### sophisticated1's blog

By sophisticated1, history, 4 years ago, , How to solve the problem from latest Codechef Long Contest :

https://www.codechef.com/FEB16/problems/SEAPERMS  Comments (8)
 » 4 years ago, # | ← Rev. 3 →   Let's think of a way construct a permutation: Initially, it's empty. One by one, we add an element. Assume, after that we have permutation p1, p2, ..., pk and pi is recent element. So, we must have: p[i-1] + d > p[i], p[i] + d> p[i+1]. If the sequence we add is non-decreasing, then we just care about the condition p[i-1] + d > p[i]. Clearly, the way we add pi equals to the number of element before pi (in sorted non-decreasing sequence) that greater than p[i]-d and plus one for adding at first position. (*) Therefore, we have a solution in O(m * n * log(n)). In order to improve that complexity, in each query, we maintain the number (*) for each element. And in each query, there are O(D) element is affected. Complexity reduce to O((n + m) * d).
•  » » But how are we counting the number of valid permutations using the above logic, Could you please elaborate your solution ??
•  » » » It's the product of them. Example, we have sequence: 1, 3, 4, 7, 9 and d = 3. There is no number less than 1 and greater than 1 — 3. There is one number less than 3 and greater than 3 — 3. There is one number less than 4 and greater than 4 — 3. There is no number less than 7 and greater than 7 — 3. There is one number less than 9 and greater than 9 — 3. So the answer in this case is: (0 + 1) * (1 + 1) * (1 + 1) * (0 + 1) * (1 + 1).
•  » » » » I got the approach, But what is the intuition behind this counting approach ??
•  » » » » Thanx got it :)
•  » » » My solution is complicated. Let's maintain cnt[i] count the number of sequence equals to i in each query. Because, each element can be large as 10^9, so we must decompress them. After having cnt[i] for each i, we will be able to count the number of way add cnt[i] numbers value 'i' in the above way.
 » How to solve WRDSUM?
•  » » 4 years ago, # ^ | ← Rev. 4 →   Use inclusive-exclusive principle: Almost always F(n) = n. So we start with 2 + ... + n. Let's look how to fix our sum we have to subtract and put instead. we have to subtract and put instead. we have to subtract and and put , and instead.And so on. If you calculate this properly you will get sth like this The formula 2d + ... + xd is a polynomial of degree d + 1 so we can use Lagrange interpolation to get coefficients, or precompute Bernoulli numbers and use Faulhaber's formula. Notice that max D is around . Depending on implementation we should get sth like O(D2) or even faster solution. Use python or java to avoid bignum implementation.