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By RAD, 7 years ago, translation, ,

271A - Beautiful Year

This is a very straight forward problem. Just add 1 to a year number while it still has equal digits.

271B - Prime Matrix

Precalculate the next prime for every integer from 1 to 105. You can do that in any way. The main thing is to test all the divisors up to square root when you are checking if a number is prime.

Now for each aij (element of the given matrix) we can easily calculate addij — how many do we have to add in order to make aij prime. After that all we need to do is to find row or column with minimal sum in this new matrix.

271C - Secret

If 3k > n there is no solution (because each of the k sets must have at least 3 elements). Otherwise we can divide first 3k words in the following way:

1 1 2 2 3 3 ... k k 1 2 3 ... k

For each of the k sets, the difference between the first and the second elements will be 1. And the difference between the second and the third elements is definitely not 1 (more precisely, it is 2k - i - 1 for the i-th set). So each set doesn't form an arithmetic progression for sure.

For this solution it doesn't matter how we divide the rest n - 3k words.

271D - Good Substrings

At first, build a trie containing all suffixes of given string (this structure is also called explicit suffix tree). Let's iterate over all substrings in order of indexes' increasing, i. e. first [1...1],  then [1...2], [1...3], ..., [1...n], [2...2], [2...3], ..., [2...n], ... Note, that moving from a substring to the next one is just adding a single character to the end. So we can easily maintain the number of bad characters, and also the "current" node in the trie. If the number of bad characters doesn't exceed k, then the substring is good. And we need to mark the corresponding node of trie, if we never did this before. The answer will be the number of marked nodes in the trie.

There is also an easier solution, where instead of trie we use Rabin-Karp rolling hash to count substrings that differ by content. Just sort the hashes of all good substrings and find the number of unique hashes (equal hashes will be on adjacent positions after sort). But these hashes are unreliable in general, so it's always better to use precise algorithm.

271E - Three Horses

It could be proved, that a card (x, y) (x < y) can be transformed to any card (1, 1 + k·d), where d is the maximal odd divisor of y - x, and k is just any positive integer. So every (ai - 1) must be divisible by d, i. e. d is a divisor of gcd(a1 - 1, ..., an - 1), and we can just iterate over all possible divisors. Let's take a look at all the initial cards (x, y), which have have d as their maximal odd divisor: these are cards with y - x equal to d, or 2d, or 4d, 8d, 16d, ... Don't forget that the numbers x and y must not exceed m. It means that the total number of cards with some fixed difference t = y - x is exactly m - t.

The resulting solution: sum up (m - 2ld), where d is any odd divisor of gcd(a1 - 1, ..., an - 1), and l is such, that 2ld ≤ m.

• +39

 » 7 years ago, # |   0 Is there any background in problem E?
 » 7 years ago, # |   0 In 4rth q after so many TLE's I have drawn an inference that generally map is slower than set though both being dynamic memory allocation....
 » 7 years ago, # |   0 I'm quiet new in Java,I keep getting TLE and use trie data structure,any advises?(Though the same code written in C++ got AC) code
•  » » 7 years ago, # ^ |   +4 It's not because you use Java, it's because you make a lot of substring() calls.
•  » » » 7 years ago, # ^ |   0 Thanks,so is there a faster way to remove the first character from string?Or i should use array of chars instead of strings or something like that
•  » » » » 7 years ago, # ^ |   0 Just remember where the string really begins.
•  » » » » » 7 years ago, # ^ |   0 Yup thanks got it
 » 7 years ago, # |   +3 i m not getting the problem E' editorial can any one explain ?
•  » » 7 months ago, # ^ |   0 Here is the proof for the solution given in the editorial.
 » 7 years ago, # |   0 For problem 4, suffix array with lcp can also be used. We can first pre-compute the bad character count for all the substrings and store it in a table. Then, iterating through the suffix array and avoiding duplicate substrings using the lcp table, we can find the answer.
 » 7 years ago, # |   +5 Hello. I am having trouble with the problem D (getting TLE). I am using the trie to solve the problem. When I build it, I already keep track of the current number of bad characters. If it is bigger than k, I leave the recursion. The final answer is the number of nodes of the tree.Could anyone help me please? What am I doing wrong? Code
•  » » 7 years ago, # ^ | ← Rev. 2 →   +5 Why do you have these 2 nested loops in main() function? You are not really using i anywhere.
•  » » » 7 years ago, # ^ |   +5 OMG, thanks RAD. It was something I was trying and forgot to remove. for (int i = 0; i < n; ++i) { put(t, s + i); } 
 » 7 years ago, # |   0 Problem C: in each set Ui, its elements ui, 1, ui, 2, ..., ui, |Ui| do not form an arithmetic progression (in particular, |Ui| ≥ 3 should hold). What's the meaning of it ? Does it means when Ui < 3 there is no necessary to hold this condition ?
 » 7 years ago, # |   0 Can you tell me what's wrong of my code ? Problem D: 3145554
•  » » 7 years ago, # ^ |   +5 Complexity
•  » » » 7 years ago, # ^ |   0 I don't think so, I learned from this AC submission? Why his is ok ? 3105181
•  » » » » 7 years ago, # ^ |   0 Try to find out what is the complexity of string comparisons
 » 7 years ago, # |   0 Somebody can help me prove that a card (x, y) (x < y) can be transformed to any card (1, 1 + k·d), where d is the maximal odd divisor of y - x in problem "Three Horses"? Thank you
 » 2 years ago, # | ← Rev. 4 →   0 Problem D 271D - Good Substrings : This is my code. I've used the concept of Rolling hash/Rabin Karp algorithm. But I guess there are some collisions happening and it's failing on test case number 8. Any help is greatly appreciated. Been trying this for a long time. Code: 27820412Update: Solved! I tried many different ways. Initially, I used two hash moduli instead of one, but this exceeded the Time Limit on test #8 (Which is weird!). I ended up using a really large prime (About the size of 2^50) which just passed the time limit for Testcase #8 and passed the cases as well. :)
•  » » 22 months ago, # ^ |   0 don't use 2 ^ k type constants. it will certainly cause collisions. i solved it using two different constants and it passed perfectly fine. 33217384
•  » » 5 months ago, # ^ |   0 Can you help me in explaining why we use two modulo hash instead of 1
 » 12 months ago, # |   0 D can be solved using 2 pointer and little bit of a lcp array.This is my solution. https://codeforces.com/contest/271/submission/44768791
 » 9 months ago, # |   0 In problem D, why bruteforce cost O(n^2 * lg(n^2)) ~= 1e7 gives TLE. Can someone explain why ? I'm really confuse about it.
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 I also had the same query, apparently, it leads to memory limit exceeded (MLE). Refer this post: https://codeforces.com/blog/entry/6668
 » 8 months ago, # |   0 Can anyone explain the solution to 271A - Beautiful Year more clearly ?
•  » » 8 months ago, # ^ |   0 You can solve it by brute force. For a given year Y, start with the year Y+1 and check if it has all distinct digits, if not, keep adding one to it till you find such a year.
 » 5 months ago, # |   0 Can anyone explain me why we use 2 modulo hash here rather than 1. Thanks in advance
 » 4 weeks ago, # |   0 Problem D is like using 2 hashes of 10^9 size-TLE Using 1hash of 10^9 size -WA and using a very big hash and multiplying recursive -TLE