Uploaded to Youtube: https://www.youtube.com/watch?v=3njFJqbY5D0

This is primarily because the time may vary from place to place, as per their time zone. timeanddate.com URLs can help with directly showing time converted in the local timezone, thus saves the confusion and hassle of converting them manually.

I think only CF contests can have time-zone adjustable displays of event time. And to avoid the confusion HarshitKumarGupta mentioned, they are doing this.

I wish all event times can be correctly displayed in CF blogs soon tho.

The contest is sponsored by Panasonic (https://na.panasonic.com/us) and there is hiring advertisement on Japanese version of the contest page.

https://atcoder.jp/contests/abc186/submissions/18889820

it was sweep line basically, first assume line extending until it can for both $$$x$$$ and $$$y$$$ directions , now you must have noticed that you have done over count, so you have to remove that over count, so question reduces to given $$$n$$$ and $$$m$$$, horizontal and vertical lines respectively, find total number of intersections of them.

https://www.hackerearth.com/practice/math/geometry/line-intersection-using-bentley-ottmann-algorithm/tutorial/

You could read more about here, you could use any range sum data structure, to calculate intersection in current sweeper line. also update them.

Even, if there was a brute force approach, it won't work for the sample test case (I tried for 10^7 iterations) and one of the testcases was giving the wrong answer.

This is what I did.
(k.x)%n=(-s)%n
Let g=gcd(k,n).
We know that (k.x)%n takes all (i*g)%n values.
If ((-s)%n)%g is non zero than answer is -1.
For other case just divide k, n, and (-s)%n by g.
Now the equation changes to (a*b)%m=c where gcd(b,m) is 1.
It has a unique soln for a in range [0,m). Its a=(inv(b)*c)%m. You can use extended gcd to find that inverse.

Use extended euclid's algo to just find any solution to k.x-n.y=-s, then use shifting to find for finding minimum x>0

E:https://cp-algorithms.com/algebra/linear_congruence_equation.html

Here is how I solved problem E.

So assume that position of the throne is at position n and we are at position s.

So to reach the throne following equation can be formed —

s + x*k = y*n where(x is the lowest positive integer)

So this equation can interpreted as ax + by = c which is standard linear dopamine equation.(There are many good resources for this on internet).

I have a solution for F involving policy-based DS(indexed set).
First, you calculate the number of cells reachable by using steps down and then right.
Then you need to calculate (for every column) how many rows are there such that their minimum y is less than the current column number. Add that number to the answer. This can be done using a Policy-based DS.

One solution for F is as follow :

Let us find 2 arrays, call them R[H], C[W] and initialize R[i] = {w+1}, and C[i] = {h+1} for each valid i. So, the information we store in R[i] = first column which has a block in ith row, similarly , C[i] = first row which has a block in ith column.

Now in first move we can go to anything from [1,R[1]] and [1,C[1]] and then we also have stored value, of how many blocks we can visit in front of each of them . So, we just add them up.

Are we considering all the blocks which can be visited ? Yes. But, we also need to now remove the blocks which can be visited from both paths, going right then down or going down and then right.

To count these cases, here is one approach 
for each row, the maximum such blocks which can be repeated are min(R[1],R[i]), and we will not be repeating the cases, if there exists a row above them, which has any of these columns blocked.
So, sort the blocked points by row, then for each row, insert the columns into a set, and we need to get lower bound for min(R[1],R[i]). 

You can use pbds, segment tree or anything you like to answer these queries.

Code

You can have a look at my code.

I tried using merge sort tree and policy based trees to but code failed for 15 test cases and they dont share test cases I don't know what is the error now shifting on BIT lets see the results.

I used segment tree with two operations: point update and (sum of range, number of zeroes in a range).

Link to code

we can use DIGIT-DP for modulo 8 and modulo 10 separately, but i am not sure about how to calculate intersection of both the sets.

I think you have transfromed the problem into get the minimal positive integer $$$y$$$ that $$$N \cdot x + K \cdot y = N - S \text{ mod } N$$$ stands.

The next step of my solution is to use EXGCD to get the exact value of $$$y$$$, which is a classic use of EXGCD.

Maybe you can have a look at my code first, I am trying to write the editorial for F now.

First, calculate $$$a[i]$$$ which means the number of squares in the $$$i$$$-th row that you can reach (from $$$(i,0)$$$ in one move) and $$$b[i]$$$ for the $$$i$$$-th column, respectively.

Set $$$c=a[1]$$$ and $$$d=b[1]$$$, so that you should only consider the first $$$d$$$ rows and the first $$$c$$$ columns. The only problem we need to solve now is to calculate the number of sqaures $$$(x,y)$$$ such that $$$1 \le x \le d$$$, $$$1 \le y \le c$$$, $$$y \le a[x]$$$ and $$$x \le b[y]$$$.

We solve this problem in the ascending order of $$$b$$$. When we consider $$$i$$$, we need to put the first $$$b[i]$$$ elements of $$$a$$$ in BIT so that we can only calculate the number of elements which are not less than $$$i$$$ in BIT, which is called $$$p[i]$$$.

Finally, the answer is $$$\sum_{i=1}^c(b[i]-p[i])+\sum_{i=1}^da[i]$$$.

While I am not aware of the Linear Congruence Equation method — I am reasonably sure it should be if(b%g!=0) in your lce method.

(b%g!=0) is one, also i was getting garbage values until i realised that modular exponentiation can only be used to find inverse if the given "mod" is a prime. Else it'll give wrong answers. i applied the same to your code and it gets right answers.I also changed MOD=n just in case.

AC

You should take the product modulo M as well, and $$$42 \equiv 2 \pmod {10}$$$.