The editorial of C made me angry...

E: It's easy to see that there is no solution when $$$n = 3$$$ or $$$n \mod 3 = 2$$$. For $$$n \mod 3 = 2$$$ the total number of edges is not divisible by $$$3$$$. For $$$n = 3$$$ the only configuration with equal number of edges per color is a triangle, whch is prohibited.

What about the other possible values of $$$n$$$?

One of the possible solutions is to output matrix with equal numbers in each row. If we do that, condition 2 holds automatically. The reason: for each triangle there is a row having two of its edges, and we have equal numbers in each row. So the three numbers cannot be all different.

Now we need to match condition 1. The number of symbols in rows are 1, 2, 3, ..., n — 1. So the question is whether we can split these numbers into three groups with equal sums. Turns out that it's always possible for $$$n \mod 3 \ne 2, n \ge 6$$$. Examples:

6: {1 4}, {2 3}, {5} — sum of numbers in each subset is 5.

7: {1 6}, {2 5}, {3 4}

9: {1 2 3 6}, {4 8}, {5 7}

10: {1 2 3 4 5}, {6 9}, {7 8}

Any larger number $$$x$$$: partition for $$$x - 6$$$, and remaining 6 numbers are split into pairs similar to {1 6}, {2 5}, {3 4}

Problem D:

In the optimal solution, there is an arrow in $$$[0, d-1]$$$.

If you fix the position $$$p$$$ of the arrow in $$$[0, d-1]$$$, you can get the solution greedily by always choosing coordinates as close to $$$0$$$ as possible. Let $$$a_p$$$ be the optimal sum with an arrow in position $$$p$$$.

You can calculate answers, in order, from $$$a_0$$$ to $$$a_{d-1}$$$, after preprocessing: each $$$r_i$$$ contributes to at most $$$2$$$ differences $$$a_p - a_{p+1}$$$.

D: It's easy to see that the optimal configuration is a sequence of shots where the distance between consecutive shots is exactly $$$D$$$. Moreover, we can always assume that one of the shots is located at some point $$$r_i$$$ (otherwise we can always slightly move all shots and answer won' change). Therefore it is sufficient to consider only $$$D$$$ placements uniquely defined by the remainder of dividing a position of shots by $$$D$$$, and find the best placement among them.

We can evaluate one such placement naively (e.g. for remainder $$$0$$$ it is straightforward). We need to count number of shots in each of segments, and sum up the largest $$$n$$$ numbers. To process the other placements, we can use a sweepline. Each border $$$r_i$$$ corresponds to an event where you add one number and remove another number. We can sort all events and process them by a sweepine, managing a set of $$$n$$$ largest numbers.

Time complexity: $$$O(m \log m + n \log n)$$$ or $$$O(D + m + n \log n)$$$, if instead of sorting all events we just fill $$$D$$$ vectors of events.

RWB victim also... QQ

In problem A have you taken account the case where B is 1?

it is failing when b is equal to 1

try this test case: 1 1

You can read my comment or the editorial, it was published by AtCoder: https://atcoder.jp/contests/arc131/editorial/3057

My solution is based on a fact that, If N is odd, player making first move always wins. I am going to prove that later and for now let us assume that it is true.

If N is even and XOR of all elements(let, X) is exist in the array, then I can win in my first move by removing the element which is equal to X(arr[i] = X) and game ends. Otherwise, I have to remove an element but the game continues and the game becomes where (N — 1)(which is of course odd) element exists and other player makes the first moves and so the other player wins.

Ok. Now let's prove, If N is odd, player making first move always wins.[Reminder: All elements are distinct]

Let's assume I started the game by making the first move and after some moves the other player got a move where removing one element makes XOR of remaining elements 0. If that happens I always could swap my previously taken element with any one of the currently remaining element which would make remaining XOR not-zero because all elements are distinct.

This process continues until the array is empty. That makes player with first move always win.