### chokudai's blog

By chokudai, history, 23 months 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

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 » 23 months ago, # |   -21 Looking forward to facing some great problems
 » 23 months ago, # |   0 OK, seems greedy is wrong for F....What's the correct solution?
•  » » 23 months ago, # ^ |   +1 How to solve E ?
•  » » » 23 months ago, # ^ |   0 I didn't solve it neither...there's little time after solve 4 problems, so I just implement F
•  » » » 23 months ago, # ^ |   +4 I solved E using diophantine equations... u can see my sol here
•  » » » » 23 months ago, # ^ |   0 Can you explain how to handle the segments correctly? I mean, with diophantine in first place we calculte single integer values.
•  » » » » » 23 months ago, # ^ |   0 just brute force Y * Q times....
•  » » » » » 23 months 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.
•  » » » 23 months 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.
•  » » » » 23 months ago, # ^ |   +8 It should be "...In other words $X \leq t \mod (2X + 2Y) < X + Y$". Insn't it?
•  » » » » » 23 months ago, # ^ |   0 (2X + 2Y) = (2 * (X + Y)), so because of this we can take mod as (X+Y)?
•  » » » » » » 23 months ago, # ^ |   0 I don't think that's always true, consider a simple example of $X = 1, Y = 1, 6 \mod (2X + 2Y) = 2, 6 \mod (X + Y) = 0.$
•  » » » » » 23 months ago, # ^ |   0 Good catch, fixed.
•  » » » » 23 months ago, # ^ |   0 Nice explanation, thanks.I think you made a small typo in the first sentence: it should be $k(2X+2Y)$ instead of $n(2X+2Y)$.
•  » » » » » 23 months ago, # ^ |   0 Must have not been fully awake when I wrote that :D. Thanks for catching it.
 » 23 months ago, # |   +6 Couldn't catch the train till the end!
 » 23 months ago, # |   0 F, what strategy works, how to prove it?
•  » » 23 months ago, # ^ |   0 can't we do it via dp ?
•  » » » 23 months ago, # ^ |   0 Most solutions I inspeced created a max flow graph.
•  » » » » 23 months ago, # ^ |   0 can you elaborate ?
•  » » » » » 23 months 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.
•  » » » » » » 23 months 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?
•  » » » » » » 23 months 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!
•  » » » » » » 23 months 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.
 » 23 months ago, # |   +1 How to solve E and F?
•  » » 23 months 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).
 » 23 months ago, # |   0 I don't know why using doubles is giving WA in D!
•  » » 23 months 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
•  » » » 23 months ago, # ^ | ← Rev. 4 →   0 .
 » 23 months ago, # |   +8 I had some idea how to do F with Maximum Flow. But couldn't complete it. Anyone can share their approach?
 » 23 months 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 (:
 » 23 months ago, # |   +4 My usual screencast with explanations, but only A-E this time, sorry guys.
 » 23 months ago, # |   0 How do you solve C?
•  » » 23 months 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.
•  » » 23 months 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))
•  » » 23 months 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 :)
 » 23 months 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.
 » 23 months ago, # |   +25 A more interesting ABC than usual. Kudos to problemsetters.
•  » » 23 months ago, # ^ |   0 Yes best round in recent history. Hope this becomes a trend!
 » 23 months ago, # |   0 In problem F, how do I extract the chosen colors for all '?' cells (from the min-cut/max-flow)?