I tried to make it a bit more balanced this time :)

How to download Test #15 of 650C Table Compression?

Are You Retarded?

Thnx :D

When it starts in your country?

if you cheat in this contest, this contest is unrated for you too, and if nobody discover it, you can have a high rating:)

Good Luck!) Hope you will be a Candidate

why people doesnt like this post whats wrong about it

Nice observation!

This one is for you

Look on the bright side, man, this time there are no pupil users as problemsetters :D

Learning so much that you start to understand Croatian. (Source: COCI 2016/17 Round #6 solution, p. 4, line 8)

Let Pa < Pb i k * Pa ≤ Pb < (k + 1) * Pa.

Test ur program on custom invocation and u will undrestand why got WA.

More like, CSA coincides with Codeforces Round 461.

CF round moved by 40 minutes.

Okay, this should be the last time I see this anywhere.

You cannot do that. At all.

(Well, except if the problem setter provide the full package publicly).

For convenience, all test cases are compressed in case they are too big, so large input/output files are near to complete omittance.

Usually, if you get stuck with a big tests, it can only mean a few things:

• You ran out of memory (MLE verdict).

• You ran out of time / your algorithm was too complex (TLE verdict).

• You handled the undefined behaviors poorly (WA/RE verdict).

These tests are mostly used to tests how a code works against large input streams, and not to be corner cases, so please check your own source code for places that can be optimized, not just spamming the blogs all the time.

I registered him manually, just a prank ^^

Best of luck !

you will solve nothing we will come newbie

And I thought heaven didn't exist.

One hour earlier, but yeah, I feel you, dude...

12:15 AM in UTC+7 (Hanoi, Vietnam). Uh-oh...

Happy!

You Should Behave On website like codeforces , though fun is necessary , but people seach for good comments over here . I hope you understand it ! Good Luck for contest .

Pessimism level: Notali.haidar

i think you love negative contribution.

I hope the int vector is fixed setprecision(10) after the contest end ;!

What's wrong? I thought you can lock the problem as long as you have the "Pretest passed" verdict, irrespective of whether you get hacked or not. Correct me if I am wrong.

Even you hacked me. This is first time in my life when I got 3 hacks on A

I think you are confused just like I was — summoning a bird doesn't increase your mana, it increases you maximum mana.

Take 1 bird from the first nest (you don't have enough mana to take two birds), take 10 birds from second nest (your mana capacity is 17 and you have 15 mana so you can take them all).

I just used normal sorting with a comparing function that returns numS[a] * numH[b] > numS[b] * numH[a]

For small Ks (probably <= 1000) you will find two exact values, so just implement

just do it with brute force :)

People have this idea:

Since you need to have all unique values till k. They should be obtained serially from 1,2,3,..k like this: 0,1,2,3,4,5...k-1

so, n should be of the form: 2x+1, 3x+2, 4x+3, 5x+4, 6x+5, 7x+6,... all at the same time. But if something is of the form 5x+4 .. it cannot be of the form 2x+1 .. (coz one is odd and other is even, for even x, similarly chose a pair for odd x)

Edit:

The conclusion seems flawed. Maybe we can continue the logic like this:

Number has to be of form: 2x+1, so 3,5,7,9.. 3x+2, so 5,8,11,14.. ... ....

It kind of creating a sieve of eratosthenes, which generates a prime number.

I don't what that gets us :)

sorting : )

if you donnot have a copy,you cannot get more copy

Most common one: x 1 (while x is a positive even number).

Expected answer: "NO" (as there are no initial copies to generate new ones)

0 0

0 1

1 0

3 0

2 1

2 1

output

NO

something!=0, y=0 and something!=0, y=1

it's correct

Did you use sliding window maximum?

problem is that max summoned birds numbercan be c_i * n = 10³ * 10⁴ = 10⁷, so your array will have size dp[10³][10⁷].

EDIT: I didn't read task carefully :{

ans = 0
for a = 1 .. n
  for b = a .. n
    c = a ^ b
    if b <= c <= n and a + b > c
      ans++
print ans 

can you please tell me how to solve C?

For a <= n, b <= n, let c = a ^ b, if all other conditions hold, then increment answer

a^b^c = 0 implies a^b = c. So bruteforce for a, b, and check if the 3 numbers can form sides of a triangle.

You can iterate through all possible cases of a and b. c can be calculated by bitwise xor:

As a xor b xor c = 0, we can agree that c = a xor b.

If c is not lower than a or b, and not greater than n, and a,b,c forms a non-degenerate triangle (simply check that if all 3 inequalities of triangles are proved right), then you can add 1 to the answer.

Simply iterate over all a, b values . Check a + b > c, a + c > b and b + c > a

I actually passed it with brute forcing which is O(N^3). It gave me 780ms even though it doesn't seem to be the intended solution.