Educational Codeforces Round 122 [Rated for Div. 2]

3 years ago, # |

In problem D, Ceil(log2(bi)) is star I think because this many steps is required to change 1 to the particular number.It's just a theory based on 1 + (1/1) => 2 + (2/2) or 2 + (2/1) etc.. for each number.

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prabhav_

3 years ago, # ^ |

No this fails for b[i] = 7

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codemastercpp

3 years ago, # |

Video Solutions for A-D for anyone looking

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osource

3 years ago, # |

← Rev. 2 →

D was a nice question. 0-1 Knapsack DP hidden behind a good observation — max operations on any index i can be atmax 12. So if k>=12*n, ans would be sum of array c otherwise apply standard DP for 0-1 knapsack.

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VaibhaveS

3 years ago, # ^ |

Is it possible to get the cost to transform a number to b[i] using greedy?

something like,

k=1
while(a<b and k<=a):
   while(a+a/k > b) k+=1
   steps+=1
   a+=a/k
if(a==b)
   return steps

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arun07

3 years ago, # ^ |

Tried the same and got WA on 2nd test case. So, I'm assuming it will not work. BFS is the way, I think

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bluescorp

3 years ago, # ^ |

I thought of BFS too but implemented array of sets approach along with DP. My submission https://mirror.codeforces.com/contest/1633/submission/144766502

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quiet_one

3 years ago, # |

+11

problem d approach is as knapsack dp problem ?

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lazy_loop

3 years ago, # |

someone pls help! https://mirror.codeforces.com/contest/1633/submission/144748512

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Gary2005

3 years ago, # |

← Rev. 2 →

Problem F is nice .

But I wrote the dfs order as index ,and debug for about 40 minute (how can this pass the first 12 tests ???).

And I found this silly mistake in the last 30 seconds ..

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Gary2005

3 years ago, # ^ |

+39

Share my solution here :

Spoiler

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Apoorv_Rathore

3 years ago, # |

Problem A is exact copy of this TOPCODER ROUND. Even the number 7 is same.

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codersattu

3 years ago, # |

Solution for C ?

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aksht

3 years ago, # ^ |

Notice that the sum of k<2*10^5.

That means the solution should be O(n).

To Check who wins- find number steps needed by both to win (that means if the character kept attacking how many rounds are needed to defeat the monster and vise-versa)

character_steps = ceil(monster_health/character_damage)

monster_steps = ceil(character_health/monster_damage)

if character_steps<=monster_steps then character wins (First chance is of character so <=) Checking who wins is o(1)

To find all possible combinations in the way we can use the coins- we can go from 1 to coins (i) and use i as the coins used for damage boost and use coins-i as the coins for health boost and add the boost to the damage and armor. We can do this in O(n)

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rsg23

3 years ago, # |

← Rev. 2 →

It seems that D can be passed in $$$\mathcal{O}(nk)$$$ time

Maybe it will be better if the time limit is changed to 1s ...

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3 years ago, # ^ |

+11

It's not actually $$$O(n\cdot k)$$$ where $$$k\approx 10^6$$$.

Since the number of operations needed for any $$$b_i$$$ to get constructed is at max $$$12$$$, if $$$k\ge 12\cdot n$$$, you can simply print $$$\sum_{i=1}^n c_i$$$. Otherwise you can use a knapsack dp solution so solve the problem in $$$O(n\cdot k)$$$ time where $$$k\lt 12\cdot n$$$.

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3 years ago, # ^ |

why?, if k >= 12*n then we can take all coins?

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3 years ago, # ^ |

If $$$k\ge 12\cdot n$$$, it is possible to make all $$$a_i$$$ equal to $$$b_i$$$. Hence we can collect the value $$$c_i$$$ for all $$$i$$$'s.

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3 years ago, # ^ |

I mean i Know that but how, some example or proof/link?

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Ray_Striker

3 years ago, # ^ |

Simply use DP to calculate that.

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3 years ago, # ^ |

← Rev. 2 →

Compute the answer for each $$$1\le i\le 10^3$$$ and check for the maximum. It will come out to be $$$12$$$.

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.Miikka.

3 years ago, # ^ |

But in fact,some guys solved this problem in $$$O(n\times k)$$$ and I cannot hack them. Are there any stronger cases than this? QaQ

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3 years ago, # ^ |

I can't open the link you gave, it says "access denied".

Can you provide me with such a solution?

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.Miikka.

3 years ago, # ^ |

Sorry,that is my problem.

This!

$$$n=1000,k=10^6,b_i=1,c_i=10^6$$$ cannot hack him. QaQ

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vrintle

3 years ago, # ^ |

← Rev. 2 →

Try this one: https://mirror.codeforces.com/contest/1633/submission/144717276

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3 years ago, # ^ |

← Rev. 4 →

Actually the total number of operations is $$$\approx 10^9$$$, so it might be possible for a solution to pass under 2 second time constraint. Plus the operations done inside the test case loop are simple enough. The only thing that user needs to optimize is the space used which cannot be $$$O(n\cdot k)$$$.

I've tried making test cases with max constraint and the program runs perfectly within 2 second time limit.

DitaMirika,rsg23

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3 years ago, # ^ |

Hey, you stole my comment! Just kidding.. =)

The time constraint is $$$2$$$ seconds by the way ))

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3 years ago, # ^ |

Um, I hadn't read your comment until now actually.

The time constraint is 2 seconds by the way ))

Thanks fixed!

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rsg23

3 years ago, # ^ |

Yes I understand that, but what I want to express is you can pass this problem even if you use the 01-pack dp for every $$$k$$$ but not just when $$$k\le 12 * n$$$

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3 years ago, # |

+32

How to solve E?

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Quicksvolex

3 years ago, # |

What is the hacking phase??

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3 years ago, # ^ |

There might be some solutions which either shouldn't pass according to the given constraint or are actually wrong but the systems tests don't cover the cases where they fail.

Codeforces gives the users, an opportunity to identify such solutions and hack them (submit a valid test where they fail). Obviously it does not guarantee $$$100\%$$$ strength to the new test set formed after hacking.

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joshuajy03

3 years ago, # ^ |

I hacked my own solution for B rip.

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3 years ago, # |

Really nice contest.

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3 years ago, # |

Today is the day I will learn Knapsack, Thanks for the round

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Crypto

3 years ago, # |

In problem D how do I find the number of operations for each number I tried to do it but got WA here is my code: https://ideone.com/Fm0aPE

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fonmagnus

3 years ago, # ^ |

I precompute them using BFS. Find the shortest path from 1 to all nodes

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3 years ago, # ^ |

You can write a brute force dp for this. Just make sure you have an idea about dp solutions.
Consider $$$dp_i$$$ stands for the minimum number of operations required for making the number $$$i$$$ starting from $$$1$$$.

Then, it can be seen that before forming the number $$$i$$$, we might have had a number $$$j$$$ such that the value of $$$\lfloor \frac{j}{x} \rfloor$$$ was equal to $$$i-j$$$ for some $$$x$$$ (because that's the only way $$$j+\lfloor \frac{j}{x} \rfloor$$$ will become equal to $$$i$$$). Let's iterate on $$$j$$$ too. We will then need to find the value of $$$x$$$ (if it actually exists).

Finding the value of x

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3 years ago, # ^ |

I will understand and try this thanks.

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nyet

3 years ago, # |

← Rev. 2 →

In problem D , I assumed x should be a divisor of a[i] out of nowhere, and now I won't reach expert because of this.

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3 years ago, # ^ |

Well it says ai / x floored (note the floor function brackets) which indicates that x doesn't necessarily divide ai.

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nyet

3 years ago, # ^ |

Yeah, I read it initially, but while implementing missed it completely.

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Mksiyak

3 years ago, # ^ |

Same Here, but i figured it out after 1 hour and 3 wrong attempts :(

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MrOtter

3 years ago, # ^ |

← Rev. 2 →

Same thing bro

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DJDD01

3 years ago, # |

← Rev. 3 →

Help needed in Problem D. This is my solution What I did was calculate operations for every b[i] and then select those with best (c[i]/operations) in decreasing order. But this gives WA. Any ideas why?

EDIT: The way I calculate minimum operations is probably wrong. I'll just wait for the editorial.

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3 years ago, # ^ |

I think you answered your own question. You can't solve 0-1 knapsack with greedy.

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DJDD01

3 years ago, # ^ |

I am actually taking the entire points and not just a fraction of it. I wrote greedy because I sorted the numbers in decreasing order based on their worth (points divided by operations) and selected them greedily. Maybe I should remove greedy knapsack from the comment if it is confusing.

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3 years ago, # ^ |

What I am understanding from your comment and code is that you are trying to solve this knapsack which is impossible with greedy.

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ershivasoni

3 years ago, # ^ |

Never be greedy :p

c — 10, 4 operations — 2, 1 k = 1

according to you 10/2 is better than 4/1 but we cannot take 10/2 as limit is k=1.

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DJDD01

3 years ago, # ^ |

I am checking for that as well, if operations are greater than what is allowed I skip it.

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seraphimstreets

3 years ago, # ^ |

No the point is that you cannot solve the knapsack problem with a greedy approach. Take for example values — 3, 2, 2, 0 and cost — 2, 2, 2, 1 and capacity=4. If we take elements greedily, we will take the first element because 3/2 is more than 2/2. However, the optimal solution is to take the second and third element,so this approach fails.

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abdzag

3 years ago, # |

I've wasted too much time and energy on D thinking b_i<=1e6. Sad

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magga

3 years ago, # |

I like the idea of E but its implementation is way out of my skillset :(

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justaninactivealt

3 years ago, # ^ |

Klog(m^2) passed for me with maps in 2000ms. Hack it, maybe? :p

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fonmagnus

3 years ago, # |

I have a question, does an unsuccessful hacking attempt on Hacking phase gives us any penalty in some ways?

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NavierStokes

3 years ago, # |

In Problem D, number of operations to convert 1 to x is (index of MSB in x) + (number of set bits in x) - 1

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ershivasoni

3 years ago, # ^ |

No need of fancy math. run n*n bfs as {1,1}

{i, steps} -> {i + (i/j), steps + 1} for all 1<=j<=i

states = 10^3, TC: 10^6 bfs

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3 years ago, # ^ |

← Rev. 2 →

+10

No need for fancy BFS. Just run 2 for loops =)

int w[1001];
for (int i = 1; i < 1000; i++)
    for (int j = 1; j <= i; j++) {
        const int t = i + i / j;
        if (t <= 1000)
            w[t] = min(w[t], w[i] + 1);
    }

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CarryNotKarry

3 years ago, # ^ |

How do you know this regularity?

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DankTank

3 years ago, # ^ |

Have u tried? I did the exact same thing first but got 2 WA. Then I switched to BFS.

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lazy_and_slow

3 years ago, # ^ |

I did this in contest and got wrong answer, brute force implementation after contest gave AC, are you sure this is correct? (maybe I implemented it wrong)

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aviral10

3 years ago, # ^ |

This isn't correct:

Try for 19: 10011
Index of msb: 4
setBits: 3
Your ans: 4+3-1=6
Actual answer: 5
How to reach: 1->2->4->8->16->19 [19 = 16 + 16/5] in total 5 steps

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_nathan_drake_

3 years ago, # ^ |

I did this mistake too.

15 = 1111 Here ans ≠ (3) + 4 — 1 = 6

But optimal soln is 5 (1->2->4->8->12->15)

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CarryNotKarry

3 years ago, # |

Happy Chinese New Year for every Chinese!Hope all of you will get good grades in xcpc(icpc&ccpc) in this year!

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akshitm16

3 years ago, # |

+65

What happened to rainboy??

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justTrolling

3 years ago, # |

Is standing broken?

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msarnecki

3 years ago, # |

Any hint for problem E?

I got WA on test 7, I mapped all queries, if a query occurs an odd number of times I sort all edges by abs(x — wi) and find MST with Kruskal's algorithm.

My code: https://mirror.codeforces.com/contest/1633/submission/144721913

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jakubd

3 years ago, # |

+39

When you get traumatized by interactive binary search problems xd

meme

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SagarPrasad

3 years ago, # ^ |

I didn't know 2 can be done with binary search.

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Angel_Dust

3 years ago, # |

E was too hard to code

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samyakjain677

3 years ago, # |

In problem D, Why I am getting tle on testcase 12 https://mirror.codeforces.com/contest/1633/submission/144763541

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jakubd

3 years ago, # ^ |

← Rev. 4 →

$$$k = min(k, 12n)$$$

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hwei002

3 years ago, # ^ |

after considering this condition, still got TLE under python3 .... here is my code

BTW, I found another code, which was accepted during contest, but also TLE now...

really weird ....

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jakubd

3 years ago, # ^ |

+14

codeforces.ml ? kinda sus

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tvcv901

3 years ago, # ^ |

Spoiler

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3 years ago, # ^ |

← Rev. 7 →

Actually you could solve it even without that optimization =)

I use the following justification:
1. Knapsack performs $$$k * n = 10^6 * 10^3 = 10^9$$$ operations.
2. CPU speed is 2.4Ghz or $$$2.4 * 10^9$$$ operations per second.
3. If we access the data properly, CPU will utilize it's fast cache and registers instead of waiting for data from RAM (RAM is 10 times slower than CPU cache).
4. We have 2 seconds limit on the problem, so we have some room for overhead computation and round trips to RAM.

144830143

By the way, I like the code that I got after incorporating the idea of short circuiting.
That is, we don't do $$$dp$$$ if $$$k$$$ is large enough.
This short circuiting allows us to reuse unoptimized code from above without using magic constants k = min(k, 12 * k). Everything works out by itself and the code would work even if we have a different problem with larger constraints, so we don't need to recalculate new constant for 12 * k.

Short circuiting

if (k >= accumulate(b.begin(), b.end(), 0)) {
    cout << accumulate(c.begin(), c.end(), 0);
    return;
}

144830112

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abhay.raihal

3 years ago, # |

← Rev. 2 →

same code, but different compiler, one gave wrong answer but one got accepted(after contest)

144765410 144698088

Anyone knows the reason ? Help will be appreciated :'(

Upd:My second submission also gave wrong answer later in main tests due to float, so I think

double >>>>>>>> float

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18jad

3 years ago, # ^ |

← Rev. 5 →

It's because you are using float and float is not accurate, i had the same problem from few days my solution got accepted in c++ 17 but not in c++ 20 and someone told me why this happened you can check the reason float fail sometimes to give AC here
PS: changing float to double in your code give AC, so when u want a floating number next time use double!

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abhay.raihal

3 years ago, # ^ |

thanks man !!

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tiasmondal

3 years ago, # |

Can someone clarify this? In E when I used edges as vector of vectors , it gave TLE. In some AC submission I saw they have used structure for edges. When I used structure instead of vector, it passed. Why this behaviour?

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Bliss_of_comprehension

3 years ago, # ^ |

I submitted with vector of vectors and it ran in TL. 144786518

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-Yoshimitsu-

3 years ago, # |

I didn't know there is a space optimized version of solving KnapSack

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Priyansh31dec

3 years ago, # |

← Rev. 2 →

Is O(NK + KlogK) the intended solution for E or is there something better? I am processing the queries in a sorted order and just updating the edges in the MST only when the sorted order of the edges changes in the next query. It is easy to see that the sorted order of the edges won't change more than M * M times.

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aryanc403

3 years ago, # ^ |

+31

You can use promo code CountingSortUsingBitset for O(logK) off. You may need Find_Next as well.

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NotImplemented

3 years ago, # ^ |

Thank you, how Find_Next may help?

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hwei002

3 years ago, # |

for problem D, anyone succeed to get accepted by python3? why I always got TLE? here is my code

BTW, I found another code, which was accepted during contest, but also TLE now... weird ....

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3 years ago, # ^ |

I was able to make it up until 8'th test with Python3: 144836343
Exactly the same code with PyPy 3.7 (64 bit) passes for 187ms 144836442

Probably the difference is in the input speed.

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vaibnak7

3 years ago, # |

← Rev. 2 →

In Problem D test case

4 4
1 7 5 2
2 6 5 2

cant we make [1 1 1 1] -> [7 1 1 1] in 4 steps, as conversion from 1 to 7 takes 4 operation and then answers should be 6+2+2+2 = 12. Whats wrong here

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vaibnak7

3 years ago, # ^ |

Never Mind, Seems like I have completely misinterpretted the question

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ershivasoni

3 years ago, # |

Problem E. There can be at max e*e event points where we need to re-calculate MST. giving e*e*e*log(n) as preprocessing time.

for given x we need to find y<=x in above event point list and do some calculations.

I did it with y = map.lowerbound(x) -> Gave TLE complexity O(k * log(e * e))

People did it with presorting x and using two pointer to get y for given x- TC: O(k * logk)

SInce log factor of sorting is better than log factor of map, hence i got tle.

I think problem shouldn't have such tight constraint that sorting pass but map.lower_bound() fails. What do you guys say?

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3 years ago, # ^ |

You could presort y and lower_bound for every x query (without using two pointers and presorting x), map just has terrible constant factor.

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krishnakumar

3 years ago, # |

← Rev. 2 →

Can someone tell why this method of finding the minimum number of operations to convert 1 to b[i] is wrong :-

This is what i've done — Im finding the nearest power of two (call this num) for b[i] and then convert num nearest power to b[i]. To do that first i find the difference between b[i] and num and update num greedily such that num + [num/x] <= b[i] till i get num == b[i].

This is the piece of code for finding this minimum number of operations.

Code

for(int i=0; i<n; i++)
{
    cin>>b[i];
    weight[i] = (ll)log2(b[i]);
    ll num = (ll)pow(2,(ll)log2(b[i]));

    while(b[i] != num)
    {
        ll diff = b[i] - num;
        ll x = num / diff;
        if(num / x > diff) x++;
        weight[i]++;
        num += num/x;
    }
    //weight[i] is the min moves

}

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palindroam

3 years ago, # ^ |

Greedy doesn't work because you cannot make sure that by incrementing with the largest possible move will lead into minimum possible weight.

Example

For 31
Your ans : 1 2 4 8 16 24 30 31 , 7 steps
Optimal ans : 1 2 4 8 16 21 31 , 6 steps

So apply dp here too (similar to coin change).

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Yoav

3 years ago, # |

+20

I usually don't write such stuff, but I must say I'm really disappointed with the pretests for problem C. For such a short mathy problem, You'd expect to give constraints in a way that don't give an overflow with a pretty standard solution. And if there is a standard solution that gets an overflow, it's only right to give a not-so-difficult-to-construct pretest that prevents it.

For educational rounds it's even more important to have strong pretests on the easier problems, due to the fact each problem weighs the same.

Thanks, a CM that instead of getting to master for the first time, gets a -95 :(

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weakestOsuPlayer_244

3 years ago, # ^ |

Yeah already got a contender for the worst problem of 2022.

Hey but let's rejoice we can just move ahead with int128 from here onwards and forget about these silly problems the setters come up with.

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3 years ago, # ^ |

but longlong is enough here

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justaninactivealt

3 years ago, # ^ |

Tbh, most of the times there is at least one problem with ridiculously high constraints in Edu rounds.Moreover, almost always the pretests are weak to encourage hacking. Maybe you have been just lucky to not get FSTed before.

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Yoav

3 years ago, # ^ |

I haven't seen that many problems with ridiculously high constraints (that can't be handled by long long) in Edu rounds. Can you give examples? Moreover, it sounds extremely weird to me that the pretests are weak to encourage hacking. After all, the purpose of the pretests is to validate solutions... But I agree that I might have been somewhat lucky to not get FSTed until now.

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https://mirror.codeforces.com/contest/1633/submission/144683828

3 years ago, # ^ |

+13

Yoav What do you mean by "You'd expect to give constraints in a way that don't give an overflow with a pretty standard solution"? Pretty standard solution here is to use long long and it got AC here

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Yoav

3 years ago, # ^ |

Look at my hacked solution. In my opinion my approach is quite standard, and is correct but gets an overflow. After checking again the constraints it is easy to see there can be an overflow. So if the constraints are made in a way that there could be such overflow, there should be pretests that cover it.

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Sheber

3 years ago, # |

← Rev. 2 →

Good contest and interesting tasks !

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gagan_sharma

3 years ago, # |

Are editorials released?

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tgp07

3 years ago, # |

Can someone please help me figure out why my solution for D isn't working? 144791596

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Newtech66

3 years ago, # |

In my opinion, 2 seconds TL on problem D was too much. It let some unoptimal solutions pass.

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3 years ago, # ^ |

Yeah, I saw many O(n*k) solutions passing

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_wandering_brain_

3 years ago, # |

Hey There!

Can someone please tell me what stress testing is and how to use it during the contests?

I received this suggestion from some revered person :"You can try stress-testing to find the test cases you're wrong at" I often apply the right logic in contests but end up with failing a few test cases.

Best, Wandering Brain

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3 years ago, # ^ |

You can write naive solution (even exponential one), generate some small test cases and compare your solution against naive on those small cases. To make it efficient you should automate the process of generating random test cases and comparing your solutions.

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pearlz

3 years ago, # |

← Rev. 3 →

I should have done problem C in python :( (Wrong Answer on testcase 13 due to Overflow)

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nis_c_hat

3 years ago, # |

Hello guys. I'm new in codeforces and this was my first contest. I solved one problem. Will my rating increase?

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Perry_Prasad_Platypus

3 years ago, # ^ |

Yes

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3 years ago, # |

+12

FSTforces :(

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pearlz

3 years ago, # ^ |

+100 delta to negative delta

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3 years ago, # ^ |

Got FST on both C and D today.
If this had happened on Topcoder, it would not be unusual because TopCoder emphasizes on proving your solutions with your own edge cases as their questions are usually more about implementation (imho).
But this is certainly not what I used to expect from Codeforces rounds as the questions are usually more about coming up with the idea and being guided about whether that works or not with pretests.

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marvel_coder

3 years ago, # ^ |

what is FST?

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3 years ago, # ^ |

Failed System Test.

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zykkezi

3 years ago, # |

fstqwq

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mnkqc

3 years ago, # |

I figured out that in D, knapsack was the way to go, since we can interpret that weights are no. of operations to reach each element. I found the required way to reach all elements in nlogn, and since N was <=10^3, thought knapsack would work, but k was <=10^6. I do not know how to lower this. any hint would be appreciated

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3 years ago, # ^ |

Using DP on the number transitions, you could pre-compute the number of steps required for each of the numbers (1->1000), and because this was kind of logarithmic (max. 12 steps), you could effectively reduce 1e+6 to around 12000, which is manageable.
My Submission (has FSTed)

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mnkqc

3 years ago, # ^ |

thank you, I'll try again. Why wasn't this came in my head earlier

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quantau

3 years ago, # ^ |

The maximum operations required to obtain any b will be logarithm in nature so the actual max value of $$$k$$$ will be $$$nlogn $$$. Hence, the problem becomes knapsack DP with complexity of $$$n*n*logn$$$. Check out my solution, it did not FST

144759233

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Yaria

3 years ago, # |

← Rev. 3 →

+10

I suddenly realize that in Educational Rounds solving ABD with more penalty time will be ranked lower than solving ABC with less penalty time. So, in such rounds instead of solving harder problems it's important to guarantee the solved easier problems are correct.

I initially "solved" ABCD but just got FST on C which makes me feel sad. (C makes me realize "long long overflow in C++". Perhaps I should use Python 3 for all such large integer problems.)

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senR

3 years ago, # ^ |

C gave another chance to Python users to mock CPP users ;-;

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i.am.pratik

3 years ago, # |

← Rev. 2 →

IMO, This is one of the worst contests ever. Felt like the tasks as well as the test was created just as a formality. Please don't create this type of contest.

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Omja

3 years ago, # ^ |

Can't disagree with you, I always highly rate the educational round but this one was far below standard.

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devss

3 years ago, # |

-18

I'm down-voting this contest and this post , because of weak testcases for problem C and problem D , that causes me negative delta! From ~2800 to ~8000 ! This content damaged my rating a lot. But actually the problems were interesting.Thanks.

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rohitkumar0510

3 years ago, # ^ |

Couldn't agree more! Weakforces

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italianDefence

3 years ago, # ^ |

I dunno know why you were downvoted. If the pretests are gonna be this weak they might as well not include them at all and leave participants to make submissions on a gamble.

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vijay30046

3 years ago, # |

-6

Weak test cases for D!!Even an O(nk) solution passed.

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strange14

3 years ago, # |

144799151 Problem C, my submission is failing on TC 13. Can anyone please point out what's wrong?

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cybermonk

3 years ago, # ^ |

Its overflowing when you do the operation $$$(attacks - 1)*dm$$$ . Very weak testcases this time

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cybermonk

3 years ago, # |

Pretests for Problem C were very weak, I had an overflow in my code and it passed, if pretests were strong then I would've gotten a WA but usually codeforces tells us if we have an overflow and I'd have been able to AC.

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procodgok

3 years ago, # |

Two Contests back : I might become CM Now : Specialist , I am coming

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3 years ago, # |

Can someone please explain why this submission for D is failing test case 21?
The Submission
Thanks.

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MikeMirzayanov

3 years ago, # |

+65

To not keep you waiting, the ratings are updated preliminarily. In a few hours/days, I will remove cheaters and update the ratings again!

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weakforces

3 years ago, # ^ |

-21

Down with codeforces From now on weakforces

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3 years ago, # ^ |

Chill dude, it was just one mediocre contest with weak pretests.

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gagan_sharma

3 years ago, # ^ |

Can you make a post when this testing is done as well?

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italianDefence

3 years ago, # ^ |

Is there any update? Are the ratings updated?

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handleme

3 years ago, # |

The test data in question C makes the variable overflow of "long long" in some incorrect writing, which is nicely designed

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