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We can expect lots of explanation with examples from various topics of maths from Personal Injury Lawyer here. I think this site offers lots of solved examples from each of the topic. I am looking for further updates from here. Keep up the good work.
I think this problems requires using FFT to calculate convolutions in O(n*log(n)) time. Taking the kth-order difference is equivalent to a convolution of a[n] and b[n] with the combinatorial numbers.
We can expect lots of explanation with examples from various topics of maths from Personal Injury Lawyer here. I think this site offers lots of solved examples from each of the topic. I am looking for further updates from here. Keep up the good work.
I think this problems requires using FFT to calculate convolutions in O(n*log(n)) time. Taking the kth-order difference is equivalent to a convolution of a[n] and b[n] with the combinatorial numbers.
Aren't you trailing a red herring with that upper bound on k?
Don't worry. If k>=n-1, any two sequences of length n would be similar. So the real upper limit is that of n, which is 10^5.
That's exactly what I meant by the upper bound of 10^9 on k being a red herring.