Will be announced later as usual.

but when "later" is?

UPD: Educational round!! nileshjha19

I think div3 is a good idea

1e9+7

0x3f3f3f3f I think.

Are there rooms in this contest?

I really wonder who that person is.

ER features trivial/classic problems, don't be surprised if you have seen some of them before

Ordinary contest:

• Different points on different problems

• Hacks in-contest (you can't hack after round)

Educational contest:

• Problems have the same points (like ACM — ICPC rules)

• Hacking phase opened after round. Usually it's 24 hours (but it's 12 this time).

Ordinary contest: Computer is used to test solutions.

Educational contest: a human named halyavin is used to test the solutions.

This has also been happening to me sometimes in contests

clear your caches and cookies, maybe your csrf token is not refreshing

First, note that we are essentially just solving .

So iterate over n, check if n² - X is a square, and check whether m exists that satisfies the condition. The appropriate bounds for n are easy to find. Each TC will run in .

Given x, first You have to find n, k such that n^2-k^2 = x, and second check and find whether n/m = k is possible.

Since n^2 <= 4/3*x, the range of n is not that big, and you can easily match k for all n which satisfies x <= n^2 <= 4/3*x by check whether n^2-x is square. Then by calculate candidate of m = n/k, you can know whether n/m = k is possible

Using Greedy & Priority queue, expand the routes by increasing order of cost of arriving concert.

Since number of road is also small, you don't have to care about multiple visit.

You can take a path through multiple cities that might be cheaper.

Say the concert costs are 100 100 1 And the edges are 1 2 1 2 3 1

Then it's cheaper to take both edges from city 1 to city 3. Your code only considers taking 1 edge.

Это уравнение, из которого нужно найти m, перебрав n (оценив как корень * 2 + 1000, например)

You may maintain the base of cycles by Gaussian elimination. Since the size of each base is not larger than 30, and each edge adds at most one number to the base, we can easily rollback the base in dynamic connectivity.

I didn't completely go over your code yet, but using int for x seems a bit dangerous.

You are calculating x * x sometimes, where x may run up to 10⁵. Yet you used int for x...

Try 2 600000 800000

Here's my code: https://pastebin.com/attptiBj

Basically, the key thing for B is to

1. Make sure that if you are moving simultaneously
2. In order to make sure you don't over or undercount intervals, try each interval.

Find the minimum distance from current start to the next interval greater, and current end to the previous interval smaller, then shift whatever is closer.

This prevents you from moving a large interval and missing certain intervals.

So for start = 1, end = 10^6. If the intervals are 1, 500000, 600000, you don't want to be double counting 400000+500000, because if you moved 400000 on both sides, then you only have to move an additional 100000 from start to get to 500000.

Constantly update start and end with the new closest intervals to count correctly.

You also want to break out of the program and return when start and end intersect, which signals that you have seen all the intervals so you should exit.

Main tests will handle that issue.

For E. First by sorting, compress the array as (number, occurrences).

Now, we will calculate how much time each number will by added in our summation.

Let's look at this number x — say there are u occurrences of x, and there are v numbers that are larger than x. Then, x appears in our summation if and only if the v numbers that are larger than x appear AFTER at least one x appears. The "probability" for this is , so the number of permutations that add x to our summation is . Now it's just some partial sums, easy calculation. Time Complexity just for sorting.