### Hello Codeforces!

$$$\newline$$$

**Do you know that eating cheese can make you code faster than the running speed of a Tyrannosaurus REX?**

It's been a while, today GlowCheese and I are delighted to invite you to participate in Codeforces Round 963 (Div. 2). This round will be **rated** for all participants with a rating lower than **2100**.

Our round will start on Aug/04/2024 17:35 (Moscow time). You will be given **6 problems** and **120 minutes** to solve them.

This contest is brought to you by Code Mely, a Vietnamese community for Computer Science. Reach out to us here.

Special thanks to:

- Akulyat for his wonderful coordination and translating our statements.
- Our testers: Umi, lmqzzz, danghuyhau, _Fake4Fun, AkiLotus, LetterC67, McPqndq, Evirir, Lyde, Hihihah, Yugi.Hacker, hhoangcp, t1r3d_, mp1309, phungthienphuoc, Etohari, aeonary_will_win_VOI25, hydroshiba, zeena., bibimoni, kodomo_tachi, StenSoSmol, Habaxl, and DOCUTEE.
- DeMen100ns, shine_, Asamai, Miko_UwU and nohandle_5 for joining with us.
- MikeMirzayanov for amazing platforms Codeforces and Polygon.
- KFC for their chicken nuggets.
- You for participating. $$$\newline$$$

**Score distribution:** $$$500 - 1000 - 1500 - 2000 - 2750 - (2500 - 1000)$$$

Hope to see you in the final standings!

**UPD1.** Editorial is out!

**UPD2.** Congratulations to the winners:

Div.2:

Div.1 + Div.2:

As a tester, TrendBattles orz

DA GOD, ORZ

My birthday is on contest day :)

Amazing!

Happy Birthday in Advance.

Me too

Happy Birthday :)

Happy Birthday!

Happy Birthday :D

Any idea when will points distribution updated

Orz

As a participant, I am here before the testers.

Oops, I'm late.

No, I'm late.

As a tester, I was too busy doing last-minute testing that I run out of good pick-up lines for contribution pleadings.

Also, I wanna say a lot but the authorities will shut me dow-

As a tester, danghuyhau orz

Hello, I don't have much time to tell this but as a tester I've known quite some things. Specifically for problem D you can solve usingdajpoiiadaodald;/;

Ahem...

coughYes my lord, as a tester I am programmed to prioritize security and integrity of Codeforces rounds, and problem spoiling heavily damages the experience and trust of the majority of round participants, which can be considered inappropriate, or even harmful to the Codeforces community as a whole. I can offer suggestions for alternative positive problem solving methods that can be practiced for enhancing one's ability to compete in such Codeforces rounds instead. Please let me know if you would like me to assist you with any further requests.

No, I'm just having a flu.

get well

currently working on decoding dajpoiiadaodald

1 Hint: It's first query in first test

lookes like string problem + dp, it's my weakness

as a tester, danghuyhau orz

as a tester, danghuyhau orz

As a tester, I did nothing but still be mentioned.

as a participant, Asamai orz

em chao dai ca

orz Asamai, nga`i

Orz

As a participant,

SpoilerI wasn't involved in any problem, thank you for noticing :D

As a participant, TrendBattles orz

as a random person on cf, TrendBattles orz

i thought it's

SpoilerChatGpt solved 0 probelm

As a testers, I tested all over the problems

Hope for reaching Master after after after this round.

I think you need a top-30 finish to gain this much. Good luck!

Good luck!

As a participant, danghuyhau orz

Hope to reach master before i die !

yo wtf

What extension are you using to differentiate everyone’s comments in your image?

I think everybody's just screen captioning one after another.

yo wtf

Imagine if one day your wish were granted, and codeforces permanently shut down the next day. You'd be immortal, but you probably wouldn't be immune to aging. That sounds like hell on Earth.

Hope to Solve 3 problems

As the tester, I tested.

wooow impossible i don't believe that i didn't notice that testers test

orz

as a participant, bibimoni orz

SpoilerCan you spoil 1 problem? <(")

no.

=))

omg based frieren the slayer pfp

Best wishes to my purple friends, hope they become orange!

As a tester

As a participant, I know you are a tester

Wow,Marisa

patchy my beloved

https://i.imgur.com/Ym6joaX.png (imgur links not working D:)

Seems that nobody like my joke

And what's the joke?

In the spoilerThe T-REX are d.e.a.d.

Oh, I got that joke about trex. I thought there's some hidden joke in that picture, you've posted earlier.

as a dino, :skull:

did you know that eating cheese can't make you code faster than the running speed of Jeffrey Epstein?

as a tester, i did literally nothing...

ngl it was the shittiest joke I've ever read

as the weakest tester who tested the roundhydroshiba orz

why target him?

we love him

why does the round number doesn't show up in cf contests page? image

Amaterasu!!!!!!

Hmm "GlowCheese and I are", not " GlowCheese and me are".

Watch your 6 homeboy I'm coming for your as-

Hoping for a top 7000 finish :'(

So cool!

As a Faze fan, I can confirm karrigan is the World No.1 IGL

Hope the best round! ^_^

Make you code? Shouldn't be make your code?

Hope to reach Pupil after this

As a participant, i'm on hopium :D

i want to be tester

as a participant i wish to be a tester someday.

As a tester, I currently have no contributions.

ORZ

What is ORZ can you explain?

orz, lowercase, looks like a guy bowing

ORZ uppercase doesn't mean anything but people like to capitalize orz. If you want a capital version of orz, OTZ works.

oh my gad glow cheese round!!!!

As a participant, chikien2009 orz

What do you mean?

orz

Mely orz

As a GlowCheese fan, I will participate

will there be pics of the authors ?

sorry but... GlowCheese and *I =))

Oops, sure

As a participant, hhoangcp orz

As a participant, khoad orz

I hope I can achieve green after this contest if I have time to participate.Good luck to everyone!

As a contestant, I wish i can change be master

Announcement with a meme at the beginning? Something new... (for me)

Still preparing cheese for the contest

As a participant , I hope everyone get an increase in rating after this contest ;)

Done eating 5 kg of cheese...

SpoilerA faster submission doesn't need to be an AC ;)

hope to solve C and get +ve delta

I think it will be Permutation_force :(:(

Do you know that eating cheese can make you code faster than the running speed of a Tyrannosaurus REX?is it some kinda hint or wot??

spoilerLoveSquirrel pls comeback cf im begging u

Hoping you guys will make this contest fantastic

I hope I get a 1000+ rating after this contest (●'◡'●)

orz

GlowCheese, thanhchauns2 oooooorzzzzzzzzzzzzzzz

oh, Vietnamese contest!!!

Hope to reach Candidate Master after this contest

All the best

should increase rating at least by a bit

rating distribution when?

Hope to get 1250+ after this contest!

score distribution?

As a participant,

confidentialI don't know what the heck ppl r talking about hereAs a participant, I hope I'm not choking in this contest

I hope everything goes well. I can play as well as I should in this game.And....stO Akulyat Orz.

Hoping to reach pupil today

Score distribution ??

As not a tester, how to become tester?

As a Newbie, hope, I'll become

Pupil.CM go go?

I wish will turn green today :)

Will solve 5 problems today

newbie to codeforces. any advices?

don't rely on the codeforces IDE

wish work out E today.

hope for Salah7_a <3

This made me happy <3

XD

great round!

good A B C, never again ^_^

may be problems are easy or difficult compare to other div2 round but i find it much difficult need to practice more

d and e are much harder than most div 1 and 2, let alone a div 2

My dislike to this contest is addressed not to the authors, who did a great job, but to Akulyat, who completely fucked up problem difficulty estimation. Awful work as a contest coordinator.

I'm sorry that the gap between C and D turned out to be that huge

Based on the testers' performance C seemed harder and D easier, so we assumed that the gap was reasonable

No luck. Unfortunately, it happens quite often to div2 rounds. Probably something is wrong with the way problems are being assessed.

I think it is a bit harsh to blame the coordinator for that as the problem difficulty assessment is based on the testers feedback (and the tester pool seems diverse enough). It turned out that D was maybe a little bit on the hard side (but it still got around 600 solves in contest which is reasonable imo). The issue might be that C is "too easy" as it has 7k AC so the round feels speedforces for some contestants. However, I believe this is more of an issue with the div2 format rather than problem selection.

Problem C is not "too easy". it has 7k AC out of which at-least 3k are cheated solutions.

SpeedForces!

I'm seeing a lot of people who were not even able to solve 3 problems in div3 or 2 in div2 in their past few contests have solved 1500-rated problem in this one. Are they suddenly becoming intelligent or what? If they are cheaters, will CF do something for them, or are they also silent like LC? I'm not aware of what CF does to catch them. Do anyone know, if they have a strong plag check or not?

C is easy for a 1500

yes it was slightly easy for a 1500 rating but still no that easy.

There are obviously a lot of cheaters. I think problem C is harder than usual, but a large number of people passed it in an instant.

that is why I'm suspecting

someone like me?

Actually, You have the strength of IGM.

actually, i'm an IGM

Multiaccount actually

casework orz

C and D are too far apart.

I was very mad...

Is D binary search?

Yes.

Dang, I kept getting WA and wasn't sure if it was correct. How does the check function work? I took all numbers greater than the one we're checking and did two pointers.

I binary searched on the answer and verified with dp. I kept getting MLE and it would've probably TLEd but basically create a new array $$$b$$$, $$$b[i] = (a[i] >= mid ? 1:-1)$$$ and now you have to find the maximum sum of the elements outside the intervals where you placed $$$ n/k - (n mod k == 0) $$$ disjoint intervals on the array $$$b$$$. If that sum is greater than 0, the mid is valid.

I think problem C has extremely weak pretest,

I passed pretest it without binary search O(n * k)

even wrote hack myself but anyway I didn't hacked :)

UDP: it actually passed systest wtf I'm so lucky

You don't even have to binary search, O(n+k) is possible, so yeah

i think the main tests is a lot harder

is binary search the intended solution? I solved it without binary search in O(n). At least I passed the pretests, I hope it doesn't FST.

Uphacked as per your wish :)

Thank you for the great animations in F!

Ordered sets got TLE in D, I suppose BIT is the intended solution? Didn't have time to implement at the end

UPD: nvm, in worst case for values $$$k$$$ close to $$$n$$$ and all unique values of $$$a_i$$$ this approach is both quadratic time and memory, regardless of DS used ;c

think about binary search

Binary search + DP.

Sounds evil, ngl

I did that only. https://codeforces.net/contest/1993/submission/274414638

The only thing that screwed me is I guess, the constraint mentionining final array should be more than 'k'.

I Please elaborate how to handle that edge case ?

the speed of solving B, C is normal or bcos of cheating?

plenty of cheaters these days. Just compete with yourself, learn new concepts and upsolve. Dont bother too much about the rankings or ratings.

seeing alot of people solve very fast/difficult problems (not normal) while u still didn't get it yet, not good feeling

F2 = F1 + CF710D

D is so hard that only 400 (unofficial excluded) solve it!

What a joke! 7500+ solved C — 500+ solved D

Is problem D DP? Whether to retain current element or not and then update median?

I really liked problem D, the fact that the $$$i$$$-th element of the resulting set must be in a position $$$j \equiv i (\mod k)$$$ is a really really cool invariant, and the resulting DP to check if a median is achievable is really nice as well.

Can you tell me the mistake in the C problem. vector =vl rep=for loop I = cin<< ; ~~~~~ //this is the code ll solve() { ll n,k; I n>>k; vl arr(n); rep(i,n){ I arr[i]; } ll l=0,r=k; sort(arr.begin(),arr.end()); ll at=arr[0]; rep(i,n){ if((arr[i]-at)%(2*k)>=k){ r=min((arr[i]-at)%k,r); }else{ l=max((arr[i]-at)%(k),l); } } if(l>=r){ return -1; }else{ ll l2=0; if(((arr[n-1]-at)%(2*k))<k){ l2=(arr[n-1]-at)%k; } ll addd=0; if(l2>=l){ addd=0; }else{ addd=l-l2; } return arr[n-1]+addd; }

} ~~~~~

ARGHH i think i solved F2 but i was about 10 mins too slow

basically from (0,0) -> (a,b) -> it will visit some nodes [k = 1] -> (x,y) -> (a+x, b+y) -> more nodes [k = 2] -> (2x,2y)

it's a (0,0) if it's a multiple of (2w,2h) or x'%2w = 0 AND y'%2h = 0

let W = 2w, H = 2h

let's say you're working with some coordinate (u,v) -> it's a (0,0) iff (u-kx)%W = 0, (v-ky)%H = 0

kx%W = u

ky%H = v

if gcd(x,W) not 1 then kx%W = -u eqv to k(x/gcd)%(H/gcd) = (-u/gcd) (same for the y case)

solve for k for each one (assuming there is a solution) -> k = u*lcm(H,W) + w -> then just check how many k exists between [1..k]

repeat for each coordinate (there can only be at most 10^6 of them)

In D, I tried doing binary search and converting the arrays into 1s and 0s based on if element >= mid. Now we basically need to choose more 1s and 0s. What do we do beyond this? I though of dp but thats n*k

Hint 1Which elements of $$$a$$$ can be the first element of the resulting array after performing the operations.

Hint 2More generally, which elements of $$$a$$$ can be the $$$i$$$-th element of the resulting array?

Hint 3Using your "0/1" formulation, if you pick the $$$i$$$-th element of the original array as the $$$(i \mod k)$$$-th element of the resulting array, can you calculate the minimum number of "0s" till now using dp?

You are AWESOME!!!!

nvm.

Didn't read the question carefully.

In your states, if you try to minimize the number of remaining $$$0$$$, you don't need to remember how much elements you've taken (you will have to take at least $$$n \bmod k$$$ and there is no point in taking more). You just need to remember if you took at least one element in the case $$$k$$$ divides $$$n$$$. So you have $$$2n$$$ states.

I think it's a cool observation, congrats to the problem author :)

Damnnn thats actually cool

That is probably sufficient to pass, but FYI it can be optimized to a linear DP with $$$n$$$ states. We want to pick, sequentially, an element that is in a index with

`1 mod k`

,`2 mod k`

, ...,`n mod k`

(using 1-indexed and considering the case where`n mod k!=0`

). The problem arises only when we "wrap around" and start taking`1 mod k`

,`2 mod k`

, ...,`n mod k`

, ...,`0`

,`1 mod k`

and start taking more than the limit. This can be avoided by simply preventing the transition from`0`

to`1 mod k`

and instead setting`dp[i]=(a[i]>=x)`

for every index $$$i\equiv1 \pmod{k}$$$. SubP.S. It's even possible to optimize it to $$$k$$$ states.

can anyone tell why this failed, this is hurting my brain 274401938

Take the case

`{1,2,10}`

If you have to make two moves why not make it on the largest even element and in that way you can take all other element in one move

you need to run backward 1 more time cause there are some case where running backwards have smaller moves

Oh okay, i get it now, when we need to perform 2 operation of swapping then we will do it on the max even number, this would nullify the need for (+=2 moves ) for rest of even, hence atmost we will repeat only one move

Think in this example: 1 2 6 Your code gives answer=4 but the answer is at most number of evens+1, because you could turn the greatest even number into an odd number and it would be bigger than the rest of evens. If the greatest odd is less than the greatest even you can turn the greatest even into odd in 2 moves, otherwise it will take only 1 move.

anyone explain B and C solution

How C?

Denote $$$\text{ans} = \max(a_i)$$$, iterate all lights and find the next $$$x \ge a_i$$$ that the light in question will be on, then assign $$$x$$$ to $$$\text{ans}$$$. Check if $$$\max(a_i) + k < \text{ans}$$$.

I tried using this and its giving a slight wrong answer in the test case of +1 in some test , i dont know why. ~~~~~ ll solve() { ll n,k; I n>>k; vl arr(n); rep(i,n){ I arr[i]; } ll l=0,r=k; sort(arr.begin(),arr.end()); ll at=arr[0]; rep(i,n){ if((arr[i]-at)%(2*k)>=k){ r=min((arr[i]-at)%k,r); }else{ l=max((arr[i]-at)%(k),l); } } if(l>=r){ return -1; }else{ ll l2=0; if(((arr[n-1]-at)%(2*k))<k){ l2=(arr[n-1]-at)%k; } ll addd=0; if(l2>=l){ addd=0; }else{ addd=l-l2; } return arr[n-1]+addd; }

} ~~~~~

.

I don't mean any offence towards anyone! Just a joke

Congratulations!

you might get boosted to CM

I couldn't finish C but I was thinking of segment tree with range update... was thinking of sorting input, use smallest value to find all start points until its greater than largest element. Use lower_bound to find where each element is compared to the ranges of the smallest element. Determine the offset of this value to the nearest first element start ranges

eg

if the min offset and max offset diff was >= k, then it was probably invalid. Could only think of inserting all of first element's active ranges using segtree range updates then do the same for all following elements, then query which time appear n times. Not sure what i'm talking or thinking about anymore, fairly new to CF this was tough

What is idea B?

Except for when all elements are even at first, the common parity of every final array is odd.

if parity of any two num is different then can only be solved by making every element odd..rest think yourself

mathforces :(

A-D din't need any crazy math.(I don't know about the other problems) :)

A: sum(c)(max(0, cnt(c)-a)) for 'A'<=c<='D'.

B: Assume there are both odd and even numbers in a[i] (otherwise answer is trivially 0), because in each operation we can reduce count of odd numbers by at most 1 in each operation, and when there's only 1 odd number we can't change its parity, we need to change all numbers to odd. So we can add the greatest odd number to smallest even number repeatedly, until there's no even number. If we can do this, the answer is (count of even numbers). Otherwise, we can add the greatest even number to an odd number, and add this odd number to all even numbers, then answer is (count of even numbers)+1.

C: Let A = max(a[i]), the number of lights turned on will change with period 2*k after A minutes, and the answer cannot be small than A, so we only need to check range [A, A+2*k-1] using prefix sums.

D: Let's find the answer by binary search. So for certain M we need to change a[i] to 1 when a[i]<M or 0 otherwise. Then we need to choose some remaining indexes {i0=0, i1, i2, ..., i_r, i_(r+1)=n+1} where i_(j+1)-i_j-1 is multiple of k (0<=j<=r), and sum(a[i_j]) is minimized. We can solve this by dp: let dp[i] = the minimum sum if we let i_r=i. Then we have dp[0]=0, dp[i]=a[i]+min(0<=j<i, j%k==(i-1)%k)(dp[j]), the minimum valid sum is min(1<=i<=n, i%k==n%k)(dp[i]).

for c : The formula used to adjust the installation times and determine when each room's light will be on is:

$$$adjustedtime[i]$$$ $$$=$$$ $$$a[i]+k×$$$ $$$⌈ \dfrac{maxa−a[i]}{k} ⌉$$$

Where:

$$$a[i]$$$ is the installation time for the $$$i-th$$$ room.

$$$k$$$ is the interval at which the light toggles.

$$$maxa$$$ is the maximum installation time among all rooms.

$$$⌈x⌉$$$ denotes the ceiling function, which rounds $$$x$$$ up to the nearest integer.

D is somehow similar to https://codeforces.net/problemset/problem/1486/D

E: Thinkless $$$O((n+1)^2(m+1)^2 (2^n+2^m))$$$ cooked me (I need one more optimization to get AC)

btw, in this kind of problems, why not limit $$$t \le 10$$$ (small constant) or use single test, instead of complicated bound for one input?

Single test leads to having too many tests. Since the resources are limited, it leads to worse coverage.

If we do $$$t \le 10$$$, then we can have $$$10$$$ large tests or $$$10$$$ small tests, meanwhile we want to have just a couple of big tests (otherwise the complexity would be multiplied by $$$t$$$) or lots of small tests (for better coverage)

TBH, $$$\sum (n^2+m^2) \le 500$$$ seems very heuristic and I think it's difficult to determine which complexity is acceptable.

This problem is D2E, so the limit of #tests isn't have to tight I guess.

The limit was set in such a way so as not to let determine the complexity based on it

Is judging cooked? Some early submissions are stuck in queue (including my A)

i think so i have same issue with problem B still in queue

same, all problems i passed pretest are in queue

any idea why it doesn't work for B ? https://codeforces.net/contest/1993/submission/274399672

1

3

1 20 40

your code returns 4

correct answer is 3

1 20 40 -> 41 20 40 -> 41 61 40 -> 41 61 101

Thanks, should've either 1) hardcoded the answer in case i find the bigger one and just subtracted from the number of evens the ones i ve already cleared and made it +1 2) when i add a bigger number take it from the end so it is bigger than anything in the middle

why are some submissions skipped

Cheaters must be stopped before it runs into Cheatforces

Problem C by 7k+ people. It's a sure shot cheating case. Please do look at it :)

Either that or just really weak pretests on C (lots of O(nk) sols passed)

I do agree that the pretest were weak, but still 7k+ is huge. I saw several almost same solutions, so ig it was leaked.

Can anyone tell why O(N*K) working for problem c https://codeforces.net/contest/1993/submission/274391283

Problem D: Find error in my logic, it is giving WA (176th number) on test case 2

274418415

Hey everyone

can someone please tell me why my solution for Problem B — Parity and Sum is WA on test case 2

My c++ solution:

I think the intution is same as per the editorial

my submission

whenever you would do ans+=2, you can just break and print out the amount of even numbers+1, because if that ever happens, then its optimal for it to happen to the biggest even element, after which you will always be able to just do it in 1 step. i hope i explained it well

Consider this case :

4 ----> 2 6 10 1

according to your code answer will come 5, but answer is 4 1 + 10 = 11, 11 + 10 = 21, 21 + 6 = 27, 27 + 2 = 29

Sorry I didn't get it why ans will be 4 (I think you're missing updating 4, we just updated 1,10,6 and 2)

4 is the array size.

Link of editorial don't work

update : it is working now : )

The test of Problem C is too weak. Many people accept this problem with a solution which has time complexity of O(nk).

Problem B too https://codeforces.net/contest/1993/submission/274429127 This code runs upto 10^16

There are already more than 100 fails on my uphack on C. I can't even imagine how many on-contest solutions would've failed if this was on systest.

Thank you the authors and

`Code Mely`

for the awesome contest! I enjoyed it a lot!I think the cases were too weak for problem B and C. Many submissions of O(n*k) for C are accepted and for problem B https://codeforces.net/contest/1993/submission/274429127 this code runs for 10^16 how? MikeMirzayanov thanhchauns2

It's Gcc optimization

it gets compiler optimized...

Omg Dominater069 Orz. Can you please tell me more about this. I don't know what GCC optimisation is

Those loops don't make any actual effect on anything, and the compiler can notice it just by analysing the code. So therefore it just entirely removes the loops so they become non-existent.

Thanks...i got it

Why does it work, i can't understand.

Can't open tutorial. I'm getting this error:

`You are not allowed to view the requested page`

The same

thanhchauns2 GlowCheese please make it public.

Fixed!

I solved ABC under an hour as a 1000 rated. C definitely seemed easier than almost all Div 2s I’ve done.

Wow I gained over 200 elo from this. Crazy

2 pupils in top 5 beating all candidate masters sounds fishy. Moreover they aren't new accounts either. The guy at rank 1 had 10k rank in last contest and 8 in last to last. Don't you think it's too strange of a variation?

that's because he tried the rainboy style solving in the reverse order

still 2 pupils having lgm perf just screams alt accounts

B, D, and E are great problems <3

Problem C was not qualityfull at all whereas B was enough good

Thank you to the authors and coordinator for the nice problems!

D is a great problem, but F2's trick is trivial :(

i think it'll be better if swap(E,F), E is much harder than F for me.

Funny, while testing I keep baffling myself at "why don't people solve E? it's straightforward as heck".

Actual contest seems to show the opposite of my opinion, heh...

As a random dude, phungthienphuoc orz :)

I wasn't able to solve D in contest because I didn't think in terms of binary search and then converting the array to binary and then solving it using DP.

Is it a known technique, which I am not aware of or are there more questions like this?

I want to solve them and learn as it isn't intuitive for me yet.

I think the data of problem C is a bit weak. I used the method of O(nk) with a little trick and then passed problem C.

.

Bruh, it is skipped because he submitted it again, and only the last submission is judged in the contest

However, my newly submitted methods are all O (nk) in complexity, but with an optimization, such as removing duplicate elements

I hate Median

7k people solve C? That's impossible. I almost become expert, but I didn't solve C and get -80 scores, that really hurts : (

Anyone please tell my c ques passed in n*k loop. Is it correct or wrong.

I wrote a code earlier for light switches ques i made a approach of using arithmetic progression formula a+(n-1)d to get the n for every element and if it is even then i break the loop....it was working but there are large testcases where i dont know at what limit to break the loop to return -1...please help and tell if it is even possible to use this approach or is it only noobish...

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ //this is code

## include <bits/stdc++.h>

using namespace std; typedef long long ll; typedef vectorvl; typedef vector vvl; int main() { ios::sync_with_stdio(false); cin.tie(0);

ll t;cin>>t; while(t--){ ll n,k; cin>>n>>k; vl v(n,0); ll maxi=0; for(int i=0;i<n;i++){ cin>>v[i]; maxi=max(maxi,v[i]);

}

} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ it works for input given but for last one it return 100 i don know why....