### chokudai's blog

By chokudai, history, 7 weeks ago,

We will hold Caddi Programming Contest 2021(AtCoder Beginner Contest 193).

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

We are looking forward to your participation!

• +42

 » 7 weeks ago, # |   -21 Looking forward to facing some great problems
•  » » 7 weeks ago, # ^ |   -7 *painful problems. Especially D and E .
 » 7 weeks ago, # |   0 OK, seems greedy is wrong for F....What's the correct solution?
•  » » 7 weeks ago, # ^ |   +1 How to solve E ?
•  » » » 7 weeks ago, # ^ |   0 I didn't solve it neither...there's little time after solve 4 problems, so I just implement F
•  » » » 7 weeks ago, # ^ |   +4 I solved E using diophantine equations... u can see my sol here
•  » » » » 7 weeks ago, # ^ |   0 Can you explain how to handle the segments correctly? I mean, with diophantine in first place we calculte single integer values.
•  » » » » » 7 weeks ago, # ^ |   0 just brute force Y * Q times....
•  » » » » » 7 weeks ago, # ^ | ← Rev. 2 →   0 yes then we can check the values of k in " equation: ax + by = c. let x0,y0 be integers which satisfy the following: a⋅x0 + b⋅y0 = c Then x = x0 + k⋅b/g, y = y0 − k⋅a/g are solutions of the given Diophantine equation. g: gcd(a,b)"which give us x >= 0 && y >= 0. and calculate the values at border values of k.
•  » » » 7 weeks ago, # ^ | ← Rev. 4 →   +35 We want to find some $t$ such that $X + n(2X + 2Y) \leq t < X + Y + n(2X + 2Y)$ and $P + m(P + Q) \leq t < P + Q + m(P + Q)$. In other words, $X \leq t \mod (2X + 2Y) < X + Y$ and $P \leq t \mod (P + Q) < P + Q$. Because $Y$ and $Q$ are sneakily only at most 500, we can naively brute force $t \mod (2X + 2Y)$ and $t \mod (P + Q)$ pairs and check the existence of a solution with Chinese remainder theorem. There's conveniently a crt implementation in Atcoder library under math header.
•  » » » » 7 weeks ago, # ^ |   +8 It should be "...In other words $X \leq t \mod (2X + 2Y) < X + Y$". Insn't it?
•  » » » » » 7 weeks ago, # ^ |   0 (2X + 2Y) = (2 * (X + Y)), so because of this we can take mod as (X+Y)?
•  » » » » » » 7 weeks ago, # ^ |   0 I don't think that's always true, consider a simple example of $X = 1, Y = 1, 6 \mod (2X + 2Y) = 2, 6 \mod (X + Y) = 0.$
•  » » » » » 7 weeks ago, # ^ |   0 Good catch, fixed.
•  » » » » 7 weeks ago, # ^ |   0 Nice explanation, thanks.I think you made a small typo in the first sentence: it should be $k(2X+2Y)$ instead of $n(2X+2Y)$.
•  » » » » » 7 weeks ago, # ^ |   0 Must have not been fully awake when I wrote that :D. Thanks for catching it.
 » 7 weeks ago, # |   +6 Couldn't catch the train till the end!
 » 7 weeks ago, # |   0 F, what strategy works, how to prove it?
•  » » 7 weeks ago, # ^ |   0 can't we do it via dp ?
•  » » » 7 weeks ago, # ^ |   0 Most solutions I inspeced created a max flow graph.
•  » » » » 7 weeks ago, # ^ |   0 can you elaborate ?
•  » » » » » 7 weeks ago, # ^ |   +40 Let's suppose you had to minimize the number of adjacent pairs of cells with different colors. For a final coloring of the grid, this number can be calculated by dividing the nodes of the grid-graph (graph formed by adding edges among all pairs of cells which share a side) into 2 parts based on their colors and then counting the edges which connect nodes in different parts. To minimize this value, you can just find the minimum-cut of the graph. But you also have to ensure that cells that were already colored and had different colors are in different parts of the cut. For this, you can add an edge with infinity capacity from the source to each cell which is already colored black and from each cell which is already colored white to the sink. These edges can never be part of a minimum cut dividing source and sink and hence they will ensure that edges with fixed but different colors are in different parts. To solve the original problem, we will minimize the number of adjacent pairs of cells with same colors in a valid final coloring and subtract this value from the total number of edges in the grid graph. Notice that the endpoints of all edges in the grid graph have different value of (row no. + col no.) mod 2. So, if we flip the color of every node with (row no. + col no.) mod 2 == 0, then edges whose endpoints had different colors will now have same colors and vice-versa. This transformation reduces our problem to the problem discussed in the previous paragraphs which can be easily solved with Dinic's algorithm.
•  » » » » » » 7 weeks ago, # ^ |   0 How can we flip the colour of vertices that have a fixed colour? Won't that violate the min-cut condition as the new value would be infinity?
•  » » » » » » 7 weeks ago, # ^ |   +16 I probably read the editorial 3-4 times but still couldn't make sense with it, you just made it a lotttt clear.Thanks!
•  » » » » » » 6 weeks ago, # ^ |   +3 Why are we flipping 'B' and 'W' in problem F only when i+j is odd in the editorial ? I am unable to understand it. If anyone else understood it, please explain in simpler terms.
 » 7 weeks ago, # |   -12 I found D very frustating because of overflow. What is the point ?
•  » » 7 weeks ago, # ^ |   0 Integers (Longs) were supposed to be used instead of Doubles. In order to attain a precision of 10^-5, I simply multiplied the number by 100000, performed integer division, and split the answer into two parts using a formatting function. 1st part -> (long)result / 100000 2nd part -> (long)result % 100000 The link to my submission: Submission
•  » » » 7 weeks ago, # ^ |   0 Yes, the output is a ratio a/b of long long. So your method is fine.But a simpler way is to output double(a)/b as done in the exercise editorial.
•  » » » » 7 weeks ago, # ^ | ← Rev. 2 →   0 Well, that was what I initially did and it didn't work (failed one of the 32 test cases). The Original Submission (Java)
 » 7 weeks ago, # |   +1 How to solve E and F?
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   0 My solution for E:Among the first 2(X+Y) seconds, the train stops at Y of those seconds. Label these seconds 0,1,2...Y-1. Also, if second i is labeled, apply the same label to second i+2(X+Y).Similarly, among the first P+Q seconds, label the seconds where Takahashi is awake, and if second i is labeled, apply the same label to second i+P+Q.The problem is to find the earliest second that was simultaneously labelled in both steps. We can brute-force this by enumerating all Y*Q pairs of labels and considering when is the earliest second labelled with those specific pairs of labels. For each pair, the answer is an intersection of APs (https://math.stackexchange.com/questions/1378976/computing-the-intersection-of-two-arithmetic-sequences-a-mathbbz-b-cap).
 » 7 weeks ago, # |   0 I don't know why using doubles is giving WA in D!
•  » » 7 weeks ago, # ^ | ← Rev. 4 →   +3 For D I simulated all possible card selections (2 for loops, skipping invalid combinations that pass the card limit) then calculating the sums after picking those cardsif its bigger add number of ways to get that card combination to num(if the cards picked by both are the same, let cnt = num of remaining cards with i written on them not picked. Its cnt * cnt-1 ways, else its num of remaining cards numbered i * number of remaining cards numbered j) Add the same to tot then output num/tot. btw i used doubles. You can check my solution here:https://atcoder.jp/contests/abc193/submissions/20554741
•  » » » 7 weeks ago, # ^ | ← Rev. 4 →   0 .
•  » » » 7 weeks ago, # ^ |   0 Interesting... When I used doubles, my solution failed 1 of the 32 test cases. Had to use longs instead... Submission (Java)
 » 7 weeks ago, # |   +8 I had some idea how to do F with Maximum Flow. But couldn't complete it. Anyone can share their approach?
 » 7 weeks ago, # | ← Rev. 2 →   +8 For D, i used ll x=n-mp1[i]+mp2[i],y=n-mp2[j]+mp1[j]; instead of ll x=n-mp1[i]-mp2[i],y=n-mp2[j]-mp1[j]; . and tried to find the bug for 1 hour+. (: fuc* my luck (:
 » 7 weeks ago, # |   +4 My usual screencast with explanations, but only A-E this time, sorry guys.
 » 7 weeks ago, # |   0 How do you solve C?
•  » » 7 weeks ago, # ^ |   0 Note that all numbers usable as $a$ are smaller or equal to sqrt(n), so max ~1e5.Then we simply find all numbers (collect them in a set) we can build by all possible $a^b$, and subtract this number from n.
•  » » » 7 weeks ago, # ^ |   0 Ah I see, thank you!
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   0 run two for loops i and j to calculate all i^j <= N, storing values in a set, the first loop runs from 2 to sqrt(N), while the second runs starting from 2 till i^j > N (which is in worst case logN)ans = N — Set sizehttps://atcoder.jp/contests/abc193/submissions/20535157works in O(log N * sqrt(N))
•  » » 7 weeks ago, # ^ |   0 As you can see clearly, the maximum base number is not more than $10^5$, and the maximum power does not exceed to $\left\lfloor\log_{2}10^{10}\right\rfloor=33$. So we just directly list all numbers within n that satisfy the b-th power form of a, and then remove the multiplicity then get the answer :)Hope it'll help you :)
•  » » » 7 weeks ago, # ^ |   0 Thank you! That makes a lot of sense
 » 7 weeks ago, # |   +10 I have realized how powerful the AtCoder library it is... Look at this AC submission of E if you want to know reasons.
 » 7 weeks ago, # |   +25 A more interesting ABC than usual. Kudos to problemsetters.
•  » » 7 weeks ago, # ^ |   0 Yes best round in recent history. Hope this becomes a trend!
 » 7 weeks ago, # |   0 In problem F, how do I extract the chosen colors for all '?' cells (from the min-cut/max-flow)?