Also clashes with CodeChef Lunchtime.

like who cares to be honest?

I try to hack the problem B&&D , but I failed

last minute smart move: i'm not going to take part in this contest because i know from experience it's going to be a lot of math. not worth taking the risk.

I don't know why, but at last contests problems are easy or hard, there is no medium ones.

There is ceil function, so ceil(i/x) gives 1.

https://uva.onlinejudge.org/index.php?option=onlinejudge&page=show_problem&problem=1500

Almost the same logic XD

Mine's using sliding window. Keep extend our window when the string in the next row is equal to the string in the current row. Keep track all size of the windows and find the GCD of all windows' size

Basically we need to check over all x such that x is factor of n,for a given x we need to check all the [n/x]*[n/x] sub-matrix whether they contains equal element or not, if a submatrix contain equals elements either all of them equal to 0 or 1 i.e sum is either 0 or x*x,we can easily check sum of a submarix with the help of 2d-prefix sum.

Don't know if this is the indended solution, but I did this:

Convert hex to binary and store it into a matrix. Then you try each divisor of N as x in O(N*N). At most, N has 60 divisors (N = 5040). I've executed this solution with the worst case in codeforces custom test and it seems to pass, but who knows...

Find all divisors of n. Then for each divisor, divide the matrix into the dXd submatrices and check all the submatrix has similar elements (1 or 0). Find the maximum divisor that satisfies this.

I think, (sum_of_current_length, left_index, right_index), but I'm little scary about time limit of my solution

It's

mine is O(n*n*sqrt(n))

I have a much neater solution.

• Find all lengths of consecutive segments with same character. (like 0001111110 then you count 3,6,1)

• Do this for both vertical and horizontal.

• Answer is the gcd of all the lengths.

O(N^2), because you can just determine whether a segment of this length exists, and so gcd would be called N times at most.

lengthy code? for which problem?

I don't actually understand why are people so angry about the statements. There are two main mistakes in our statements, one is not telling about 1-indexation in D and another one is writing that paragraph about skipping hits in C too ambiguous. Can you point out any other mistakes in the statements?

from sys import stdin

The trick is to use PyPy and read the input with sys.stdin.read(), this will read the entire input as a string. From my testing all other ways of reading input is significantly slower.

Here is my solution running in 1123 ms. The main thing slowing the code down is that I'm calling a function to calculate gcd. By inlining the gcd call my solution runs in 826 ms.

EDIT: If I additionally add a check for if the gcd hits 1, and then print 1 and exit, it runs in 390 ms.

EDIT2: Played around with it a bit more, was able to get it down to 343 ms by pretty much removing the gcd calls.

You can try to input 12 rows of 0FC(000011111100). The answer is 2 but your output is 1.

I think it's safer to calculate the GCD of lengths of those substrings.

u found it on google?

me: brute forcing 1 to 100 :P

You can see that each problem worths a — c.

Let's fix gap value. That is, for each i from 1 to n — 1, binary search to find l and r satisfying l <= i <= r, and all x and y satisfying l <= x <= i <= y <= r the subsegment from x to y have the gap value equal to (di — di+1)^2. Now the only thing you have to do is find the maximum subsegment containing i in l...r. It's can be done with a segment tree storing maximum prefix and maximum suffix (very similar technique to GSS1 spoj)

AFAIK you must not mix cin/cout and scanf/printf if you use fast I/O. Your solution passes if the corresponding command is removed.

Just make a vector and push_back values as long as it's the same letter then sort the vector and get the largest possible values

i think you also misunderstand the problem.. your ans for test 5 = 68, same with me, check again the problem page, they add explanation and example, the consecutive count will not reset after skip character,

for example: 4 2 1 1 1 1 aaaa

you cant pick 1,2,4 because you press 3 button 'a' in a row

When you do k1 = se.size() — k, it gives you error. Since s.size() is an unsigned value, k gets typecasted to unsigned and gives error.

String in C++ has a null character on the end of it.

Tried to make a nice solution of E which should be pretty readable and self-explanatory.

(Instead of using DP-states involving triple indices I just play around with strings and do recursion).

EDIT: fixed the link

I was about to comment the exact same thing. Fake account

This is Educational round who cares about get successful hacks?

Because system testing is not over. After three hours you can see the change.

System testing is over. Waiting for rating change

Get a good night sleep and wakeup, It will be updated.

same

The rank list mechanism is different. The educational round is more focused on concepts and logic whereas normal round is more focused on the right/optimal way of implementation of your logic.

The hacking mechanism is also different.