### maroonrk's blog

By maroonrk, history, 7 weeks ago,

We will hold Tokio Marine & Nichido Fire Insurance Programming Contest 2021（AtCoder Regular Contest 122).

The point values will be 400-500-600-700-800-1200.

We are looking forward to your participation!

• +99

 » 7 weeks ago, # |   +35 Reminder that contest starts in 5 minutes
 » 7 weeks ago, # |   +14 Reminder that contest starts in 4 minutes
 » 7 weeks ago, # |   +14 Reminder that contest starts in 3 minutes
•  » » 7 weeks ago, # ^ |   +7 Reminder that contest starts in 2 minutes
 » 7 weeks ago, # |   0 Reminder that contest starts in 1 minute.
 » 7 weeks ago, # |   +3 Reminder that contest starts in 10 seconds
 » 7 weeks ago, # |   +1 it's been 45 min and I have done NOTHING.
 » 7 weeks ago, # |   +21 F*ckin' precision errors >:O
 » 7 weeks ago, # |   0 its been 1:30 hours ..and I have done only B :(
 » 7 weeks ago, # |   +54 I used to treat ARCs as Div. 2 contests.I was wrong.
•  » » 7 weeks ago, # ^ |   +11 rng_58 says they are more of a Div 1.5 https://codeforces.com/blog/entry/75163?#comment-592623
 » 7 weeks ago, # |   +1 Can we solve Problem B — Insurance , when the scenarios don't happen with equal probability?
•  » » 7 weeks ago, # ^ |   -9 I can't think of any possible solution for that case.
•  » » 7 weeks ago, # ^ |   +17 I think the solution will be pretty much the same.
•  » » 7 weeks ago, # ^ |   +17 It would be exactly the same, but when you are computing the expected value you multiply by the probability, not by $1/n$.In this code, in function f you can just replace E += p * v; with E += p[i] * v;.
•  » » » 6 weeks ago, # ^ |   0 bro i have one question i did it this way. Am i doing the same thing like your code aor its differnt. Because when i saw the function is first increasing and then decreasing i just check the previous x — 0.00001 value and if it is smaller than current x. we surely have to go to the right = mid — 0.00001. So is it similar to yours or should i learn your algorithm too For future questions?? HELP !! https://atcoder.jp/contests/arc122/submissions/23687736 MY CODE
 » 7 weeks ago, # | ← Rev. 2 →   -8 C had a cancer implementation.
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   +8 My (unproven) solution is quite easy to implement. Just apply Euclidean algorithm with pairs $(n, b)$ such that $b$ is close to $n / \varphi$, until you find a solution with less than $130$ operations.
•  » » » 7 weeks ago, # ^ |   0 what is n/φ and how did you arrive at this conclusion?
•  » » » » 7 weeks ago, # ^ | ← Rev. 3 →   0 $\varphi = (1 + \sqrt 5) / 2$. The last operations (of type $3$ and $4$) are executed only once (alternately) for each side, so $a, b$ grow "fast". The only bottleneck is that there may be many initial operations of type $1$ and $2$ (because of the error propagation of $b - n / \varphi$), but the total operations are about $200$ on average with a very high deviation.
•  » » » 7 weeks ago, # ^ |   0 that was my first solution, then I couldn't find a way to get it down below 150 moves.
•  » » » » 7 weeks ago, # ^ |   0 For which input did your code reach $150$ moves?
•  » » » » » 7 weeks ago, # ^ |   0 1e18 gave exactly 150 moves (for my initial code)
•  » » » » » » 7 weeks ago, # ^ | ← Rev. 2 →   0 b = 618033988749894686 works in 106 moves. b = 618033988749894689 works in 127 moves. b = 618033988749894693 works in 119 moves.The density of $b$ that work seems quite high.UPD: actually it's not necessary to choose $b \approx n / \varphi$, just using a random $b$ seems to work.
 » 7 weeks ago, # |   +25 Problem D: Try and fail to write the solution by yourself. Remember there's a XOR MST problem in codeforces, google it and copy a solution from therehttps://codeforces.com/contest/888/submission/32165432
•  » » 7 weeks ago, # ^ |   0 I don't get the relationship between today's problem and MST. Is "the graph with edges $\leq x$ is connected" equivalent to "Bob can get a score $\leq x$"? (Why?)
•  » » » 7 weeks ago, # ^ |   0 It isn't. We take only the edges that merge components of not both even size. Think about how things work in a trie, if you have both subtrees of even size then you'll solve the subtrees independently otherwise there'll be one number going to the other subtree and that'll dominate everything.
 » 7 weeks ago, # | ← Rev. 2 →   +8 How To Lose Rating search on Google "pair with min xor" find this solution on gfg: "sort the given array, traverse and check xor for every consecutive pair" get wa debug the solution for $30$ minutes $2$ minutes to the end, realize that the solution in 2) can't be generalized if you want to find $\min(a_i \oplus b_j)$ (i.e. using, for each $a_i$, lower_bound() and upper_bound() to find the closest elements in $b$ doesn't work)
•  » » 7 weeks ago, # ^ |   0 How to prove that the function in B is convex? I can't get it?
•  » » » 7 weeks ago, # ^ |   0 It's a sum of convex functions.
•  » » 7 weeks ago, # ^ |   +3 There should've been a red alert going in your head when you tried the first step and realised you don't have a given array, but 2 given arrays.
•  » » » 7 weeks ago, # ^ |   0 My thinking flow ok, wait, we have 2 arrays oh nvm, I can just do upper_bound() and it works
•  » » » » 7 weeks ago, # ^ |   0 Same by sorting the two given arrays and get WA. Luckily I came up with a counterexample soon and implement Trie after that.
•  » » » » » 7 weeks ago, # ^ |   0 Can you share that counter-example? I got WA using that too lol
•  » » » » » » 7 weeks ago, # ^ | ← Rev. 2 →   +8 For example if $a=[3],b=[4,7]$, by only checking consecutive pairs you get the answer $7$, but the real answer is $4$.I haven't come up with counter-example for upper_bound ones yet :P
•  » » » » 7 weeks ago, # ^ |   0 That takes an array as argument. Which array and why not the other way around? Hmm.It's why I want to at least intuitively understand the algorithms I copy. If I get wrong results, might as well throw it away and start over otherwise.
 » 7 weeks ago, # |   0 how to solve C?
•  » » 7 weeks ago, # ^ |   +3
•  » » » 7 weeks ago, # ^ |   +3 Its a very nonintuitive way (atleast for me) , if you have solved it , can you tell me your intuition or how u solved it.
•  » » » » 7 weeks ago, # ^ |   +14 I didn't solve it :P
•  » » » » 7 weeks ago, # ^ |   +1 While exploring the possible solutions, I've decided to see what happens in some easy scenarios. One of those scenarios was to start from $x=1$, $y=0$ and then do operations $x=x+y$ and $y=x+y$ alternatively. I've noticed that in this case $x$ and $y$ after some steps are always Fibonacci numbers, so we need just to represent $N$ as a sum of Fibonacci numbers.
•  » » » » » 7 weeks ago, # ^ |   0 Okkk thanks .
 » 7 weeks ago, # |   0 I used Fibonacci triangle rows in Ahttps://oeis.org/A144154https://atcoder.jp/contests/arc122/submissions/23371678With so many solves probably not the intended solution, but I was happy to have found the relation
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   0 can you please explain the theory?is it pascal triangle?
•  » » » 7 weeks ago, # ^ | ← Rev. 2 →   0 I drew some small examples of how many + and — we can have at each position. For example with 4 signs we have: 5 6 6 5 3 2 2 3 for 5 signs: 8 10 9 10 8 5 3 4 3 5 Assuming the 1st and last columns always contain fibonacci pairs, how to generate the ones in between? Well at each step each of the — factors will turn into + on the next turn since we cannot have two consecutive -, and also we can notice from drawing smaller examples that the number of + factors also break into fibonacci pairs — 8+ breaks into 5+ 3-, 5+ breaks into 3+ and 2- and so on.Then we can see in the above example 8/5, the second column is generated as8+ breaking into 5+ and 3-, and 5- turning into 5+therefore second column contains 2*5 + and 1*3 -similar transition happens for the next column, 2*5+ breaks into 2*3+ and 2*2-and 1*3- turns into 1*3+, combined that is 3*3+ and 2*2-and we can generalize to1*8 2*5 3*3 5*2 8*11*5 1*3 2*2 3*1 5*1which are also fibonacci numbers!So in my solution I generate these two rows, compute each index's coefficient and add it to my result.This "linearity" type solution is used in the video https://codeforces.com/blog/entry/91672?#comment-803151
 » 7 weeks ago, # |   +11 I wish I had more clarity of mind while implementing problems like C :D
 » 7 weeks ago, # |   +1 Can someone please explain me the the implementation of dp in problem A?
•  » » 7 weeks ago, # ^ |   +7 let dp[i][0] be sum of values of equation end with +a_i, dp[i][1] = end with -a_i, and cnt[i][0] is the number of equations end with +a_i, cnt[i][1] = end with -a_ithen dp[i][0] = (dp[i-1][0] + dp[i-1][1]) + (cnt[i-1][0]+cnt[i-1][1]) * a_idp[i][1] = dp[i-1][0] — cnt[i-1][0] * a_icnt[i][0] = cnt[i-1][0] + cnt[i-1][1]cnt[i][1] = cnt[i-1][0]the answer is dp[N][0] + dp[N][1]
•  » » » 7 weeks ago, # ^ |   0 thanks
•  » » » 7 weeks ago, # ^ |   0 Hello, i have read your code, your code is amazing, but i have some questions. what's the meaning of "#include" in line 2. Why did I remove it and change the "mint" into "long long" and get the wrong answer
•  » » » » 7 weeks ago, # ^ |   +5
•  » » » » » 7 weeks ago, # ^ |   0 oh thanks
 » 7 weeks ago, # |   0 B was easier for me than A as B was simple math.But I approached B very late sadly
 » 7 weeks ago, # |   0 Managed to solve C only after the contest :( Here's my Randomised solution:
 » 7 weeks ago, # |   +8 When you realize, for E, that arbitrarily choosing the last element satisfying the condition works, and you spent last half hour thinking how to select last element if multiple element satisfies condition.
 » 7 weeks ago, # |   +16 I implemented a different solution for B than the one in the official editorial. Hopefully it will prove interesting for some:We should find minimum after x and y for: (n * x + PS[n] - PS[y] - 2 * x * (n - y)) / n, where PS[i] is the sum of the first i elements of the sorted input array. Basically, y is an index such that all elements 1 <= i <= y will provide min(A[i], 2x) = A[i] and all elements y < i <= n will provide min(A[i], 2x) = 2x. This also suggests a constraint for a fixed y: V[y] / 2 <= x <= V[y + 1] / 2.Minimizing the formula above is the same as minimizing: n * x - PS[y] - 2 * x * (n - y) = x * (2 * y - n) - PS[y]. In order to minimize this, we can iterate 1 <= y <= n and find a suitable x. In fact, if 2 * y - n is negative, x should be as large as possible, so x = V[y + 1] / 2. Otherwise (2 * y - n is positive), than x should be as small as possible, so x = V[y] / 2.
•  » » 7 weeks ago, # ^ |   0 can you give a link to your submitted solution?
 » 7 weeks ago, # |   +53
 » 7 weeks ago, # |   0 How much can we rate A in terms of CF ratings ?
•  » » 7 weeks ago, # ^ |   +4 I suppose 1600/1700.
 » 7 weeks ago, # |   0 In B, I first wrote a binary search, but the ans was increasing with decreasing precision. I think, the side it was going was not correct when precision was small and numbers were large. But, ternary search was consistent on this. Can someone clarify more on this?
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   0 I did it using binary search on the x Solution. The loss vs x function is initially non-increasing and then non-decreasing, so we'll have to find x corresponding to the trough in the loss function [Start with low = 0 and high = max(Ai), then check the losses for mid, mid-1, mid+1 to half the search space] Later checked the editorial, turns out this is just equivalent to finding the median.
•  » » » 7 weeks ago, # ^ |   0 I was checking on mid and mid + e(the error term), I think that caused the issue
 » 7 weeks ago, # |   +9 For problem B my solution tries to perform binary search to find when the derivative flips from negative to positive.$f(x) = \sum\limits_{i=1}^N (x + A_{i} - min(A_{i}, 2x)) = \sum\limits_{i=1}^N (x + A_{i} - \frac{Ai + 2x - abs(A_i - 2x)}{2})$$f'(x) = \sum\limits_{i=1}^N \frac{2x - A_i}{abs(2x - Ai)}$Points where we get NaNs because of division by zero are skipped.
•  » » 7 weeks ago, # ^ |   +7 Thanks.This just proves that median observation in editorial since that function is sum of signum functions derivate will change sign at median only.
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   +8 Nice observation! That way to write $min(a,b)$ is very clever.
 » 7 weeks ago, # | ← Rev. 2 →   0 ARC E -If one didnt realises second observation in E. Then $LCM_{j \neq i}(A_j)$ can be computed in $O(n \log max(A_i))$ using prefix and suffix LCMs with bigints.Python SubmissionTime complexity $O(n^3 \log max(A_i))$ remains same.More specifically how did I come up with this particular solution was -Since there exists one last no from index $i$ such that $LCM_{j \neq i}(A_j) \neq LCM_{j}(A_j)$. This python solution just brute forces to find one such index which can be last. Then recursively solves for the rest of $n-1$ numbers.
 » 7 weeks ago, # |   0 In D I thought it was ok to remove all pairs of equal elements initially, so that all the elements are distinct but it was failing on one test case and on removing this it passes all test cases, can someone point out a case where it fails(the test cases are not available on the dropbox yet)
•  » » 7 weeks ago, # ^ |   0 ${1,19,19,20}$Answer is 18 (1 xor 19 = 18, 19 xor 20 = 7) but removing 19 makes it (1 xor 20=21).
•  » » » 7 weeks ago, # ^ |   0 Thanks!
 » 7 weeks ago, # |   +8 C:spend hour to write solution which uses up to 250 operations, delete it, write proper solution in 15 minutes
 » 7 weeks ago, # |   0 Can anybody explain the curve in Question B
 » 7 weeks ago, # | ← Rev. 2 →   -17 .