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

- Contest URL: https://atcoder.jp/contests/abc193
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20210227T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 6
- Writer: evima, kyopro_friends, QCFium, tatyam
- Rated range: ~ 1999

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

We are looking forward to your participation!

Looking forward to facing some great problems

OK, seems greedy is wrong for F....

What's the correct solution?

How to solve E ?

I didn't solve it neither...

there's little time after solve 4 problems, so I just implement F

I solved E using diophantine equations... u can see my sol here

Can you explain how to handle the segments correctly? I mean, with diophantine in first place we calculte single integer values.

just brute force Y * Q times....

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.

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.It should be "...In other words $$$X \leq t \mod (2X + 2Y) < X + Y$$$". Insn't it?

(2X + 2Y) = (2 * (X + Y)), so because of this we can take mod as (X+Y)?

I don't think that's always true, consider a simple example of $$$X = 1, Y = 1, 6 \mod (2X + 2Y) = 2, 6 \mod (X + Y) = 0.$$$

Good catch, fixed.

Nice explanation, thanks.

I think you made a small typo in the first sentence: it should be $$$k(2X+2Y)$$$ instead of $$$n(2X+2Y)$$$.

Must have not been fully awake when I wrote that :D. Thanks for catching it.

Couldn't catch the train till the end!

F, what strategy works, how to prove it?

can't we do it via dp ?

Most solutions I inspeced created a max flow graph.

can you elaborate ?

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.

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?

I probably read the editorial 3-4 times but still couldn't make sense with it, you just made it a lotttt clear.

Thanks!

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.

How to solve E and F?

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).

I don't know why using doubles is giving WA in D!

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 cards

if 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

.

I had some idea how to do F with Maximum Flow. But couldn't complete it. Anyone can share their approach?

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 (:My usual screencast with explanations, but only A-E this time, sorry guys.

How do you solve C?

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.

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 size

https://atcoder.jp/contests/abc193/submissions/20535157

works in O(log N * sqrt(N))

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 :)

Code using set

Code using map

Hope it'll help you :)

I have realized how powerful the AtCoder library it is... Look at this AC submission of E if you want to know reasons.

A more interesting ABC than usual. Kudos to problemsetters.

Yes best round in recent history. Hope this becomes a trend!

In problem F, how do I extract the chosen colors for all '?' cells (from the min-cut/max-flow)?