/submission/282371555

I have performed dijsktra with two distance arrays and two visited arrays (with horse and without horse), once from 0 and once from n-1. I am getting WA on test case 3 and not able to debug the mistake.

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    2 years ago, hide # ^ |
     
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    Realized overflow from the test cases. It was a typo — used long instead of long long :'(. Thanks a lot if you have taken a look at my submission in the mean time.

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2 years ago, hide # |
 
Vote: I like it +4 Vote: I do not like it

Dijkstra problem was amazing!

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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

How to do the range query for H efficiently?

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2 years ago, hide # |
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H << E

H can be easily solved by Zobrist hashing 282353218

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2 years ago, hide # |
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I can't figure out why my C++ solution is timing out on problem E.

I'm duplicating the graph with the half weights and having edges between the two copies at all the horse locations. Then I run dijkstra from either end and check each vertex to see who gets to it last.

I'm thinking it should be O(E*log(V)) and while I've doubled E and V, that shouldn't be enough to time it out. I've tried tweaking things to account for any extra time that might have been spent copying or allocating, but no luck. Anyone else have an idea?

https://codeforces.me/contest/2014/submission/282381456

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2 years ago, hide # |
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In $$$H$$$ answer is YES for a query if in that subarray all the numbers have even frequency otherwise NO but how to check that?

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2 years ago, hide # |
 
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Does using float types in problem C will get hacked? sorry for my poor english

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2 years ago, hide # |
 
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My best contest so far

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2 years ago, hide # |
 
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For Problem E , I think of a solution Its giving TLE

so for person to meet at k  ,  1 -> k   n -> k  are two disjoint path with just one element common = k
so we can binary search over it to find first path where the common element is just one

logn * (nlogn)*mul  -> mul is ig in my implementation is 2 or 3 which leads to TLE

Accepted Version : Same idea with Binary search but reducing Complexity in Check function which i just realised when i am writing this Comment

Instead of computing sets which can be visited from 1 and n again and again , just do once and count
complexity :  logn*n   -> Accepted

Link : https://codeforces.me/contest/2014/submission/282393166

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2 years ago, hide # |
 
Vote: I like it -12 Vote: I do not like it

HOW CAN THE PROBLEM D. Robert Hood and Mrs Hood test case 1 9 2 4 7 9 4 8 1 3 2 3 has output 3 4. Shouldn't it be 1 4 because valid days are 1 to n-d (2 2 2 1 1 2 2)?

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2 years ago, hide # |
 
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Please help, I am getting TLE even after using Dijkstra in E. Submission https://codeforces.me/contest/2014/submission/282399768

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2 years ago, hide # |
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pls put the announcement link with the contest i could not find it

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2 years ago, hide # |
 
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I don't know if it is a formatting issue, but showing the first lines of the statements in italics was very confusing. I assume the italics at the top are just part of the plot, but in this it seems the first paragraph of the actual statement was in italics

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2 years ago, hide # |
 
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Could someone please tell me why am I getting a TLE in this Submission

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2 years ago, hide # |
 
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The Dashboard was so Hard as Steelboard!!!

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2 years ago, hide # |
 
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Why is it unrated?

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2 years ago, hide # |
 
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I have a friend who wants to ask why unrated.

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2 years ago, hide # |
 
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why is system testing frozen?

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2 years ago, hide # |
 
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what is vjudge btw tons of submission from named vjudge unrated chinese accounts are making trouble i guess

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2 years ago, hide # |
 
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Nice picture.

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2 years ago, hide # |
 
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estimated ratings

A — 800

B — 800

C — 1000

D — 1600

E — 1800

F — 2000

G — 2200

H — 2000

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23 months ago, hide # |
 
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As a test, I have been tested

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