Have a strong heart, dear.

So why are we downvoting the tester >_<

A tester leaves a rated contest for us, Have some decency please

what we get if we have some contribution.i often see people are asking for contribution ??

I want positive contribution instead of negative contribution!

I think it is already highlighted.

Are you son of Time_tears ? :v

What I improved is following:

In the task E, there was one more constraint "It is guaranteed, that an integer $$$p>1$$$ such that $$$x$$$ is divisible by $$$p^2$$$ does not exist. You want to find $$$x$$$."

I noticed that we can erase this constraint. If you couldn't solve E, it's good that thinking this version first!

The contest is ICPC style not IOI style,so there will be no subtasks.

We used to have subtasks, but we found it meaningless to have subtasks on our problems. So we deleted them in the end.

Div1 people: Are we a joke to you?

They are written, but not published.

.

hpbd :3

happy Birthday. good luck.

Today is my birthday :) I was born on 9/11

Its useful too. Isn't it?Trgt_2021_explosion

No.In fact most of the CF contests a few years ago had 5 problems.

We know Jiangly, because he is the best one in the city,but we are not so good as him so he definitely doesn't know us.

your rating graph is quite interesting!

with strong pretest.

How many times have you solved more than 3-4 questions lol?

On a serious note, you and I can't even imagine the amount of work it goes into setting a problem and preparing the statements, test cases, etc. So, simply saying things is very easy, when you don't know what is going on behind the scenes

T̶u̶r̶n̶ ̶b̶a̶c̶k̶ ̶w̶h̶i̶l̶e̶ ̶y̶o̶u̶ ̶s̶t̶i̶l̶l̶ ̶c̶a̶n

Good luck on your first contest!

No.

It's a platform that helps make problems, created by Mike

fixed, thanks for pointing out

Thanks!

you just raised a flag, 10 mins delayed

YES

haa its really good

wtf dude, this is codeforces not a nrop website.

umm... okay

Go for virtual . :p

yup!!!

umm , we actually know..

Thanks for the info :)

Just curious to know, what was the last minute change because of which score distribution was changed? I think present scoring distribution was accurate.

why though? why to you specifically?

good luck!

We're very sorry. But please, still enjoy the contest.

Didn't passed system test yet but my solution idea is :-

Hints:-

ans is

tot = a[0]+d where d = sigma( max( 0, a[i+1]-a[i] ) );

ans = (tot+(tot>=0))/2;

so only edges L and R in queries can change the value of d.

I guess now you can figure out the solution

My solution has not yet passed sys tests but here is how I solved it. Suppose we let initial value of non decreasing array be c so initial value of non increasing array will be a[1]-c.

Now for each next element if is greater than previous value we can keep the value in non increasing array constant therefore the other array remains non decreasing. We can similarly keep the value in non decreasing array constant if the next element is less than the previous one.

Therefore we get the final value of non decreasing array as c+(sum of all positive a[i]-a[i-1]). Now we just have to minimise max(a[1]-c, c+sum of +ve dif). We can see that it will be minimum if both elements are equal.

After each query only 2 differences are changed at max so we can store values of each element by Fenwick Tree and recalculate the optimal c. Upd: passed sys tests.

Observe that since $$$b_{i}$$$ is non-decreasing and $$$c_{i}$$$ is non-increasing, the max is just $$$max$$$$$$(c_{0}, b_{n - 1})$$$.

We can notice that $$$b_{i} \gt b_{i - 1}$$$ will be required when $$$a_{i} \gt a_{i - 1}$$$ and $$$c_{i} \lt c_{i - 1}$$$ will be required when $$$a_{i} \lt a_{i - 1}$$$.

Clearly $$$b_{0} + c_{0} = a_{0}$$$ is a required condition. Let $$$b_{0} = u$$$ and $$$c_{0} = v$$$.

If $$$sum_{inc}$$$ is the sum of $$$a_{i} - a_{i - 1}$$$ whenever $$$a_{i} \gt a_{i - 1}$$$, $$$b_{n - 1}$$$ is clearly $$$u + sum_{inc}$$$. So we want to choose $$$u$$$ such that $$$max$$$($$$a_{0} - u$$$, $$$u + sum_{inc}$$$) is the minimum possible. This is clearly $$$\frac{a_{0} - sum_{inc}}{2}$$$ which gives an answer of $$$\lceil \frac{a_{0} + sum_{inc}}{2} \rceil$$$.

As for the queries, we can notice that only the contributions from $$$a_{l} - a_{l - 1}$$$ and $$$a_{r + 1} - a_{r}$$$ to $$$sum_{inc}$$$ can change. Maybe there is some elegant observation that makes this easy to calculate, but without that also there is an easy way to figure out the change. Subtract the contribution of those two endpoints, update the range (standard lazy prop) and re-add the contribution of them.

BE CAREFUL that you perform ceil correctly for negative values.

Talking about this ?

I think ask for all prime numbers ( ~9600 ) and make some Inclusion–exclusion principle (ex if you have n = 6 ask for x=2 => s = {1,3,5,6} and for 3 you have to see that 6 is still there (because after the first operation just 1 element %3=0 had to remain in the set) ) .

First we use the second operation to all primes < 1000, all the remaining numbers can either be a prime or $$$x$$$.

We can query A 1 and check if $$$x$$$ is a prime.

If $$$x$$$ is not a prime, we query every prime number $$$\leq n$$$, and obtain x by finding all prime divisors of $$$x$$$.

If $$$x$$$ is a prime, we can do the following: delete S prime numbers, and query A 1 to see if $$$x$$$ is in the $$$S$$$ primes numbers you just deleted. When you find the $$$S$$$ numbers that contain $$$x$$$ you do $$$S$$$ extra queries to find $$$x$$$. Let the number of prime numbers be $$$N$$$, The number of operations is $$$N + S + N/S$$$, and we can set $$$S$$$ to $$$\sqrt n$$$

code: https://mirror.codeforces.com/contest/1406/submission/92637463. It passed the pretests anyway.

перебрать все простые делители до N начиная с таких, что их квадрат больше N, проверять можно по группам, например первые 100 такие делители с помощью функции 2 типа, и проверить есть ли среди данных 100 простых чисел нужное с помощью типа 1 и так далее по сто, а в последнем оставшиеся. Затем перебрать все меньшие простые делители и его степени. Например, k — простое, k * k < n, тогда если н делиться на к, с помощью бинпоиска найдем степень вхождения это числа.(iterate through all Prime divisors up to N starting with such that their square is greater than N, you can check by group, for example, the first 100 such divisors using a function of type 2, and check whether among the data 100 primes necessary using type 1, and so on one hundred, and in the last remaining. Then iterate over all the smaller Prime divisors and its powers. For example, k is a Prime, k * k < n, then if n is divisible by K, we use bin search to find the degree of occurrence of this number.)

memset is the problem. If you have 10000 tests of n=1 then you will do 10000 (number of test) * 1e5 (maxN) operations

You should not clear array sz for each testcase since T can be sufficiently large to give TLE for your implementation. Don't use memset(sz,0,sizeof(sz)) each time. Instead, clear first n cells.

how to solve B I was getting WA in test 3