yo participated in this contest rated but i did not recive any elo? and when i go to my acounts contest all it shows codeforces round 1013 div 3 as unrated but i did rated whats the issue?

Wow, what a coincidence to see you here again, I hope you can play superbly this time:)

I hope that today I will become a specialist. Hope on this contest.

[^-^]/

Probably not. I assume (based on other such rounds) that, if the problems are too hard, some new, easier ones will be added, or some of the olympiad problems will be simplified

High five we have the same rating and the same goal!

Same

The sooner you get it, the sooner you can solve it. Because, it's just a while loop template. Yes, I'm talking about BINARY SEARCH. Problems 'D' and 'E' — I have solved both using BINARY SEARCH! And again, I am going to be CYAN...

I downvoted, let's see what will happen! ಠ_ಠ

And, u aren't a tester T.T

what happens if i vote?

UPD : Apparently during system testing right now, I got TLE at test case 27 which is unfortunate plz ignore below.

Imagine you're Gleb, and you have to cross a river to reach a point exactly s meters away. Each time you paddle, your current power (or energy) determines how far you go. BUTTT every time you decide to turn around to adjust your course, your power drops by 1 (unless you're already at the minimum power of 1). So, you want to minimize turns because every turn costs you power.

I originally considered trying every possible sequence of moves forward and backward to see which one would land me exactly at point s. this was stupid i am stupid — charles leclerc

Now here's what I actually did:

Good ol dp. The key idea was to base my dp state on a modulus (p) and, for each residue modulo p, store a pair of numbers. These numbers represent the minimum and maximum effective move counts/strokes needed to reach that residue.

Now we can build the state transitions using modular arithmetic. For each residue in the current state, calculate the new residue after a move by taking into account the effect of the stroke. I solved a linear congruence using the extended euclidean algorithm to compute a modular inverse, which was crucial to adjust the equation based on the gcd. Now with rnges I determined the valid range of multipliers (stroke counts) that would keep the move within the acceptable bounds.

To showcas ethe actual movement, I defined two functions: one for forward moves (fwd) and one for backward moves (bwd). The forward function simulates strokes that add distance, while the backward function handles strokes that subtract distance (after a turn). Now if I alternate between these, I can build a detailed map of all the positions I could reach after any number of moves.

The final step was to check if the dp state allowed me to reach exactly point s. I did this by looking at the residue corresponding to s and verifying whether the range of move counts included a valid solution. Since every turn reduces my power, the whole process was aimed at finding a sequence of moves that minimizes the number of turns, thereby preserving as much of my initial power as possible.

Sorry if this is text heavy its hard to explain in text its why I prefer pen n paper

UPDATE : This is based on number theory and there's a better solution using bitset by Edu175 below mentioned as well.

I just used bitset 312451438

I increased 1e5 to 1e6, thinking I increased 1e6 to 1e7 and got one penalty. Think about me, though. T_T

why to code miller-rabin when we can find primes in √n*log(log(n)) by sieve only.

prefixsum + dp is suffice.

you don't need segment tree bruhh , just prefix sums are needed.

prefix sum on current and previous row

My position dropped from 600 to 1100 in last 30 min.. T_T

why? :(

Sieve of eratosthenes is a typical way to identify all the primes in some range of values. That's all that's really needed for the implementation part of things, and the idea is just knowledge of gcd and lcm defintions.

I was planning to try pushing the formula, but I simulated the example and discovered the pattern

Definitely TLE

What is your logic for F

Oh no, i was solving different problem for 1 hour

sorry

try long long?

in C when you have even n you cannot interchange 1 element at a time. What i mean to say is you need just 1 element to be at its own place for example 1 on a1 or 2 on a2, etc but when we have even n we swap 2 numbers or say we need to leave 2 numbers as is to its place for example -> 1 3 2 5 4 6 -> here 1 is left on one place but now we need to swap all other numbers say swapped 6 and 4 above and it becomes -> 1 3 2 5 6 4, but now you see the real problem any two consecutive numbers together because they will take their own place in some cycle which means -> 1 3 2 5 6 4 will become -> 4 1 3 2 5 6 and it is not allowed, i hope you were satisfied with my answer

Here's a mathematical explanation wrote it quickly in notion

hey, yea i agree it was very simple for an E. Removing the constraint would make it a proper E. (BTW even i am a huge fan of Kurt <3 )

for(ll i=2;i<=n;i++){
    if(spf[i]==i){
        ll l=1;
        ll r=n;
        ll a=0;
        while(l<=r){
            ll mid=(l+r)/2;
            if(mid*i<=n){
                a=mid;
                l=mid+1;
            }
            else{
                r=mid-1;
            }
        }
        ans+=a;
    }
}
cout<<ans<<endl;

this got acc you don't have to check for i<MAXN because for any i that is larger than n you'll never find mi such that mi*i<=n

your submissions look sus

First of all, we want to distribute the k desks as evenly across the n rows to keep the maximum number of desks in a single row as small as possible. The most desks we then get in a row is occupied = ceil(k/n). In such a row, we have free = m — occupied empty slots. Once more, we want to divide/break up the occupied slots in that row as evenly via the free slots: max_bench = ceil(occupied / (free+1)) (if you have x empty slots, you can break the desks in the row up into x+1 benches).

ksun42's solution uses the fact that ceil(x/y) = int((x+y-1)/y).

You just not in offical standings table because you are untrusted. As you know, it is because of some reasons. not offical == unoffical

It doesn't means you are unrated. Just you are not in standings.

Your rating will be updated after we run with hacked submission's datas, same as others

Unless you checked as Unrated Participation...