In codeforces we do not try to claim the first comment as in youtube

We usually try to get higher on ranks :P

why?

You won't have such problems if you'll solve 7 problem. :thinking:

--- 'The greatest enemy of knowledge is not ignorance, it is illusion of knowledge'

That's like saying ACM ICPC is unfair because of it's certain rules (which are used in many contest).

There person who solved C & D instead of A & B (assuming a and b are easier) should just use a different strategy than normal contests.

I believe that the person who can solve C&D can also solve A&B very fast. So, there is nothing to worry.

it's time to change your thinking, and also no one cares what you think.

«Если вы думаете, что способны выполнить что-то, или думаете, что не способны на это, вы правы в обоих случаях» — Генри Форд.

Men's dream. Not about Tufurama of course.

Да, можно.

You can solve them after the contest, but it would be unrated

Are you using float or double in your solution?

Look at first 3 points. If they are collinear then find those points that don't belong to that line and check if they are collinear. If yes, answer is yes, otherwise no.

If 3 points are not collinear, then you need to pick 2 points (out of these 3) that will maybe belong to the same line. There are 3 combinations, and for each one check if it possible to partition that way(same way as above)

If n<=4, the answer is YES.

Else, let's check about first 5 points.

If there are no 3 points set on the same line, the answer is NO.

The points set are exist,you have to use the line because you can use only 2 lines.

Then, you remove some points that on the line and check the left points are on the same line or not.

If n < 4, then YES

You will try to fix a line and see if all the remaining points lie on the same line.

There are just 3 possible lines that can be formed with 3 different points:

.Line formed by point 0 and point 1
.Line formed by point 0 and point 2
.Line formed by point 1 and point 2

To discover the other line, just take the first 2 points that does not lie in the fixed line.

For each line, go through all the remaining points, they must be inside the fixed line or in the other line.

If any of the fixed lines succeeded, then YES

Else, NO

You can get a row from fft right not column,Can we??

I was trying some combinatorial argument but I am going wrong somewhere. Can some one help me.First thing is ingore a[i]'s. later we can multiply with sum of a[i].
Using the formula mentioned in above comment and also that we can run i from 0 instead of 1.

My argument is that we are choosing (k-1) nonempty teams from n people(need not be all people) and selecting a captain among remaining people.

So inverse idea is selecting a captain first in n ways and then selecting (k-1) teams from n-1 people(need not be all people ). Now this part is S(n,k). So answer is n*S(n,k)

Okay, so what I do is this:

To calculate in how many subsets of size j an element can be we use the formula: SS=choose(n-1,j-1)*X where X is the number of partitions of size k-1 of the rest of the elements. Well, let's think of a partition as kind of placing a divider between the elements, we have n-j elements, every divider creates 1 new set and we need k-1 sets so we put k-2 dividers, thus X = choose(n-j-1,k-2).

This approach gets WA on test 7 Could anyone tell me what is wrong with this solution?

I don't see the way with binary search.

For F, if answer for k - substring is f_k, then note that f_k - 1 <  = f_k + 2. So, for finding f_k - 1 start from length f_k + 2, coming down and find the largest length r such that length r prefix is the same as length r suffix. Complexity can be proved to be O(n), using a KMP like analysis.

, where a_ij is an indication variable for both

Number of ways to partition such that i and j are in the same partition is (consider a new set where (i, j) is merged).

It follows that the answer is

I used a merge-sort tree,
For each season i calculate the number of seasons j in the range(1,min(A[i],n)) with episodes >= i
Sum all the queries for the final answer

Just process queries offline, use bit.

STL Policy based data structures

I just use a normal segment tree. For each season I store 1 if it contains more series than current threshold or 0 otherwise. I consequently update threshold from 0 to n-1 and so need n updates of the tree total. For each threshold i I ask the sum of the first min(a_i, n) elements.

I solved it with "wavelet tree". hope it passes the system test

You'll see ranking change after the 24 hours of hacking allowed.

i'm pretty sure i did a somewhat similar problem on codeforces earlier

maybe floating number problem

I guess that only two points occupy one of two lines. So if we choose to solve the problem by a random algorithm, it will be hard draw the second randomly in the right way. We should pick up two points those weren't be visited before to draw another straight line. And don't pick them up in a random way, either, or you will get TLE.

Oh, f*ck. PS. But why no hacks on C. I checked and both solutions pass the tests.

Can be either.

It doesn't matter because answers are the same. If [a b; c d] is optimal square placement for black lower left corner then [b a; d c] will be optimal square placement for white lower left corner.

If you were both participating in ICPC. He would beat you, so...

I just solved it with segment tree of vectors.for every 'i' I checked how many numbers in the range 1 to min(a[i],n) are greater than or equal to i. 36978334

I used a pretty straight forward segment tree , no heavy data strucutre is really required , just processed the array in reverse order from n to 1 . My code if you need help -->36983745 , code i think is self explanatory but ask if needed.

I have a feeling these are fake accounts of Div1 coders. Isn't it illegal to do this? Correct me if I'm wrong. It is really sad that our ranks degrade due to this practice. What can be done?

It means that you are asked to do the impossible — hack solution that passed all tests. It is less impossible considering very small number of tests (authors need to reduce server load) in the first problems though.

Because there are no pretests in educational rounds. Those during the contest are the main tests, therefore you get Accepted instead of Pretests passed. However, the data sets are a bit weaker than usual rounds, so instead you're given 24 hours to hack any solutions.

Try copying the code and running it manually with your inputs..