Thank you for participating in this round! Problems I are authored by [user:orzdevinwang,2025-01-20] and the rest by me.↵
↵
Sorry for the weak pretest in problem B.↵
↵
hints and code are WIP.↵
↵
[problem:2061A]↵
↵
<spoiler summary="Hint 1">↵
What is the parity of $s$ after an operation?↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
After the first operation, $s$ is always odd. You can only earn points when $a_i$ is odd after this operation. However, during the first operation, you can only earn a point if $a_i$ is even.↵
↵
Let $c_0$ and $c_1$ be the number of even and odd numbers, respectively.↵
↵
If there is at least one even number, you can place it first, making the answer $c_1 + 1$. Otherwise, since no points can be earned during the first operation, the answer is $c_1 - 1$.↵
↵
Time complexity: $O(n)$.↵
</spoiler>↵
↵
[problem:2061B]↵
↵
<spoiler summary="Solution">↵
Let $a$ and $b$ be the lengths of the two bases, and let $c$ be the length of the two legs. A necessary and sufficient condition for forming an isosceles trapezoid with a positive area is $|a - b| < 2c$, as the longest edge must be shorter than the sum of the other three edges.↵
↵
To determine whether an isosceles trapezoid can be formed, consider the following cases:↵
↵
- If there are two distinct pairs of identical numbers that do not overlap, these pairs can always form an isosceles trapezoid.↵
- If there are no pairs of identical numbers, it is impossible to form an isosceles trapezoid.↵
- If there is exactly one pair of identical numbers, denoted by $c$, we will use them as the legs. Remove this pair and check whether there exist two other numbers whose difference is less than $2c$. This can be efficiently done by sorting the remaining numbers and checking adjacent pairs.↵
\end{itemize}↵
↵
The time complexity of this approach is $O(n \log n)$.↵
</spoiler>↵
↵
[problem:2061C]↵
↵
<spoiler summary="Solution">↵
For the $i$-th person, there are at most two possible cases:↵
↵
- If they are honest, there are exactly $a_i$ liars to their left.↵
- If they are a liar, since two liars cannot stand next to each other, the $(i-1)$-th person must be honest. In this case, there are $a_{i-1} + 1$ liars to their left.↵
\end{itemize}↵
↵
Let $dp_i$ represent the number of possible game configurations for the first $i$ people, where the $i$-th person is honest. The transitions are as follows:↵
↵
- If the $(i-1)$-th person is honest, check if $a_i = a_{i-1}$. If true, add $dp_{i-1}$ to $dp_i$.↵
- If the $(i-1)$-th person is a liar, check if $a_i = a_{i-2} + 1$. If true, add $dp_{i-2}$ to $dp_i$.↵
\end{itemize}↵
↵
The final answer is given by $dp_n + dp_{n-1}$.↵
↵
Time complexity: $O(n)$.↵
</spoiler>↵
↵
[problem:2061D]↵
↵
<spoiler summary="Solution">↵
It can be hard to determine how to merge small numbers into a larger number directly. Therefore, we approach the problem in reverse.↵
↵
Instead of merging numbers, we consider transforming the number $b$ back into $a$. For each number $x$, it can only be formed by merging $\lceil \frac{x}{2} \rceil$ and $\lfloor \frac{x}{2} \rfloor$. Thus, the reversed operation is as follows:↵
↵
\begin{itemize} \item Select an integer $x$ from $b$ and split it into $\lceil \frac{x}{2} \rceil$ and $\lfloor \frac{x}{2} \rfloor$.↵
\end{itemize}↵
↵
We can perform this splitting operation on the numbers in $b$ exactly $n - m$ times. If the largest number in $b$ also appears in $a$, we can remove it from both $a$ and $b$ simultaneously. Otherwise, the largest number in $b$ must be split into two smaller numbers.↵
↵
To efficiently manage the numbers in $a$ and $b$, we can use a priority queue or a multiset.↵
↵
The time complexity of this approach is $O((n + m) \log (n + m))$.↵
</spoiler>↵
↵
[problem: 2061E]↵
↵
<spoiler summary="Solution">↵
For a fixed $p$, let $f(S)$ represent the number obtained after applying the operations in set $S$ to $p$. Define $g(i)$ as the minimum value of $f(S)$ for all subsets $S$ with $|S| = i$, i.e., $g(i) = \min_{|S| = i} f(S)$.↵
↵
---↵
↵
**Lemma**: The function $g$ is convex, that is for all $i$ ($1 \leq i < m$), the inequality $2g(i) \leq g(i-1) + g(i+1)$ holds.↵
↵
**Proof**: Let $f(S)$ be the value corresponding to the minimum of $g(i-1)$, and let $f(T)$ correspond to the minimum of $g(i+1)$. Suppose the $w$-th bit is the highest bit where $f(S)$ and $f(T)$ differ.↵
↵
We can always find an operation $y \in T \setminus S$ that turns the $w$-th bit in $g(i-1)$ into $0$. In this case, we have: $$g(i - 1) - f(S \cup \{y\}) \geq 2^w.$$↵
↵
Moreover, since the highest bit where $f(S \cup {y})$ and $g(i+1)$ differ is no greater than $w-1$, we have: $$f(S \cup \{y\}) - g(i + 1) \leq 2^w.$$↵
↵
Combining these inequalities gives: $$2 g(i) \leq 2 f(S \cup \{y\}) \leq g(i - 1) + g(i + 1),$$ proving the convexity of $g$.↵
↵
---↵
↵
For each number $a_i$, we can use brute force to compute the minimum value obtained after applying $j$ operations, denoted as $num(i, j)$. From the lemma, we know that the difference $num(i, k) - num(i, k - 1)$ is non-increasing.↵
↵
Thus, when performing operations on this number, the reductions in value are:↵
↵
$$num(i, 0) - num(i, 1), num(i, 1) - num(i, 2), \dots, num(i, m - 1) - num(i, m).$$↵
↵
You need to perform $k$ operations in total. To minimize the result, you can choose the $k$ largest reductions from all possible operations. This can be done by sorting all operations by their cost or by using std::nth_element.↵
↵
Time complexity: $O(n2^m + nm)$。↵
</spoiler>↵
↵
[problem: 2061F1]↵
[problem: 2061F2]↵
↵
<spoiler summary="Solution">↵
↵
First, we divide the string $s$ into blocks.↵
↵
Let's analyze the properties of the blocks in $s$ that do not move (to simplify corner cases, we can add a block with a different number at both ends of $s$):↵
↵
- These blocks must alternate between $0$ and $1$.↵
- Between two immovable blocks, the $0$s will shift toward the same side as the adjacent block of $0$s, and the $1$s will shift toward the same side as the adjacent block of $1$s.↵
For example, $\mathtt{0\ 10010110\ 1}$ will become $\mathtt{0\ 00001111\ 1}$ after the shifting process.↵
↵
This property can be proven by induction.↵
↵
--- ↵
↵
**Easy Version**↵
↵
For the easy version, since the target string $t$ is known, we can greedily determine whether each block in $s$ can remain immovable.↵
↵
Specifically:↵
↵
For each block and the previous immovable block, check if the corresponding digits are different. Also, ensure that the numbers in the interval between the two blocks meet the conditions (i.e., the $0$s and $1$s in this interval must shift to their respective sides).↵
This can be solved efficiently using prefix sums.↵
↵
If there are $x$ blocks between two immovable blocks, then $\frac{x}{2}$ moves are required.↵
↵
Time complexity: $O(n)$.↵
↵
--- ↵
↵
**Hard Version**↵
↵
For the hard version, we use dynamic programming to determine which blocks can remain immovable.↵
↵
Let $dp(i)$ represent the minimum cost for the $i$-th block to remain immovable. We have:↵
↵
$$dp(i) = \min_{0\leq j <i, (j, i) \text{ is valid}} (dp(j) + (i - j - 1) / 2).$$↵
↵
Without loss of generality, assume that the $i$-th block is composed of $0$s. ↵
↵
Let the distance between the $i$-th block and the nearest preceding $1$-block be $p$. The number of $0$s between blocks $j$ and $i$ cannot exceed $p$. There is a restriction: $j \geq l_i$.↵
↵
Similarly, for $j$, we can derive a symmetric restriction: $i \leq r_j$.↵
↵
We can use a segment tree to maintain the values of $2dp(j) - j$ for all valid $j$. Specifically:↵
↵
- For a position $j$, if the current $i > r_j$, update the corresponding value to $+\infty$.↵
- For each $i$, query the segment tree over the valid interval to compute the minimum efficiently.↵
\end{itemize}↵
↵
Time complexity: $O(n \log n)$.↵
</spoiler>↵
↵
[problem: 2061G]↵
↵
<spoiler summary="Solution">↵
↵
Using graph theory terminology, let edges labeled as $0$ be represented in red and edges labeled as $1$ be represented in blue.↵
↵
---↵
↵
We can prove that if a graph has $3k + 1$ vertices, the maximum matching will be at most $k$. To see this, you can construct a blue clique of size $2k+1$, with all other edges colored red. It is not hard to verify that the maximum matching in both colors is $k$.↵
↵
So the maximum matching you can find is $\lfloor \frac{n+1}{3} \rfloor$.↵
↵
---↵
↵
To construct a matching, we can maintain a chain of edges of the same color. Suppose we currently have a red chain $p_1, p_2, \dots, p_k$. Now, we want to add a new vertex $q$. We check the color of the edge between $p_k$ and $q$:↵
↵
- If the edge is red, we add $q$ to the chain.↵
- If the edge is blue, $p_{k-1}, p_k, q$ will form a "mixed-color triplet." In this case, we remove $p_{k-1}$ and $p_k$ from the chain.↵
\end{itemize}↵
↵
After at most $n-1$ queries, the graph will be divided into several "mixed-color triplets" and one chain of edges of the same color.↵
↵
For the chain of length $k$, we can find $\lfloor \frac{k}{2} \rfloor$ matchings. For each "mixed-color triplet", we can always find one matching corresponding to the color.↵
↵
We can construct a matching of size $\lfloor \frac{n+1}{3} \rfloor$.↵
</spoiler>↵
↵
[problem: 2061I]↵
↵
<spoiler summary="Solution">↵
↵
First, there is a straightforward dynamic programming solution with a time complexity of $O(n^2)$.↵
↵
Let $dp(i, j)$ represent the minimum cost for Kevin to achieve $j$ victories in the first $i$ games. However, optimizing this dynamic programming solution directly is difficult because it lacks convexity.↵
↵
Therefore, we use a divide-and-conquer approach to solve it (which is more commonly seen in certain counting problems).↵
↵
--- ↵
↵
Key Idea: To compute $dp(r, *)$ from $dp(l, *)$ efficiently, the following observations are used:↵
↵
At time $l$, if the difference between Kevin's and Nivek's number of victories exceeds $r-l$, then:↵
↵
- For Type 2 matches, Kevin will either always win or always lose.↵
- For Type 1 matches, Kevin should prioritize matches with the smallest cost $a_i$ in ascending order.↵
\end{itemize}↵
↵
Let $g(i)$ denote the minimum cost of selecting $i$ matches in this range, and let $t$ represent the number of Type 2 matches in the interval.↵
↵
When $j - (l - j) > r-l$, the following transition applies: $$dp(l,j)+g(i) \to dp(r,j+t+i).$$↵
↵
Similarly, when $j - (l - j) < -(r-l)$, a similar transition is used.↵
↵
Since $g(i)$ is convex, the best transition point is monotonic. This property allows us to optimize this part using divide-and-conquer, achieving a time complexity of $O(n \log n)$.↵
↵
For the case where $|j - (l - j)| \leq r-l$, brute force is used to compute the transitions explicitly, with a time complexity of $O((r-l)^2)$.↵
↵
By dividing the sequence into blocks of size $O(\sqrt{n \log n})$, the overall time complexity can be reduced to $O(n \sqrt{n \log n})$.↵
↵
--- ↵
↵
To further optimize the problem, a more refined strategy is applied. Specifically, for the case where $|j - (l - j)| \leq r-l$, instead of computing transitions explicitly with brute force, we recursively divide the problem into smaller subproblems. The goal in each subproblem is to compute the transitions from a segment of $dp(l)$ to $dp(r)$, ensuring that the segment length in $dp(l)$ does not exceed $O(r-l)$.↵
↵
Let $m = \lfloor\frac{l + r}{2}\rfloor$. First, compute the transitions from $dp(l)$ to $dp(m)$. Then, use the results from $dp(m)$ to compute the transitions from $dp(m)$ to $dp(r)$. For each subinterval, the strategy remains the same:↵
↵
- For differences in the number of victories that exceed the interval length, apply monotonicity to handle the transitions efficiently.↵
- For the remaining part, recursively compute transitions using the same divide-and-conquer approach.↵
\end{itemize}↵
↵
The recurrence relation for this approach is: $T(n)=2T(n/2) + O(n \log n)$ So the time complexity is $T(n)=O(n \log^2 n)$.↵
↵
</spoiler>↵
↵
↵
Sorry for the weak pretest in problem B.↵
↵
hints and code are WIP.↵
↵
[problem:2061A]↵
↵
<spoiler summary="Hint 1">↵
What is the parity of $s$ after an operation?↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
After the first operation, $s$ is always odd. You can only earn points when $a_i$ is odd after this operation. However, during the first operation, you can only earn a point if $a_i$ is even.↵
↵
Let $c_0$ and $c_1$ be the number of even and odd numbers, respectively.↵
↵
If there is at least one even number, you can place it first, making the answer $c_1 + 1$. Otherwise, since no points can be earned during the first operation, the answer is $c_1 - 1$.↵
↵
Time complexity: $O(n)$.↵
</spoiler>↵
↵
[problem:2061B]↵
↵
<spoiler summary="Solution">↵
Let $a$ and $b$ be the lengths of the two bases, and let $c$ be the length of the two legs. A necessary and sufficient condition for forming an isosceles trapezoid with a positive area is $|a - b| < 2c$, as the longest edge must be shorter than the sum of the other three edges.↵
↵
To determine whether an isosceles trapezoid can be formed, consider the following cases:↵
↵
- If there are two distinct pairs of identical numbers that do not overlap, these pairs can always form an isosceles trapezoid.↵
- If there are no pairs of identical numbers, it is impossible to form an isosceles trapezoid.↵
- If there is exactly one pair of identical numbers, denoted by $c$, we will use them as the legs. Remove this pair and check whether there exist two other numbers whose difference is less than $2c$. This can be efficiently done by sorting the remaining numbers and checking adjacent pairs.↵
\end{itemize}↵
↵
The time complexity of this approach is $O(n \log n)$.↵
</spoiler>↵
↵
[problem:2061C]↵
↵
<spoiler summary="Solution">↵
For the $i$-th person, there are at most two possible cases:↵
↵
- If they are honest, there are exactly $a_i$ liars to their left.↵
- If they are a liar, since two liars cannot stand next to each other, the $(i-1)$-th person must be honest. In this case, there are $a_{i-1} + 1$ liars to their left.↵
\end{itemize}↵
↵
Let $dp_i$ represent the number of possible game configurations for the first $i$ people, where the $i$-th person is honest. The transitions are as follows:↵
↵
- If the $(i-1)$-th person is honest, check if $a_i = a_{i-1}$. If true, add $dp_{i-1}$ to $dp_i$.↵
- If the $(i-1)$-th person is a liar, check if $a_i = a_{i-2} + 1$. If true, add $dp_{i-2}$ to $dp_i$.↵
\end{itemize}↵
↵
The final answer is given by $dp_n + dp_{n-1}$.↵
↵
Time complexity: $O(n)$.↵
</spoiler>↵
↵
[problem:2061D]↵
↵
<spoiler summary="Solution">↵
It can be hard to determine how to merge small numbers into a larger number directly. Therefore, we approach the problem in reverse.↵
↵
Instead of merging numbers, we consider transforming the number $b$ back into $a$. For each number $x$, it can only be formed by merging $\lceil \frac{x}{2} \rceil$ and $\lfloor \frac{x}{2} \rfloor$. Thus, the reversed operation is as follows:↵
↵
\begin{itemize} \item Select an integer $x$ from $b$ and split it into $\lceil \frac{x}{2} \rceil$ and $\lfloor \frac{x}{2} \rfloor$.↵
\end{itemize}↵
↵
We can perform this splitting operation on the numbers in $b$ exactly $n - m$ times. If the largest number in $b$ also appears in $a$, we can remove it from both $a$ and $b$ simultaneously. Otherwise, the largest number in $b$ must be split into two smaller numbers.↵
↵
To efficiently manage the numbers in $a$ and $b$, we can use a priority queue or a multiset.↵
↵
The time complexity of this approach is $O((n + m) \log (n + m))$.↵
</spoiler>↵
↵
[problem: 2061E]↵
↵
<spoiler summary="Solution">↵
For a fixed $p$, let $f(S)$ represent the number obtained after applying the operations in set $S$ to $p$. Define $g(i)$ as the minimum value of $f(S)$ for all subsets $S$ with $|S| = i$, i.e., $g(i) = \min_{|S| = i} f(S)$.↵
↵
---↵
↵
**Lemma**: The function $g$ is convex, that is for all $i$ ($1 \leq i < m$), the inequality $2g(i) \leq g(i-1) + g(i+1)$ holds.↵
↵
**Proof**: Let $f(S)$ be the value corresponding to the minimum of $g(i-1)$, and let $f(T)$ correspond to the minimum of $g(i+1)$. Suppose the $w$-th bit is the highest bit where $f(S)$ and $f(T)$ differ.↵
↵
We can always find an operation $y \in T \setminus S$ that turns the $w$-th bit in $g(i-1)$ into $0$. In this case, we have: $$g(i - 1) - f(S \cup \{y\}) \geq 2^w.$$↵
↵
Moreover, since the highest bit where $f(S \cup {y})$ and $g(i+1)$ differ is no greater than $w-1$, we have: $$f(S \cup \{y\}) - g(i + 1) \leq 2^w.$$↵
↵
Combining these inequalities gives: $$2 g(i) \leq 2 f(S \cup \{y\}) \leq g(i - 1) + g(i + 1),$$ proving the convexity of $g$.↵
↵
---↵
↵
For each number $a_i$, we can use brute force to compute the minimum value obtained after applying $j$ operations, denoted as $num(i, j)$. From the lemma, we know that the difference $num(i, k) - num(i, k - 1)$ is non-increasing.↵
↵
Thus, when performing operations on this number, the reductions in value are:↵
↵
$$num(i, 0) - num(i, 1), num(i, 1) - num(i, 2), \dots, num(i, m - 1) - num(i, m).$$↵
↵
You need to perform $k$ operations in total. To minimize the result, you can choose the $k$ largest reductions from all possible operations. This can be done by sorting all operations by their cost or by using std::nth_element.↵
↵
Time complexity: $O(n2^m + nm)$。↵
</spoiler>↵
↵
[problem: 2061F1]↵
[problem: 2061F2]↵
↵
<spoiler summary="Solution">↵
↵
First, we divide the string $s$ into blocks.↵
↵
Let's analyze the properties of the blocks in $s$ that do not move (to simplify corner cases, we can add a block with a different number at both ends of $s$):↵
↵
- These blocks must alternate between $0$ and $1$.↵
- Between two immovable blocks, the $0$s will shift toward the same side as the adjacent block of $0$s, and the $1$s will shift toward the same side as the adjacent block of $1$s.↵
For example, $\mathtt{0\ 10010110\ 1}$ will become $\mathtt{0\ 00001111\ 1}$ after the shifting process.↵
↵
This property can be proven by induction.↵
↵
--- ↵
↵
**Easy Version**↵
↵
For the easy version, since the target string $t$ is known, we can greedily determine whether each block in $s$ can remain immovable.↵
↵
Specifically:↵
↵
For each block and the previous immovable block, check if the corresponding digits are different. Also, ensure that the numbers in the interval between the two blocks meet the conditions (i.e., the $0$s and $1$s in this interval must shift to their respective sides).↵
This can be solved efficiently using prefix sums.↵
↵
If there are $x$ blocks between two immovable blocks, then $\frac{x}{2}$ moves are required.↵
↵
Time complexity: $O(n)$.↵
↵
--- ↵
↵
**Hard Version**↵
↵
For the hard version, we use dynamic programming to determine which blocks can remain immovable.↵
↵
Let $dp(i)$ represent the minimum cost for the $i$-th block to remain immovable. We have:↵
↵
$$dp(i) = \min_{0\leq j <i, (j, i) \text{ is valid}} (dp(j) + (i - j - 1) / 2).$$↵
↵
Without loss of generality, assume that the $i$-th block is composed of $0$s. ↵
↵
Let the distance between the $i$-th block and the nearest preceding $1$-block be $p$. The number of $0$s between blocks $j$ and $i$ cannot exceed $p$. There is a restriction: $j \geq l_i$.↵
↵
Similarly, for $j$, we can derive a symmetric restriction: $i \leq r_j$.↵
↵
We can use a segment tree to maintain the values of $2dp(j) - j$ for all valid $j$. Specifically:↵
↵
- For a position $j$, if the current $i > r_j$, update the corresponding value to $+\infty$.↵
- For each $i$, query the segment tree over the valid interval to compute the minimum efficiently.↵
\end{itemize}↵
↵
Time complexity: $O(n \log n)$.↵
</spoiler>↵
↵
[problem: 2061G]↵
↵
<spoiler summary="Solution">↵
↵
Using graph theory terminology, let edges labeled as $0$ be represented in red and edges labeled as $1$ be represented in blue.↵
↵
---↵
↵
We can prove that if a graph has $3k + 1$ vertices, the maximum matching will be at most $k$. To see this, you can construct a blue clique of size $2k+1$, with all other edges colored red. It is not hard to verify that the maximum matching in both colors is $k$.↵
↵
So the maximum matching you can find is $\lfloor \frac{n+1}{3} \rfloor$.↵
↵
---↵
↵
To construct a matching, we can maintain a chain of edges of the same color. Suppose we currently have a red chain $p_1, p_2, \dots, p_k$. Now, we want to add a new vertex $q$. We check the color of the edge between $p_k$ and $q$:↵
↵
- If the edge is red, we add $q$ to the chain.↵
- If the edge is blue, $p_{k-1}, p_k, q$ will form a "mixed-color triplet." In this case, we remove $p_{k-1}$ and $p_k$ from the chain.↵
\end{itemize}↵
↵
After at most $n-1$ queries, the graph will be divided into several "mixed-color triplets" and one chain of edges of the same color.↵
↵
For the chain of length $k$, we can find $\lfloor \frac{k}{2} \rfloor$ matchings. For each "mixed-color triplet", we can always find one matching corresponding to the color.↵
↵
We can construct a matching of size $\lfloor \frac{n+1}{3} \rfloor$.↵
</spoiler>↵
↵
[problem: 2061I]↵
↵
<spoiler summary="Solution">↵
↵
First, there is a straightforward dynamic programming solution with a time complexity of $O(n^2)$.↵
↵
Let $dp(i, j)$ represent the minimum cost for Kevin to achieve $j$ victories in the first $i$ games. However, optimizing this dynamic programming solution directly is difficult because it lacks convexity.↵
↵
Therefore, we use a divide-and-conquer approach to solve it (which is more commonly seen in certain counting problems).↵
↵
--- ↵
↵
Key Idea: To compute $dp(r, *)$ from $dp(l, *)$ efficiently, the following observations are used:↵
↵
At time $l$, if the difference between Kevin's and Nivek's number of victories exceeds $r-l$, then:↵
↵
- For Type 2 matches, Kevin will either always win or always lose.↵
- For Type 1 matches, Kevin should prioritize matches with the smallest cost $a_i$ in ascending order.↵
\end{itemize}↵
↵
Let $g(i)$ denote the minimum cost of selecting $i$ matches in this range, and let $t$ represent the number of Type 2 matches in the interval.↵
↵
When $j - (l - j) > r-l$, the following transition applies: $$dp(l,j)+g(i) \to dp(r,j+t+i).$$↵
↵
Similarly, when $j - (l - j) < -(r-l)$, a similar transition is used.↵
↵
Since $g(i)$ is convex, the best transition point is monotonic. This property allows us to optimize this part using divide-and-conquer, achieving a time complexity of $O(n \log n)$.↵
↵
For the case where $|j - (l - j)| \leq r-l$, brute force is used to compute the transitions explicitly, with a time complexity of $O((r-l)^2)$.↵
↵
By dividing the sequence into blocks of size $O(\sqrt{n \log n})$, the overall time complexity can be reduced to $O(n \sqrt{n \log n})$.↵
↵
--- ↵
↵
To further optimize the problem, a more refined strategy is applied. Specifically, for the case where $|j - (l - j)| \leq r-l$, instead of computing transitions explicitly with brute force, we recursively divide the problem into smaller subproblems. The goal in each subproblem is to compute the transitions from a segment of $dp(l)$ to $dp(r)$, ensuring that the segment length in $dp(l)$ does not exceed $O(r-l)$.↵
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Let $m = \lfloor\frac{l + r}{2}\rfloor$. First, compute the transitions from $dp(l)$ to $dp(m)$. Then, use the results from $dp(m)$ to compute the transitions from $dp(m)$ to $dp(r)$. For each subinterval, the strategy remains the same:↵
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- For differences in the number of victories that exceed the interval length, apply monotonicity to handle the transitions efficiently.↵
- For the remaining part, recursively compute transitions using the same divide-and-conquer approach.↵
\end{itemize}↵
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The recurrence relation for this approach is: $T(n)=2T(n/2) + O(n \log n)$ So the time complexity is $T(n)=O(n \log^2 n)$.↵
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</spoiler>↵
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