The problem setting team (well codicon and 3366 exclusively) put their heads together to present to you the ultimate anime-themed problem! You will have to output the name of the best waifu.

Bossologist is slowly losing his sanity! Bossologist starts off with $$$S$$$ sanity points, for every email Chessbot sends from polygon, he loses $$$1$$$ sanity point. The problem writing process lasts $$$N$$$ days, where each day Chessbot sends between $$$0$$$ and $$$1000$$$ emails (inclusive). Given that Bossologist will gain $$$1$$$ sanity point for every $$$K$$$ days in a row that he doesn't get an email, calculate the number of sanity points he has at the end of $$$N$$$ days.

Note: It is guaranteed that Bossolgist's sanity will always be non-negative for the entire problem writing process.

Davin is a very busy man, and he likes to get his schedule set on the day before a new year begins. The new year will be $$$2023$$$, and Davin has $$$N$$$ different events he needs to attend. He has those $$$N$$$ events represented by $$$A, B, C$$$, where $$$A$$$ and $$$B$$$ represent the month and day of the start date of the event, and $$$C$$$ represents how many days it takes to repeat that event. For example, if an event starts on August $$$6$$$th and repeats every $$$11$$$ days (August $$$6$$$, August $$$17$$$, August $$$28$$$, September $$$8$$$...), it would be represented by "$$$8$$$ $$$6$$$ $$$11$$$".

Davin would like to go on a vacation sometime during $$$2023$$$; thus, he wants to find the largest pocket of time where there are no events. Given the events Davin has during $$$2023$$$, output the largest pocket of time, in days, that he could go on a vacation.

Thomas is doing a contest! He is facing many different problems he wants to solve, for a chance to be able to game again. The contest itself is structured like a $$$n \times m$$$ grid, where each cell entry represents the points Thomas gets if he solves that particular problem. During the contest, Thomas will choose exactly one sub-rectangle of problems, and solve all of the problems inside it to earn those points. Thomas wants to make sure his points earned is as maximal as possible. However, there is one problem in this contest (let's call it problem E) that gives a negative amount of points. Help Thomas find the maximum number of points he can earn by solving one sub-rectangle of the problems! (Please note that Thomas cannot skip problems in the rectangle that he picked).

Bossologist is in a group with $$$N$$$ people (including himself) that has recently won the lottery! Because the group is very fair, they deposit $$$100$$$ dollars into each person's bank account. Everyone is excited to spend their money, but they make sure first to keep track of their purchases in a spending list. The list is of length $$$Q$$$ and contains the purchases of each person and they amount they spent, in chronological order.

However, the person who has the longest consecutive streak of spending must pay a tax of $$$A$$$ dollars to each person in the group, where $$$A$$$ represents the length of the longest consecutive streak. If multiple people have the same longest consecutive streak, the person who got the longest consecutive streak first must pay the tax.

The group wants to know the difference in dollars between the person with most and least amount of money in their bank account after all the purchases are made and taxes are paid. Note that the amount of money in a bank account can be negative.

Misaka is confused. She is staring at her front yard, a garden of length $$$N$$$, and it seems to be lined with trees... and her clones? As Misaka looks at her front yard, she notices that for certain segments of her front yard there are an equal amount of clones and trees. Her memory has gotten worse so she has forgotten what her front yard looked like on that day. However, she still remembers $$$M$$$ intervals where the number of clones and trees were the same. Misaka wants you to find any possible arrangement of clones and trees such that it matches with the intervals she remembers.

Peach and Goma have taken up a new hobby recently. Sometimes they decide to get together to look at a number $$$x$$$ and compute $$$x \oplus (x + 1)$$$, where $$$\oplus$$$ is the bitwise XOR operator. Today they computed this value for all integers from $$$A$$$ to $$$B$$$ inclusive, and Peach and Goma are curious about their results. For a given number $$$M$$$ they would like you to find the number of integers $$$x$$$ ($$$A \leq x \leq B$$$) such that $$$x \oplus (x + 1) \geq M$$$.

This statement is entirely fictional and no such events will ever come to happen in real life. Since in accordance with popular belief, Waymo is in fact broke.

Waymo has decided to make the class he is TA'ing for solve Lattice MST (from the Teamscode Summer 2021 contest). To help keep the students motivated to solve the problem, he will buy each student a cup of milk tea (with aloe vera instead of boba since aloe vera $$$ \gt $$$ boba). However, Waymo does not feel the obligation to buy milk tea for someone with zero trust. Trust is calculated as follows:

Every student $$$i$$$ starts of with $$$t_i$$$ trust. Then over the course of $$$Q$$$ days, if by day $$$d$$$, a student has not solved Lattice MST, their trust will decrease by $$$x_d$$$. Note that similar to real life, milk tea is bought before trust decreases. Also, note that trust is never negative. If a person ever has negative trust, their trust is set back to $$$0$$$.

So over the course of the $$$Q$$$ days, you have to output how many cups of milk tea Waymo has to buy on each day as well as the sum of trusts over all students. Assume that none of the students can solve Lattice MST.

Timothy is the president of the triangle fan club, and they have decided to create a flag! The design they settled on was an isosceles right triangle. They want to cut it out from a $$$N \times M$$$ rectangular piece of cloth with the two legs of the triangle being parallel to the sides of the cloth. The orientation of the flag matters, one leg of the triangle must be on the bottom and the other must be on the right (more details are given in the notes).

Unfortunately, the cloth had been stored in the attic with moths, so there were holes all over it. Timothy wants to know how many places still exists on the cloth that he can a flag out of.

Timothy hasn't decided on the size of the flag yet, and he wants to explore all the possibilities. He asks you to find the number of different places he can cut a flag out of for each flag size from $$$2$$$ to $$$\min{(N, M)}$$$. A flag of size $$$X$$$ has legs of length $$$X$$$ each.

Thomas wants to game! But teacher is making him do some hard problems to stop him from gaming. In particular, teacher has forced Thomas to work on an $$$O(n)$$$ matrix mutliplication problem. Thomas currently has a matrix multiplication for two $$$n$$$ by $$$n$$$ matrices written as follows:

for(int i = 0; i<n; i++){
  for(int j = 0; j<n; j++){
    for(int k = 0; k<n; k++){
      c[i][j] = c[i][j] + a[i][k]*b[k][j];
    }
  }
}

Thomas's code is not $$$O(n)$$$ and TLE's! However, Thomas has thought of a way to make his matrix multiplation $$$O(1)$$$. He decided to handwrite all the transitions of his $$$n^3$$$ matrix multiplication! In particular, for all pairs $$$i,j$$$ such that $$$0 \leq i, j \lt n$$$, Thomas writes:

c[i][j] = a[i][0]*b[0][j] + a[i][1]*b[1][j] + ... + a[i][n-1]*b[n-1][j];

Please note that in all these examples so far, $$$a$$$, $$$b$$$, and $$$c$$$ have all been placeholders for the actual variable names Thomas uses. In practice, Thomas actually uses much more productive variable names (probably...)!

So below is an "expanded" matrix multiplication of $$$2$$$x$$$2$$$ matrices m1 and m2 into ret.

ret[0][0] = m1[0][0]*m2[0][0] + m1[0][1]*m2[1][0];
ret[1][0] = m1[1][0]*m2[0][0] + m1[1][1]*m2[1][0];
ret[0][1] = m1[0][0]*m2[0][1] + m1[0][1]*m2[1][1];
ret[1][1] = m1[1][0]*m2[0][1] + m1[1][1]*m2[1][1];

So now Thomas asks you, if you are given the length of the names of the result matrix, left matrix, and right matrix, and their dimension $$$n$$$, what is the number of non-whitespace characters in the expanded form of that matrix multiplication. Note that new lines and spaces are ignored when Thomas says "no white spaces"; all other characters contribute to the answer.

This is the easy version. The difference between the easy version and the hard version is the constraints.

Misaka Imouto $$$1$$$, Misaka Imouto $$$2$$$, ..., and Misaka Imouto $$$n$$$ are all playing rock paper scissors to determine who should go buy the cake. Since they are all connected to the same network, let's assume that all Misaka Imouto's act the same.

A game of rock paper scissors is dictated as follows:

1. If there is $$$1$$$ person, the game ends immediately (taking $$$0$$$ rounds).
2. Otherwise, each Misaka Imouto throws out a choice between rock, paper, and scissors with exactly $$$\frac{1}{3}$$$ probability. All choices are independent of each other.
3. If all options of rock, paper, and scissors are present, then the game is a tie and no one steps out of the game. If only one option is present, the game is a tie and no one steps out of the game. However, if only two options are present, then the winners and the losers are well defined, and the losers step out of the game. All people who are still in the game repeat step 1 for a new round.

One possible game between $$$3$$$ Misaka Imouto's could be as follows: RPS, RRS, SSX, SRX, and ending with the second Misaka winning after $$$4$$$ rounds. In this game, R means rock, S means scissors, P means paper, and X means that this Misaka Imouto has lost already and was not allowed to play.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, find the expected number of rounds a game with $$$i$$$ Misaka Imoutos initially lasts.

After seeing their rivalry, teacher has challenged Thomas and Waymo into the hardest team bonding activity eve. Rules are as follows:

1. The game is to be played on a graph with $$$N$$$ nodes and $$$M$$$ edges.
2. On each node there exists exactly one coin which can be picked up by reaching that node.
3. Each edge connects two nodes $$$u_i$$$ and $$$v_i$$$ and has a label $$$l_i$$$.
4. The $$$i$$$'th edge can only be crossed (in either direction) by a person if another person is standing on node $$$l_i$$$. So imagine a pressure plate on node $$$l_i$$$ that must be stood on for the entire time a person is trying to cross.

If Thomas starts on node $$$t$$$ and Waymo starts on node $$$w$$$, what is the maximum number of coins they can obtain after any finite amount of moves?

In his quest to become the better TA, Thomas suddenly received inspiration: "What if he tried to learn from someone who was already a teacher?". Thomas immediately thought of teacher and realized if he wanted to become a better TA, he had to be driving at all times. (Note that Thomas has tried biking as a greener method of transport, but that failed).

Thomas currently lives in a country with $$$N$$$ cities and $$$M$$$ roads between cities. Every road also has a parameter $$$t$$$, denoting the number of minutes to pass through a road.

Consider a road trip from some city $$$u$$$ to city $$$v$$$. Let us assume that Thomas has unlimited mileage from infinite money and the only thing limiting his drives is his patience. If Thomas's patience is $$$p$$$, he can only drive on roads that require $$$p$$$ minutes to drive through. For instance having patience $$$3$$$ can enable Thomas to travel through roads that take $$$1$$$, $$$2$$$, or $$$3$$$ minutes to cross.

Now Thomas has a list of cool cities that is initially empty but is always growing. He will make $$$Q$$$ additions to that list. After each addition, he asks what the minimum patience he needs to make a road trip between any two cool cities is. Note that Thomas is picking two cities and then traveling between them, not taking the answer over all pairs of cities.

Codicon is traversing through an $$$N$$$x$$$N$$$ grid ($$$2 \leq N \leq 1000$$$). He starts at the top left cell and goes to the bottom right, where in each step he either moves one block down or one block right. Chessbot however, has a scheme and would like Codicon to traverse a specific path (for no nefarious reason of course). He will do this by strategically placing impassable rocks in certain spots (not at the start or end cell). But rocks cost a lot of money, so being the thrifty person he is Chessbot would like to use as little rocks as possible to make his path the only valid path Codicon can take. Find the minimum amount of rocks needed to do so.

A valid path is one where Codicon starts in the top left, ends in the bottom right, only moves right or down, and does not pass any rocks.

Timothy travels between lattice points on a coordinate plane, with $$$Y \in$$$ {$$$1,2...N$$$} and $$$X \in$$$ {$$$1,2...M$$$}. He can freely move up and down across Y coordinates, but cannot move freely left and right across X coordinates. However, there are $$$K$$$ roads available to him, each of which spans from $$$(x_a, y_r)$$$ to $$$(x_b, y_r)$$$ where $$$x_a \leq x_b$$$, and allows him to move freely between any points $$$(x_i, y_r)$$$ and $$$(x_j, y_r)$$$ where $$$x_a \leq x_i$$$, $$$x_j \leq x_b$$$. We have to answer $$$Q$$$ queries, each of which asks if it's possible to travel from $$$(A, B)$$$ to $$$(C, D)$$$, traveling only across $$$y$$$ coordinates between $$$min(B, D)$$$ and $$$max(B, D)$$$ inclusive (so he may use roads with y-coordinate $$$B$$$ or $$$D$$$).

1900 years ago, a colossal pit was discovered on an island. The abyss is 1000 meters in diameter and at least 20,000 meters deep — at least because none have reached the bottom. The allure of the Abyss is inescapable, so Riko has decided to embark on the adventure of a lifetime.

Riko's journey passes through $$$N$$$ locations numbered from $$$1$$$ to $$$N$$$. The distance between the $$$(i-1)$$$th and $$$i$$$th location is $$$A_i$$$.

Additionally, there are $$$M$$$ types of roads. The roads of the $$$k$$$th type are described with $$$3$$$ integers, $$$S_k$$$, $$$E_k$$$, $$$L_k$$$. For any $$$(i,j)$$$ such that $$$S_k \leq i \lt j \leq E_k$$$ and $$$abs(j-i) \leq L_k$$$, there is a road of the $$$k$$$th type from city $$$i$$$ to city $$$j$$$. Notice that $$$i \lt j$$$ because ascending causes horrific symptoms from the curse of the Abyss.

Furthermore, the cost of traversing the road from city $$$i$$$ to city $$$j$$$ is $$$(A_{i+1} \oplus A_{i+2} \oplus \dots \oplus A_j) - 16$$$ where $$$\oplus$$$ is the bitwise XOR operator.

A valid journey from city $$$1$$$ to $$$N$$$ satisfies that the types of the roads used are in $$$\textbf{strictly}$$$ increasing order. It is guaranteed that a valid journey exists.

Riko is an explorer, but she is not a programmer. So, Riko asks you to find the minimum cost for her to travel from city $$$1$$$ to city $$$N$$$.

There used to exist a good story about someone getting food poisoning from eating at restaurants. It is now replaced by a cleaner statement.

You have an integer $$$N$$$ and a set of primes that initially contains the first $$$M$$$ primes. You have to handle $$$Q$$$ queries, where for each query, you make a modification to the set of primes. You either insert a prime that is not present in the set already or delete a prime that is already present in the set. After each query, you have to output the number of numbers $$$x$$$ $$$(1 \leq x \leq N)$$$ such that $$$x$$$ is divisible by at least one prime in the set. All operations will only consist of the first $$$M$$$ primes.

This is the hard version. The difference between the easy and the hard versions is that the hard version has higher constraints.

Misaka Imouto $$$1$$$, Misaka Imouto $$$2$$$, ..., and Misaka Imouto $$$n$$$ are all playing rock paper scissors to determine who should go buy the cake. Since they are all connected to the same network, let's assume that all Misaka Imouto's act the same.

A game of rock paper scissors is dictated as follows:

1. If there is $$$1$$$ person, the game ends immediately (taking $$$0$$$ rounds).
2. Otherwise, each Misaka Imouto throws out a choice between rock, paper, and scissors with exactly $$$\frac{1}{3}$$$ probability. All choices are independent of each other.
3. If all options of rock, paper, and scissors are present, then the game is a tie and no one steps out of the game. If only one option is present, the game is a tie and no one steps out of the game. However, if only two options are present, then the winners and the losers are well defined, and the losers step out of the game. All people who are still in the game repeat step 1 for a new round.

One possible game between $$$3$$$ Misaka Imouto's could be as follows: RPS, RRS, SSX, SRX, and ending with the second Misaka winning after $$$4$$$ rounds. In this game, R means rock, S means scissors, P means paper, and X means that this Misaka Imouto has lost already and was not allowed to play.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, find the expected number of rounds a game with $$$i$$$ Misaka Imoutos initially lasts.