Codeforces Round #323 Editorial

Revision en14, by danilka.pro, 2015-10-03 19:43:54

Adiv2

To solve the problem one could just store two arrays hused[j] and vused[j] sized n and filled with false initially. Then process intersections one by one from 1 to n, and if for i-th intersections both hused[hi] and vused[vi] are false, add i to answer and set both hused[hi] and vused[vi] with true meaning that hi-th horizontal and vi-th vertical roads are now asphalted, and skip asphalting the intersection roads otherwise.

Such solution has O(n2) complexity.

Bdiv2

It is always optimal to pass all the computers in the row, starting from 1-st to n-th, then from n-th to first, then again from first to n-th, etc. and collecting the information parts as possible, while not all of them are collected.

Such way gives robot maximal use of every direction change. O(n2) solution using this approach must have been passed system tests.

Adiv1

Let the answer be a1 ≤ a2 ≤ ... ≤ an. We will use the fact that gcd(ai, aj) ≤ amin(i, j).

It is true that gcd(an, an) = an ≥ ai ≥ gcd(ai, aj) for every 1 ≤ i, j ≤ n. That means that an is equal to maximum element in the table. Let set an to maximal element in the table and delete it from table elements set. We've deleted gcd(an, an), so the set now contains all gcd(ai, aj), for every 1 ≤ i, j ≤ n and 1 ≤ min(i, j) ≤ n - 1.

By the last two inequalities gcd(ai, aj) ≤ amin(i, j) ≤ an - 1 = gcd(an - 1, an - 1). As soon as set contains gcd(an - 1, an - 1), the maximum element in current element set is equal to an - 1. As far as we already know an, let's delete the gcd(an - 1, an - 1), gcd(an - 1, an), gcd(an, an - 1) from the element set. Now set contains all the gcd(ai, aj), for every 1 ≤ i, j ≤ n and 1 ≤ min(i, j) ≤ n - 2.

We're repeating that operation for every k from n - 2 to 1, setting ak to maximum element in the set and deleting the gcd(ak, ak), gcd(ai, ak), gcd(ak, ai) for every k < i ≤ n from the set.

One could prove correctness of this algorithm by mathematical induction. For performing deleting and getting maximum element operations one could use multiset or map structure, so solution has complexity .

Bdiv1

One could calculate matrix sized n × n mt[i][j] — the length of the longest non-decreasing subsequence in array a1, a2, ..., an, starting at element, greater-or-equal to ai and ending strictly in aj element with j-th index.

One could prove that if we have two matrices sized n × n A[i][j] (the answer for a1, a2, ..., apn starting at element, greater-or-equal to ai and ending strictly in aj element with j-th index inside last block (a(p - 1)n + 1, ..., apn) and B[i][j] (the answer for a1, a2, ..., aqn \ldots), then the multiplication of this matrices in a way

will give the same matrix but for length p + q. As soon as such multiplication is associative, next we will use fast matrix exponentiation algorithm to calculate M[i][j] (the answer for a1, a2, ..., anT) — matrix mt[i][j] raised in power T. The answer is the maximum in matrix M. Such solution has complexity .

There's an alternative solution. As soon as a1, a2, ..., anT contains maximum n distinct elements, it's any non-decreasing subsequence has a maximum of n - 1 increasing consequtive element pairs. Using that fact, one could calculate standard longest non-decreasing subsequence dynamic programming on first n array blocks (a1, ..., an2) and longest non-decreasing subsequence DP on the last n array blocks (anT - n + 1, ..., anT). All other T - 2n blocks between them will make subsegment of consequtive equal elements in longest non-decreasing subsequence.

So, for fixed ai, in which longest non-decreasing subsequence of length p on first n blocks array ends, and for fixed aj ≥ ai, in which longest non-decreasing subsequence of length s on last n blocks array starts, we must update the answer with p + (T - 2n)count(ai) + s, where count(x) is the number of occurences of x in a1, ..., an array. This gives us solution.

Cdiv1

Let's fix s for every (l, s) pair. One could easily prove, that if subarray contains ai element, than ai must be greater-or-equal than aj for every j such that i ≡ j ± od{gcd(n, s)}. Let's use this idea and fix g = gcd(n, s) (it must be a divisor of n). To check if ai can be in subarray with such constraints, let's for every 0 ≤ r < g calculate

.

It's true that every good subarray must consist of and only of

Tags editorial

History

 
 
 
 
Revisions
 
 
  Rev. Lang. By When Δ Comment
ru6 Russian danilka.pro 2015-10-05 16:27:13 13
en38 English danilka.pro 2015-10-05 16:27:03 13
ru5 Russian danilka.pro 2015-10-05 16:25:24 4 Мелкая правка: 'C[i][j]=\min\limits_{k' -> 'C[i][j]=\max\limits_{k'
en37 English danilka.pro 2015-10-05 16:25:08 4 Tiny change: 'C[i][j]=\min\limits_{k' -> 'C[i][j]=\max\limits_{k'
ru4 Russian danilka.pro 2015-10-04 18:11:51 2795
ru3 Russian danilka.pro 2015-10-04 17:50:17 2316
ru2 Russian danilka.pro 2015-10-04 17:32:41 1720
en36 English danilka.pro 2015-10-04 17:22:08 0 (published)
ru1 Russian danilka.pro 2015-10-04 17:21:49 11138 Первая редакция перевода на Русский (сохранено в черновиках)
en35 English danilka.pro 2015-10-04 16:38:30 4 Tiny change: '\n$mn_r=\min\limits_{i' -> '\n$mn_r=\max\limits_{i'
en34 English danilka.pro 2015-10-04 00:34:59 54
en33 English danilka.pro 2015-10-03 23:46:45 12
en32 English danilka.pro 2015-10-03 23:38:05 38
en31 English danilka.pro 2015-10-03 23:36:33 389
en30 English danilka.pro 2015-10-03 23:26:56 1 Tiny change: ' \cdot sr[mask]$, wh' -> ' \cdot sr[smask]$, wh'
en29 English danilka.pro 2015-10-03 23:15:16 9 Tiny change: 'nts $i+k-1\equiv j \mod{n}). Any it's' -
en28 English danilka.pro 2015-10-03 22:12:48 0 (published)
en27 English danilka.pro 2015-10-03 22:12:15 107
en26 English danilka.pro 2015-10-03 22:10:24 5 Tiny change: 'bitmask.\n()&()\nNow, we ' -> 'bitmask.\n\nNow, we '
en25 English danilka.pro 2015-10-03 22:09:58 1 Tiny change: ' in $mask ^ submask$' -> ' in $mask \^ submask$'
en24 English danilka.pro 2015-10-03 22:09:28 3238
en23 English danilka.pro 2015-10-03 21:42:17 2 Tiny change: '~~\n\n\n\n' -> '~~\n\n\n\n\n'
en22 English danilka.pro 2015-10-03 21:39:38 595
en21 English danilka.pro 2015-10-03 21:32:22 48
en20 English danilka.pro 2015-10-03 21:31:26 2313
en19 English danilka.pro 2015-10-03 20:42:49 196
en18 English danilka.pro 2015-10-03 20:33:33 413
en17 English danilka.pro 2015-10-03 20:24:17 26 Tiny change: 'alpha = \left$\n\n' -> 'alpha = \lfloor \frac{n}{p} \rfloor$\n\n'
en16 English danilka.pro 2015-10-03 20:22:20 26
en15 English danilka.pro 2015-10-03 20:22:08 990
en14 English danilka.pro 2015-10-03 19:43:54 93
en13 English danilka.pro 2015-10-03 19:39:55 44
en12 English danilka.pro 2015-10-03 19:38:49 1441
en11 English danilka.pro 2015-10-03 19:16:11 1016
en10 English danilka.pro 2015-10-03 18:56:25 419
en9 English danilka.pro 2015-10-03 18:55:30 1 Tiny change: ' inductions. For perf' -> ' induction. For perf'
en8 English danilka.pro 2015-10-03 18:55:08 8 Tiny change: 'g maximum operation' -> 'g maximum element operation'
en7 English danilka.pro 2015-10-03 18:54:42 13 Tiny change: '< i \le n$.\n\nOne c' -> '< i \le n$ from the se.\n\nOne c'
en6 English danilka.pro 2015-10-03 18:54:27 24 Tiny change: 'a_k, a_i)$.\n\nOne ' -> 'a_k, a_i)$ for every $k < i \le n$.\n\nOne '
en5 English danilka.pro 2015-10-03 18:53:52 47
en4 English danilka.pro 2015-10-03 18:52:50 22
en3 English danilka.pro 2015-10-03 18:52:04 1329
en2 English danilka.pro 2015-10-03 18:51:34 38 Tiny change: 'herwise.\n\n' -> 'herwise.\nSuch solution has $O(n^2)$ complexity.\n'
en1 English danilka.pro 2015-10-03 18:49:46 494 Initial revision (saved to drafts)