Codeforces Round #323 Editorial

Revision en38, by, 2015-10-05 16:27:03


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.

Jury's solution: 13390628


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.

Jury's solution: 13390612


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 .

Jury's solution: 13390679


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 ), 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 .

Jury's solution (with matrices): 13390660

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.

Jury's solution (with suffix and prefix): 13390666


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 . 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 . For finding all such subarrays we will use two pointers approach and for every good ai, such that is not good we will find aj such that are good and is not good. Let has k elements . Any it's subarray is superior, so it gives us arrays of length 1, 2, ..., k with count k, k - 1, ..., 1. As soon as sum of all k is not greater than n, we could just increase counts straightforward. There's a case when all ai are good, in which we must do another increases. Next we must add to the answer only counts of length x, such that gcd(x, n) = g.

Solution described above has complexity O(d(n)n), where d(n) is the number of divisors of n.

Jury's solution: 13390645


It is a common fact that for a prime p and integer n maximum α, such that pα|n! is calculated as , where pw ≤ n < pw + 1. As soon as , the maximum α for is calculated as .

One could see, that if we consider numbers n, k and n - k in p-th based numeric system, rounded-down division by px means dropping last x digits of its p-th based representation. As soon as k + (n - k) = n, every i-th summand in α corresponds to carry in adding k to n - k in p-th numeric system from i - 1-th to i-th digit position and is to be 0 or 1.

First, let convert A given in statement from 10 to p-th numeric system. In case, if α is greater than number of digits in A in p-th numeric system, the answer is 0. Next we will calculate dynamic programming on A p-th based representation.

dp[i][x][e][r] — the answer for prefix of length i possible equal to prefix of A representation (indicator e), x-th power of p was already calculated, and there must be carry equal to r from current to previous position. One could calculate it by bruteforcing all of p2 variants of placing i-th digits in n and k according to r and e and i-th digit of A, and make a translation to next state. It can be avoided by noticing that the number of variants of placing digits is always a sum of arithmetic progression and can be calculated in O(1).

It's highly recommended to examine jury's solution with complexity O(|A|2 + |A|min(|A|, α)).

Jury's solution: 13390698


One could prove that the number of binary functions on 4 variables is equal to 224, and can be coded by storing a 24-bit binary mask, in which every bit is storing function value for corresponding variable set. It is true, that if maskf and maskg are correspond to functions f(A, B, C, D) and g(A, B, C, D), then function (f&g)(A, B, C, D) corresponds to maskf&maskg bitmask.

Now, we could parse expression given input into binary tree. I should notice that the number of non-list nodes of such tree is about . Now, let's calculate dynamic programming on every vertex vdp[v][mask] is the number of ways to place symbols in expression in the way that subtree of vertex v will correspond to function representing by mask. For list nodes such dynamic is calculated pretty straightforward by considering all possible mask values and matching it with the variable. One could easily recalculate it for one node using calculated answers for left and right subtree in 416 operations: dp[v][lmask|rmask] +  = dp[l][lmask] * dp[r][rmask].

But all the task is how to make it faster. One could calculate s[mask], where s[mask] is equal to sum of all its submasks (the masks containing 1-bits only in positions where mask contains 1-bits) in 24·224 operations using following code:

for (int mask = 0; mask < (1 << 16); ++mask) s[mask] = dp[x][mask];
for (int i = 0; i < 16; ++i)
    for (int mask = 0; mask < (1 << 16); ++mask)
        if (!(mask & (1 << i))) s[mask ^ (1 << i)] += s[mask];

Let's calculate sl[mask] and sr[mask] for dp[l][mask] and dp[r][mask] respectively. If we will find s[mask] = sl[mask] * sr[mask], s[mask] will contain multiplications of values of pairs of masks from left and right dp's, which are submasks of mask. As soon as we need pairs, which in bitwise OR will give us exactly mask, we should exclude pairs, which in bitwise OR gives a submask of mask, not equal to mask. This gives us exclusion-inclusion principle idea. The formula of this will be

, where p is the parity of number of bits in mask^submask.

Such sum could be calculated with approach above, but subtracting instead of adding

for (int mask = 0; mask < (1 << 16); ++mask) s[mask] = sl[mask] * sr[mask];
for (int i = 0; i < 16; ++i)
    for (int mask = 0; mask < (1 << 16); ++mask)
        if (!(mask & (1 << i))) s[mask ^ (1 << i)] -= s[mask];

In such way we will recalculate dynamic for one vertex in about 3·24·216 operations.

Jury's solution: 13390713

Tags editorial


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