Каждый раз проще чем предыдущий :)

Due to "some Apple magic" I noticed such a letter in inbox instead of normal one...

The first idea was that CF made The First Holy Round For Hackers, but it's just Educational :D

Э-эх, пропало волшебство(

Yes!New season,new beginning!With our passion unchanged

yes!New season,new beginning!With our passion unchanged!

I don't see any way to withdraw my registration from this round. Is it because the round is unrated?

Ждём взломов. :D

Though it is not particularly relevant to this round, I wanted to point out a rather tiny glitch with the website. It is cool that we see the local time of the contests in Contests page, and not having to check it on timeanddate. However, the live countdown in the rightmost column of the table named "Current or upcoming contests" is probably still based on Moscow time. It currently shows there is 4.49hrs until the registration is closed, and it appears it will hit 0 at 18.00(Moscow time) which is the starting time of the contest.

we can not search a problem with the caregory name. Like if i search problems with category name dfs, then blog result shows us. is it possible to search problems with category?? This will be very helpful. Thanks

Auto comment: topic has been updated by Edvard (previous revision, new revision, compare).

Nice problem E, thanks

Can we have the editorials pls? :)

First problem was too easy for those who use Java, Python or have a BigInt class template in C++.
Third problem was a duplicate from USACO contest.
Fifth one's solution appears first when you google "Sigma(n mod i)"
Good thing this round is unrated.

На Java у многих не проходила первая задача? Просто для интереса пробовал на java, потом забил и сдал на C++

In Problem 616C - The Labyrinth why my solution 15292451 gets WA?!

Looks like A has quite a lot hacks. Will there be some statistics after finishing the hacking phase?

Why should someone Enjoy when a problem is rejudged?

How come the input file for the hacks are only up to some 200+ KB? I tried an array with 5*10^5 integers, and it led to 3MB+

What is the idea behind E?

At problem f I used suffix arrays with lcp precalculations,then just a stack and that's all. I still can't figure why I get wa on test 12.Is anybody who had the same problem? Can you help me fix this?

Edvard could i see the full data of test case #11 on problem 616C - The Labyrinth

It gives me WA and the Checker comment says that i wrong in the 1st line which i run this part in my machine and give the correct output!!!

What's wrong with my solution for Problem C? http://codeforces.net/contest/616/submission/15297727

Moreover, at test case 11, where it says my code fails- Here, my answer matches with the jury's answer(at least the part which we can see), however, it doesn't match with expected value. Isn't it strange? The expected value does not match with jury's solution.

saw someone using i<strlen(str) in for loop and ran it in my machine. gets TLE but when I ran it here it runs ok . No idea what's going on :v

Please what is the format for writing a Test generator for Java 7. The output file will be > 1mb. ?

I don't like a complicated C problem,but a easy d problem. :(

UPD: hacked Thanks

Is it bug or feature?

Почему показатели количества решивших задачу в меню раунда "задачи" и в его таблице "положение" различаются?

I find so hard to read code (example http://pastebin.com/AdDQDtWy) with all these defines forn, sz, pb, etc why can't the code be just as is / simple with no defines

What am I getting wrong here? I get different output for N = M = 1e13 (it matches up to 1e8 or so)

from math import sqrt

def get_sum(x):
	return int(x*(x+1)/2)
def range_sum(l,r):
	return get_sum(r) — get_sum(l-1)

if __name__ == '__main__':
	# MOD = 1000000000 + 7
	MOD = 1
	mod = 1000000000 + 7
	MOD = mod
	N = int(input())
	M = int(input())
	# N = 10000000000000 
	# M = 10000000000000

	print(N)
	print(M)

	minMN = min(N, M)
	# lim = int(sqrt(minMN))
	lim = int(sqrt(N))
	# total = (M % MOD )*N % MOD
	total = M*N
	reduce = 0

	for i in range(1, lim):
		if i > minMN:
			break

		a_pair = int(N / i )
		b_pair = int(N / (i+1))

		if b_pair+1 <= minMN:
			this_range = range_sum(b_pair+1, min(a_pair, minMN))
			div = i
			# red = (this_range % MOD * div) % MOD
			red = int(this_range ) * div

			reduce += red
			# reduce %= MOD

		reduce += int(N/i)*i # % MOD
		# reduce %= MOD

	print("Reduce is " + str(reduce))

	lim_pair = int(N/lim)
	for i in range(lim, lim_pair+1):
		if i > minMN:
			break
		reduce += int(N/i) * i
		# reduce %= MOD

	print(total % MOD)
	# total %= MOD
	print("Total is " + str(total))
	print("Getting from " + str(lim) + " to " + str(lim_pair))
	print("Reduce is now " + str(reduce))

	ans = total - reduce
	print( (total % MOD) - (reduce % MOD) )
	print( ((total % MOD) - (reduce % MOD) ) % MOD)

	print(ans)
	print(ans % mod)