### maroonrk's blog

By maroonrk, history, 11 months 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

 » 11 months ago, # |   +35 Reminder that contest starts in 5 minutes
 » 11 months ago, # |   +14 Reminder that contest starts in 4 minutes
 » 11 months ago, # |   +14 Reminder that contest starts in 3 minutes
•  » » 11 months ago, # ^ |   +7 Reminder that contest starts in 2 minutes
 » 11 months ago, # |   0 Reminder that contest starts in 1 minute.
 » 11 months ago, # |   +3 Reminder that contest starts in 10 seconds
 » 11 months ago, # |   +1 it's been 45 min and I have done NOTHING.
 » 11 months ago, # |   +21 F*ckin' precision errors >:O
 » 11 months ago, # |   0 its been 1:30 hours ..and I have done only B :(
 » 11 months ago, # |   +54 I used to treat ARCs as Div. 2 contests.I was wrong.
•  » » 11 months ago, # ^ |   +11 rng_58 says they are more of a Div 1.5 https://codeforces.com/blog/entry/75163?#comment-592623
 » 11 months ago, # |   +1 Can we solve Problem B — Insurance , when the scenarios don't happen with equal probability?
•  » » 11 months ago, # ^ |   -9 I can't think of any possible solution for that case.
•  » » 11 months ago, # ^ |   +17 I think the solution will be pretty much the same.
•  » » 11 months 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;.
•  » » » 11 months 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
 » 11 months ago, # | ← Rev. 2 →   -8 C had a cancer implementation.
•  » » 11 months 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.
•  » » » 11 months ago, # ^ |   0 what is n/φ and how did you arrive at this conclusion?
•  » » » » 11 months 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.
•  » » » 11 months ago, # ^ |   0 that was my first solution, then I couldn't find a way to get it down below 150 moves.
•  » » » » 11 months ago, # ^ |   0 For which input did your code reach $150$ moves?
•  » » » » » 11 months ago, # ^ |   0 1e18 gave exactly 150 moves (for my initial code)
•  » » » » » » 11 months 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.
 » 11 months 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
•  » » 11 months 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?)
•  » » » 11 months 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.
 » 11 months 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)
•  » » 11 months ago, # ^ |   0 How to prove that the function in B is convex? I can't get it?
•  » » » 11 months ago, # ^ |   0 It's a sum of convex functions.
•  » » 11 months 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.
•  » » » 11 months ago, # ^ |   0 My thinking flow ok, wait, we have 2 arrays oh nvm, I can just do upper_bound() and it works
•  » » » » 11 months 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.
•  » » » » » 11 months ago, # ^ |   0 Can you share that counter-example? I got WA using that too lol
•  » » » » » » 11 months 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
•  » » » » 11 months 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.
 » 11 months ago, # |   0 how to solve C?
•  » » 11 months ago, # ^ |   +3
•  » » » 11 months 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.
•  » » » » 11 months ago, # ^ |   +14 I didn't solve it :P
•  » » » » 11 months 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.
•  » » » » » 11 months ago, # ^ |   0 Okkk thanks .
 » 11 months 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
•  » » 11 months ago, # ^ | ← Rev. 2 →   0 can you please explain the theory?is it pascal triangle?
•  » » » 11 months 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
 » 11 months ago, # |   +11 I wish I had more clarity of mind while implementing problems like C :D
 » 11 months ago, # |   +1 Can someone please explain me the the implementation of dp in problem A?
•  » » 11 months 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]
•  » » » 11 months ago, # ^ |   0 thanks
•  » » » 11 months 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
•  » » » » 11 months ago, # ^ |   +5
•  » » » » » 11 months ago, # ^ |   0 oh thanks
 » 11 months ago, # |   0 B was easier for me than A as B was simple math.But I approached B very late sadly
 » 11 months ago, # |   0 Managed to solve C only after the contest :( Here's my Randomised solution:
 » 11 months 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.
 » 11 months 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.
•  » » 11 months ago, # ^ |   0 can you give a link to your submitted solution?
 » 11 months ago, # |   +53
 » 11 months ago, # |   0 How much can we rate A in terms of CF ratings ?
•  » » 11 months ago, # ^ |   +4 I suppose 1600/1700.
 » 11 months 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?
•  » » 11 months 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.
•  » » » 11 months ago, # ^ |   0 I was checking on mid and mid + e(the error term), I think that caused the issue
 » 11 months 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.
•  » » 11 months 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.
•  » » 11 months ago, # ^ | ← Rev. 2 →   +8 Nice observation! That way to write $min(a,b)$ is very clever.
 » 11 months 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.
 » 11 months 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)
•  » » 11 months 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).
•  » » » 11 months ago, # ^ |   0 Thanks!
 » 11 months ago, # |   +8 C:spend hour to write solution which uses up to 250 operations, delete it, write proper solution in 15 minutes
 » 11 months ago, # |   0 Can anybody explain the curve in Question B
 » 11 months ago, # | ← Rev. 2 →   -17 .