### chokudai's blog

By chokudai, history, 2 years ago,

We will hold AtCoder Beginner Contest 149.

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

Update: Due to the troubles in judging, this contest will be unrated. Sorry for the inconvenience.

• +24

 » 2 years ago, # |   +18 Please note the unusual start time. It starts one hour before the usual start time. :)
 » 2 years ago, # |   0 My submission for problem D is had stuck on test 58Can you please fix it
 » 2 years ago, # |   +2 Nice. Now the round is unrated. (First time solved 4 tasks);_;
 » 2 years ago, # |   0 admin:Due to the troubles in judging, this contest will be unrated.
 » 2 years ago, # |   +4 Can anyone clarify what F wants us to do, the language is unclear to me?
•  » » 2 years ago, # ^ |   0 S is the smallest subtree that contains all the black vertices. You need to calculate the expected value of the white vectices in S.
•  » » » 2 years ago, # ^ |   +3 Aaahhhh, got it, I mistook it for "S is the smallest subtree containing only black vertices".
 » 2 years ago, # |   +5 How to solve E?
•  » » 2 years ago, # ^ |   0 It can be solved using FFT, but there is a binary search solution too.
•  » » 2 years ago, # ^ |   +10 First you could binary search over the minimum value of pair sum such that count of all pairs satisfying this criteria is more than equal to M.To find count of such elements in each binary search check, you could use 2 pointers, one for maximum and the other for minimum and get the total elements satisfying the condition. After binary search we have the minimum pair sum so that total count of all pairs having more or equal sum is at least M. Say this minimum is 'Min'.So for pair sum more than equal to Min + 1, has count less than M.Sum up all such pairs with pair sum more than equal to Min + 1. And for the remaining pairs to make the count = M, we add that Min required number of times to the answer.
 » 2 years ago, # |   +8 has anyone wrote dp solution of D??
•  » » 2 years ago, # ^ |   0 No mine is greedy.
•  » » » 2 years ago, # ^ |   0 Can you tell me about your greedy approach. Even I tried greedy but after test case #12 every 1 out of 3 test cases are giving wrong answer.
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   0 So what i was doing is first win all the first 'k' games and store the answers of these games in a vector. Now starting from K+1 game check the condition that i — k does not contradict your winning hand. If it does add a '?' to the vector to make sure that this should not decrease the number of wins. We dont care what this '?' is.Code
•  » » » » 2 years ago, # ^ |   0 Greedily i think you could go for segregating all r,p, and s elements and storing their indices.Iterate over segregated indices and greedily mark the position ,say i, as winning position if the position i
•  » » » » 2 years ago, # ^ |   -10 Just check if s[i — k] == s[i] -> then you cannot add the score and assign s[i] = (something character you like).else add the score.
•  » » » » 2 years ago, # ^ |   0 My bad, just a small mistake cost we WA.
•  » » » » 2 years ago, # ^ |   0 Iterate all characters t[i], if t[i] != t[i-k] just plus scores to the answer, otherwise assign t[i] with a character which is different from original t[i] and t[i+k].
•  » » 2 years ago, # ^ |   0 I think you could generate K dp's representing dp over different indices mod K.
•  » » » 2 years ago, # ^ |   0 can you elaborate it further ??
•  » » » » 2 years ago, # ^ |   0 If you split indices on the basis of their mod value with respect to K, then we require consecutive indices of a group should not have same move.So with only a maximum 3 choices at each point, you could run a (N*3) dp for each group to get maximum answer for each group.
•  » » 2 years ago, # ^ |   0 Yep me.Whenever i-k is same as i pick maximum of dp[i-1][rock], dp[i-1][paper], dp[i-1][scissor] and change ith element to '\$' so that next i+k you can use any of rock, paper and scissor again.
•  » » 2 years ago, # ^ |   0 Pretty much you separate into k strings and consider each one independently.
•  » » 2 years ago, # ^ | ← Rev. 4 →   0 here is minethe source is pretty much self explanatory i guess. here is the main idea, if str[i]== str[i-k] //for i>k dp[i] = max(dp[i-k], dp[i-1]) str[i]='0' else dp[i] = dp[i-1] + (points gained from current win) Edit: complexity : O(N)
 » 2 years ago, # |   0 Auto comment: topic has been updated by chokudai (previous revision, new revision, compare).
 » 2 years ago, # |   0 can someone explain the solution of D. I thought it was dp but couldn't figure out the states as i am very very weak in dp. :(
•  » » 2 years ago, # ^ |   0 let dp[i][j] be the maximum score witch you can get if we will provide only i%k-th, k+i%k-th... i-th games and on i-th game you will use j-th hand(if j=1 Rock, if j=2 Scissors, if j=3 Paper). dp[i][j]=dp[i-k][1]+dp[i-k][2]+dp[i-k][3]-dp[i-k][j]. answer is: max(dp[n][1],dp[n][2],dp[n][3])+...max(dp[n-k+1][1],dp[n-k+1][2],dp[n-k+1][3]). code: https://atcoder.jp/contests/abc149/submissions/9217025
 » 2 years ago, # |   0 Why rating has not updated yet?
•  » » 2 years ago, # ^ |   0 The contest is unrated as said by the admin due to the slow judge.
 » 2 years ago, # |   0 How to solve question E? I got wrong .
•  » » 2 years ago, # ^ |   0 I used FFT to find all possible sums from array A and B and then picked up the required number of elements from the sum from last. A
•  » » » 2 years ago, # ^ |   0 can you please share the link to your solution
•  » » » » 2 years ago, # ^ |   0
•  » » » » » 2 years ago, # ^ | ← Rev. 2 →   +6 abhisharma_07, wangzimo, it can be solved with Karatsuba multiply too. Submission
•  » » » » » » 2 years ago, # ^ |   0 Thank you very much!
 » 2 years ago, # |   +1 Please explain how to solve F in more details. I'm unable to understand editorial
 » 2 years ago, # |   0 How to solve E using binary search?