problem D is hard for beginer contest.

why geometry !!!

Doing all tricks to somehow get the equation, but still cannot solve D. I am still wondering how I cleared physics and maths in higher secondary :\

Does anyone else also feel that loading standing page on Atcoder(so not only on current contest) takes a lot time?

Can anyone explain problem D ??

How do you solve F?

5th place! Happy about that :)

Here are my solutions and thoughts:

A) Check if a < 10 and b < 10.

Solution: https://atcoder.jp/contests/abc144/submissions/8148768

B) Simple loop through possible values of a and b. Solution: https://atcoder.jp/contests/abc144/submissions/8150815

C) Consider a divisor $$$d$$$ of $$$n$$$. Then we can go from $$$(1, 1)$$$ to $$$(d, n/d)$$$ which will cost $$$d + \frac{n}{d} - 2$$$. So we just find the minimum of this expression for all divisors of $$$n$$$. Solution: https://atcoder.jp/contests/abc144/submissions/8153886

D) We can ignore the depth, so let $$$y = \frac{x}{a}$$$, then we essentially have a square with sides $$$a \ \text{cm}$$$ and $$$b \ \text{cm}$$$ that is filled with $$$y \ \text{cm}^2$$$ liquid. Now there are two cases:

(1) $$$y \le \frac{ab}{2}$$$. When we reach the angle $$$v$$$ where the liquid starts pouring out it will form a triangle with sides $$$c$$$ and $$$b$$$ such that the area is $$$y$$$. That is, $$$c = \frac{2y}{b}$$$. The angle in this triangle closest to the side $$$b$$$ will have an angle $$$90 - v$$$ so we see that $$$\tan(90 - v) = \frac{c}{b}$$$. That is, $$$v = 90 - \text{atan} \left ( \frac{c}{b} \right )$$$.

(2) $$$y \ge \frac{ab}{2}$$$. Let $$$z = ab - y$$$. Similarly, there will be a triangle with sides $$$a$$$ and $$$d$$$ such with area $$$z$$$ — but this time everything but to triangle is liquid. We can calculate the angle similarly.

The time used is $$$O(1)$$$.

Solution: https://atcoder.jp/contests/abc144/submissions/8165013

E) We first make an observation: If no training is allowed ($$$K = 0$$$) it always gives the optimal result of we let the members with the lowest consumption coefficient ($$$A_i$$$) eat the most difficult food (highest $$$F_i$$$). So we start sorting $$$A_i$$$ and $$$F_i$$$ such that:

$$$A_1 \ge A_2 \ge \ldots \ge A_n$$$

$$$F_1 \le F_2 \le \ldots \le F_n$$$

So with no training the answer is $$$\alpha = \max_i A_i F_i$$$. Now we do a binary search. It is clearly possible to obtain a score of $$$\alpha$$$ and impossible to obtain $$$-1$$$. To see if it is possible to obtain $$$x$$$ we do the following: For each $$$i$$$ we will replace $$$A_i$$$ with $$$A_i'$$$ such that $$$A_i' F_i \le x$$$. So $$$A_i' = \min \left (A_i, \left \lfloor \frac{x}{F_i} \right \rfloor \right)$$$. Hence we need to train $$$\sum_i \max \left (0, A_i - \left \lfloor \frac{x}{F_i} \right \rfloor \right)$$$ in total. If this is $$$\le K$$$ it is possible and otherwise it is not.

We will need $$$O \left ( \log \left ( \max_i A_i F_i \right ) \right )$$$ steps in the binary search and each step takes $$$O(n)$$$ time.

Solution: https://atcoder.jp/contests/abc144/submissions/8159504

F) This is a DAG where $$$1, 2, \ldots, n$$$ is a topological sorting of the nodes. Let's consider the case where we remove no edges. Then we can calculate the expected number of jumps from each edge in $$$O(m)$$$ time: It is $$$1$$$ larger than the average of it's outgoing children. With a formula:

$$$dist_i = 1 + \frac{1}{\left | \text{adj_list}[i] \right |} \sum_{j \in \text{adj_list}[i]} dist_j$$$

So we start by doing this. Naively, we could try to remove every edge and do the calculations which would lead to a running time of $$$O \left ( m^2 \right )$$$ since there are $$$m$$$ edges to remove and each calculation would take $$$O(m)$$$ time. We can improve this by noting the following: If we remove an edge going out from $$$u$$$ then we want to remove the edge going to the child $$$v$$$ with the largest number of expected jumps to $$$n$$$. Hence we only need to check $$$n$$$ cases instead of $$$m$$$ which leads to a running time of $$$O(mn)$$$ instead.

Solution: https://atcoder.jp/contests/abc144/submissions/8163431

My solutions to all of the problems can be found at this link.

I had a binary search idea for D but hadn't got enough time to code it out. Did anyone solve the problem using binary search?

please clarify me why my ans is wrong for C,&D for C

ans=ceil(-2+(2*sqrt(n)));

and for D

{ y=((2*x)-(a*a*b))/(a*a);

ans=(180/PI)*(atan((b-y)/a)) ; }

but still it shows wrong answer .... why???

As a junior high student, I think Problem D is unfair because we haven't learnt the arcsin function before the contest. But I conquered it so I'm not very unhappy. :D

PS: I cannot browse the standings page. Maybe my location(in China) matters?

Problem D ruined my contest

Brute force (with randomize and optimization) can pass problem F.

Code

Why was this contest not shown in clist.by ?

https://atcoder.jp/contests/abc144/submissions/8178663

I've been doing binary search on Problem D, but I just couldn't get an AC. I cannot understand. Any help will be appreciated! Please Help

BTW I didn't miss the case when half of the capacity is tilted.

can anyone help what is the testcase no 9 , i.e. , 01-handmade-08 ? I keep getting WA on it.

I was stuck at F, as my answer was "-nan" in C++, to only realize later that the graph is actually an acyclic graph, need to read problems more clearly from now on, otherwise I did Infinite GP sum of a matrix, which may indeed sound absurd for this question.....

After solving this contest, I have realized that I actually can solve all of problems without lack of any knowledge. Before, I even didn't read statement problem F because I believe I didn't having sufficient knowledge to solve it.