У него все задачи интересные. Лично я не могу успокоиться, пока не разберу все задачи второго дивизиона от Нагина.

забей на инст. кф круче:)

Жалко, пропускаю второй подряд инст из-за соревнований:)

Берегитесь серого взлома...

Даже тех кто не решил XD

Даже если сам ничего не решил.lol

Даже себя походу взломает...

Серые наступают! Берегитесь!

Here 167 is a prime , 67 is a prime && 7 is a prime as well :D

they'd say if it was unusual (i mean it wasn't 500 1000 1500 2000 2500)

Ошибки орфографические, пунктуационные, грамматические лучше подсказывать в личных сообщениях. Не все владеют русским языком отлично, к тому же многие допускают ошибки.

А. Даже так. Я ее помнил, с какого-то из прошлогодних этапов кубка.

Правильно ли следующее решение? Отнесём всех коней к первой группе, затем по очереди будем удалять всех коней, у которых есть 2 и более врага в данной группе.

Мне вот интересно — это специально? Автор матча мог и не видеть задачу, в это я еще поверю. Но ведь Seyaua и sdya Тимус решали. И кубок вроде бы тоже. Они решили, что такую задачу можно допускать на матч СF?

Более того, решение которое получило АС на тимусе, тут получило ВА 13, уже с дорешки отправил) Но оно явно багнутое, так что кроме слабых тестов тимуса обсжудать нечего)

No, you should read the code properly

Let's sort all pairs. There can be equal ones, which can be now easily recognized. If there aren't — the answer is (n+m)!, obviously. But if there are, we should divide our (n+m)! by number of equal pairs (since 2! = 2; if there were equal triples, we might have divided by 6). Division goes using Fermat's little theorem since we're using modulo.

Edit: forget about the theorem, those 2's are from factorials, as MohallaBoy noticed.

basically you to find values like (n! / (2) ^ k) % m You can do this by removing 2's in the numbers which are used for calculating factorials. and taking then modulo m.

When we calculate factorial let's just divide each factor by 2, if there is any pairs left. Even if we use chinese remainder theorem, we can't calculate modInverse of 2 if 2 is divider of our modulo.

long f(long n) {
	long res = 1;
	for (int i = 2; i <= n; i++) {
		int k = i;
		while (k % 2 == 0 && badCount > 0) {
			badCount--;
			k /= 2;
		}
		res = (res * k) % MOD;
	}
	return res;
}

The answer is integer so when we got our answer badCount == 0

Edited : I was wrong !

Randomly distribute horses in 2 parties. Now if each horse has at most 1 enemy in its party, this distribution is correct. Otherwise, there exists horse A in party 0 that is enemy with horses B and C in the same party. Now move horse A to party #1 and it'll have at most 1 enemy in its new party. To prove this solution works, just define an invariant as count of enemy pairs in same party. Observe that after each move, this invariant decreases and hence our procedure is finite.

Везёт, я её вообще сдать не сумел. Кто-нибудь сталкивался с ошибкой на 14 претесте?

Та же фигня. Искал в этой задаче тупую багу весь контест, в результате за 5 минут до конца нашел)

PFFFFFF)

ans++ , seriously , it will TLE

You shouldn't continue on a[i] == a[j] at line 40.

And you use insertion sort which is O(n^2). I think it will timeout before finishing input.

well, i used BIT

submission in queue...

I used Fenwick Tree for that problem :)

No, you weren't :/ Segment tree consumed all of my time — I should have read D first...

lol, i wanted to code it, but then i saw a number of acceptions for this task.

I use segment tree, too. I've got some MLE after some TLE and finally I forgot to use long long.

same here :) I got it Accepted and also I apologize to Segment tree for using it in this easy problem

I was fool too :(

I didn't notice that the stairs are sorted

I also didn't notice that boxes fall from the left until I coded it

anyway I was fooler because my code was fail on test 45 because of overflow :(

The same http://community.topcoder.com/stat?c=problem_solution&rd=14422&pm=11179&cr=10574855

I solved this problem after read this about one year ago ... > < .. I lost some of my code after a hard-drive broken and can't log-in to TC to access my solution. (because there is a SRM 2 hours early .. ) This is not a hard dp problem, but need implementation carefully. Even if I solved it before can't ensure I can solve it on the contest within the time. (I even didn't have a Java Complire on my laptop ... so only change the code under a text-editor ..

.. PS: The DIV 1 C is also a original problem ...

There are more than 2: 3122346

3121856

3121632 (change variable names?)

3120914

3119084

and there might be more...

Эх, а я не заметил. Хорошо, что поломали достаточно быстро.

Да ладно, все равно было куча взломов по А на переполнении. Мои, например, все на этом.

Да я тоже об этом подумал. Наверное многие, как и я, дописали 64 перед тем как сделать сабмит)

http://acm.timus.ru/problem.aspx?num=1128 It is here.

I have seen this so many times :D. I have even coded exactly the same task and did this problem with 2 of my pupils as a classic problem for variants :P.

You can easily check it on the status or standings page

i was about to ask the same question !!!!

An untidy way.. but if u really really need it you can make a hack.. Just print say 20 tokens once (or whatever size fits in the screen), then next 20 and so on..