After scoring fn has been updated.
My submission page shows score of 0.046pts \approx 4.6 point on scoreboard. But it seems scoreboard is not updated. And when I visit Ranking Your Rank: 106 Your Score: 340.00002 is showing which correspond to 0.00002 pts.
Can you please check and remove conflicts.
hey one test case is showing tle in TABGAME else all the other test case are passing in time.
only top 10 will get the laddus???
Hey guys, so the contest draws to a close.
The editorials for 7 of the problems are ready. Editorials were quite ahead of the schdule, but sadly there was an unexpected personal problem from 9 to 14th september which messed the schdule up :(. The setter’s solution for those 4 editorials will be released so so yous dont have to wait for official editorials to explore the solutions.
Assign some pattern or ordering to the 2-D array such that P_i and P_{i+1} are adjacent. Example, we can assign an ordering like-
1, 2,3,..., N
2N...N+3,N+2,N+1
and so on. Convert this into a 1-D array and try to minimize the sum of maximum and minimum subbaray in it. Trying various orderings and various moves on 1-D array can give a good score.
From what I comprehended from setter’s notes, it says that collisions are of no use. Use matrix exponentiation to compute contribution of each Selina individually. Details of how to construct the matrix can be seen from some of the AC solutions
write phi(n) = 2^s * m where m is odd , let a be a random integer between [2 , n-1] , if gcd(a , n) is nonzero , we found a nontrivial factor , otherwise we have that (a^m-1)(a^m+1)(a^2m+1)(a^4m+1)...(a^{2^{s-1}} + 1) = a^{\phi n} - 1 which is a multiple of n , but from among the factors on left , with probability 3/4 , none of them divide n , hence taking their gcd with n , we will find a factor of n with high probability , then we recursively factor x and n/x where x is the factor found (Note that we can use the same phi for factors as we dont really need phi but any multiple of exponent of Zn* which is a factor of phi).
The permutation we’re looking for is in fact a perfect maching in the complete bipartite graphs represented by rows x columns.
Suppose some of these edges are black, while the others are white.
To check if we have a perfect matching with odd number of black edges we want the sum of even coefficients of
f(z) = det(M(z)) where M_{i, j} = x_{i, j} * z (if {i, j} is black) or = x_{i, j} (if {i, j} is white) to be non-zero.
To go about finding the sum of odd coefficients, we’ll compute f(1) - f(-1).
Let’s generalize a bit. Let’s say we want to see whether there is a matching with odd number of edges of colors K_1, K_2, ..., K_M.
We’ll consider a polynomial in F^{M[z_1, z_2, ..., z_M]}.
f(z_1, z_2, ..., z_M) = det(M(z_1, z_2, ..., z_M)) where M_{i, j} = x_{i, j} * z_{i_1} * z_{i_2} * ... * z_{i_p} (if {i, j} consists of colors i_1, i_2, ..., i_p).
We’re interested to see if there is at least one coefficient of this multinomial in which all exponents of z_i are odd.
This can be done using a generalized approach of the former solution, computing
f({1, 1, ..., 1})
- f({-1, 1, 1, ..., 1}) - f({1, -1, 1, ..., 1}) - f({1, 1, -1, ..., 1}) - ... - f({1, 1, ..., 1, -1})
+ f({-1, -1, 1, ..., 1}) + f({-1, 1, -1, ..., 1}) + ... +
+ ... + (-1)^M f({-1, -1, -1, ..., -1}).
We will start from the most significant bit and determine whether we can add a new bit to our configuration using the above mentioned algorithm.
The time complexity is T(N) = T(N/2) + O(N * W) = O(N * W), where N is the maximum value and W is the time to compute the determinant.
I apologize for the delay, the required editorials will be put soon. Meanwhile, please enjoy the rest. 
Why are the Junior Ratings calculated only till SEPT18A?
When will our Junior Ratings get updated?
how many participant are eligible for job opportunities at ShareChat?
Agreed. This problem statement leaves a lot to be desired.
Not only near perfect, but perfect score, as all other users are getting exactly zero points, not something like 0.000 something.
@admin, please look into this.
Ironically name of problem CHEFZERO = Chef will give you Zero. xD.
Score of @mcfx1 0.000017 pts. His own soln with 0.642936pts fetches him zero.
We need a change in scoring fn. And rest of scores are of order 1e9.
A small change, like (max-min+1) in place of (max-min) will ensure avoidance of non zero score, though there might be some better scoring function.
Can you please explain the third condition in a way that does not reveal the solution. I am still not able to get it completely. Thanks.
XD…
relatively his solution’s score is 10^9 times less than other’s … there’s a lot of optimization he did… great work by him… XDDDDD
there has been many challenge problem’s where bf or random answer fetches 10-50 points but this one is completely reverse of it…
Tie Breaker is a tie breaker only b/w 1st and 2nd ranks xD
disappointed.
@codebreaker123 I would like to, but since I am not associated with the problem setting team it is probably not my place to do so.
No response or action till now, incredible.
Seems like they updated it, now people with double digit scores are getting some points. Not big enough to be visible on ranklist page tho.
@vijju123 how long does it take to rejudge? The scores of old submissions have not been updated yet.