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Clashing with Hourstorm 11
HourStorm is tomorrow , Cook off is today!
Almost TLE :O 7.99 out of 8s
Does Problem Bitsetbaba and Power Grid based on Gaussian Elimination to find the subset with given xor?
Gaussian Elimination for finding rank of binary matrix of all numbers as rows.
In Bitsetbaba and Power Grid, why is the answer not number of elements divided by number of unique poisitive elements
For any positive number, the numbers will make pairs. So after one number (say x) we get pairs say (a, b), (c, d), (e, f), ...
Now for second number (say y) if we have another pairing. If it a pairs up with c then a ^ c = y Claim is that b ^ d = y. a ^ b = x and c ^ d = x so a ^ b ^ c ^ d = 0. Now a ^ c = y then b ^ d = y.
So with every unique number the number of component gets divided by 2.
I have only read first sentence.
If you something like $$$x=[1,2,3]$$$, you can obtain only numbers smaller than $$$ \lt 4$$$. So answer is divided by $$$4$$$.
I think the component gets divided by 2 only when the current number inserted lets say a[i] is not the xor of the subset of already inserted i-1 elements otherwise we will get an edge which will be in the same component. Correct me If I am wrong
So If I am right how can we check whether the number we inserted is the xor of subset of already inserted elements.
what will be the approach for Expected xor?
Maintain the probabilities, that number will have bit i.
Can you please elaborate a little?
Let X be a random variable denoting the resultant beauty of the exhibition.
Then X can be written as: $$$X = 2^{31} * X_{31} + 2^{30} * X_{30} + ... + 2^1 * X_1 + 2^0 * X_0$$$, where $$$X_i$$$ denotes the $$$i$$$-th bit of X.
Now, $$$E(X) = 2^{31} * E(X_{31}) + 2^{30} * E(X_{30}) + ... + 2^1 * E(X_1) + 2^0 * E(X_0)$$$, by linearity of expectation.
and also $$$E(i)$$$ = Probability of the $$$i$$$-th bit being equal to 1. To compute $$$E(i)$$$, you may write a simple $$$dp$$$ as $$$dp[i][j][k]$$$ which is the probability of obtaining a subset of the first $$$j$$$ numbers where the $$$i$$$-th bit is having a value equal to $$$k (k = 0/1)$$$.
and $$$E(i) = dp[i][n][1]$$$.
when will the MOSS of the contest get over
Does anyone have a solution to Nearest Color?
Hi. All the editorials are published here :).
Amazing tests in BICLIQUE
Solution: lol lol lol
I did note that problems with YES/NO answers need a LOT of tests to not be very weak. *shrugs*
Could someone explain GCDSET? Isn't it just counting the number of multiples in the range $$$[l,r]$$$ ?
Didnt solve it . But what it seemed to me was — divide l and r by the desired gcd. Suppose we get l' and r'. Now there is a set made out of numbers between l' and r' . These numbers must be mutually co prime. We have to report the maximum number of elements in this set.
There is one corner case. What if there is only one multiple in range [l,r], then answer maybe 1 or 0 depends on this multiple.