Testing Round #6 starts on May 11, 2013, 20:00 (UTC). Our goal is to test the platform after recent improvements. All of them are in the Codeforces backend, but they affect many lines of the code.

I invite you to take part in the round. It will be Div. 2 + unofficials from Div. 1. It will contain four-five obsolescent problems. But I think it will be interesting for many of you. The problems contain very weak pretests to force more hacks. It will be unrated round.

Many thanks to participants!

P.S. As it is *testing* round, we do not guarantee stable work and so on.

Good to see that resource is developing. What I can't say about topcoder.

So if this round is unrated for both divisions then what the difference between official participation of this round and unofficial participation? as you say "Div. 2 + unofficials from Div. 1"

Problems are for Div2 participants, I think.

development of resource is good for cf.

nice...hope it will work

what is the difference between before and after improvements ?

Hope to see those recent improvements as soon as possible!

Pleased to see that Codeforces is improving. :)

Success! =)

After weak pretests, there were some strong enough testcases to BLOW you..

Time for a nap, already 3:40 in the morning here ..

for problem C how can the answer for test 3 3 1 be 12 ? isn't it 6?

When will have only three days first day will be always good, second day will be always bad and third day will be always good.

The only bad event in the test will take place during the second day.

Now we have 3 events. There must be at least one event in each day, so there

`3`

options to choose first event of the first day and`2`

options to choose first event of the third day. Now there is only one good event left. It must happen on the first day or on the third day. This is`2`

more options.`3 * 2 * 2 = 12`

Aha thanks, I misunderstood the problem

Can anyone give a hint for problem D? :)

Such polygon exists for all n >= 5. You can take a regular polygon and move its sides preserving the angles. Or knowing the angle pi*(n-2)/n, take some set of distinct segments and try to construct a polygon of them, going counter-clockwise.