Check if $$$r$$$ is less than $$$1200$$$: if true, output Noob; otherwise, output Not a noob.

Check if $$$r < 1200$$$: if true, output $$$1200$$$

If $$$r = 10000$$$ or $$$r = 9999$$$, output $$$100000$$$

Otherwise, output $$$r + 1$$$, which guarantees the conditions are met

Iterate over $$$k$$$ from $$$1$$$ to $$$X$$$. For each $$$k$$$, check if $$$X - k = S(Y + k)$$$ (where $$$S$$$ is the sum of the digits)

Output the smallest $$$k$$$ that satisfies the condition or $$$-1$$$ if no such $$$k$$$ exists

Calculate the total scores for all players. Check if the highest score is unique by counting its occurrences. If the count is 1, print YES; otherwise, print NO.

Short Solution Explanation

• We have $$$P_m = 2^{\frac{2^m - 1}{k}}$$$.
• $$$P_m$$$ is integer $$$\iff \frac{2^m - 1}{k} \in \mathbb{Z} \iff 2^m \equiv 1 \pmod{k}.$$$
• If $$$k=1$$$, answer $$$m=1.$$$
• If $$$k>1$$$ is even, then $$$2^m-1$$$ is odd and cannot be divisible by $$$k,$$$ so $$$m$$$ does not exist ($$$-1$$$).
• Otherwise, we find the minimal $$$m$$$ such that $$$2^m \equiv 1 \pmod{k}.$$$
• Iterate $$$m=1,\dots,k.$$$
• Check if $$$2^m \bmod k=1.$$$
• The first such $$$m$$$ is the answer; if none, output $$$-1.$$$

Explanation with MathJax

Let’s maintain a function $$$f(n, m)$$$ to decide if a pair is good or not.

Case 1: Odd values of $$$n$$$

For odd values of $$$n$$$, we can re-express the string $$$s$$$ as:

The sum:

and the middle digit can be denoted as $$$h$$$.

Thus, the total sum is:

Since $$$h$$$ can be any digit, the sum $$$2f + h$$$ lies between the minimum and maximum possible values, inclusive.

Conditions for a "good" pair: [ \begin{cases} n \leq m \leq 9n, \ n \mod 2 = 1, \end{cases} ] then the pair is YES.

Case 2: Even values of $$$n$$$

For even values of $$$n$$$, we re-express the string $$$s$$$ as:

Using symmetry:

Let $$$f$$$ denote the left and right segments. The sum of medians is $$$2h$$$, and the total becomes:

For even $$$n$$$, the sum must also satisfy the condition of being even.

Conditions for a "good" pair: [ \begin{cases} n \leq m \leq 9n, \ (n + m) \mod 2 = 0, \end{cases} ] then the pair is YES.

Complexity

For the easy version, we can iterate through all pairs $$$(a, b)$$$, and the complexity will be: [ \mathcal{O}(n^2). ]

Optimized Solution

We can iterate for each $$$a$$$ from $$$1$$$ to $$$n$$$. This allows us to count valid answers:

• For odd $$$a$$$, all answers in $$$[a, 9a]$$$ are valid.
  Thus, there are $$$8a + 1$$$ valid answers for odd $$$a$$$.

• For even $$$a$$$, only even indices in $$$[a, 9a]$$$ are valid.
  Thus, for even $$$a$$$, we add $$$4a + 1$$$ valid answers.

Complexity

The total complexity is:

Optimized Solution with Closed Formula

From the Medium version, we obtained the following conditions: [ \begin{cases} 8a + 1, & a \mod 2 = 1 \ 4a + 1, & a \mod 2 = 0 \end{cases} ]

Step 1: Sum for Odd Numbers

For odd numbers, there are $$$\left\lceil \frac{n}{2} \right\rceil$$$ odd integers. We want to calculate:

Using the formula for the sum of the first $$$m$$$ odd numbers ($$$m^2$$$), we find that the count for odd numbers is:

Step 2: Sum for Even Numbers

For even numbers, there are $$$\left\lfloor \frac{n}{2} \right\rfloor$$$ even integers. We want to calculate:

Using the formula for the sum of the first $$$m$$$ even numbers ($$$m(m + 1)$$$), we find that the count for even numbers is:

Final Formula

Combining the results for odd and even numbers, the final formula is:

Complexity

This solution has a complexity of:

To solve the problem of transforming an array into an "almost sorted" array with minimal operations, the solution uses a dynamic programming approach combined with a balanced structure for optimization. Here's a breakdown:

Explanation:

1. Dynamic Programming Approach:

• We iterate over the array while maintaining a dp array, where dp[i] stores the minimal operations needed to fix the array up to index i.
• For each element, calculate the operations needed to ensure it is at least as large as the previous element. If it's smaller, add the difference to the operations count.

1. Efficient Tracking of State:

• To minimize redundant calculations for future indices, we maintain a data structure (set) that efficiently tracks previous adjustments and allows merging ranges.

1. Optimization with Sliding Window:

• The algorithm uses a sliding window over the array while calculating the total adjustments needed (now) for transforming the remaining elements into valid states.
• By combining the dp values with the adjustments tracked in the set, the solution finds the minimal operations.

Key Functions:

• get(l, r): Calculates the sum of integers from l to r. This helps in efficient range operations.
• maxs and mins: Utility functions to update values efficiently while iterating.

Complexity:

• Time Complexity: $$$O(nlogn)$$$, due to the use of a balanced structure (set) for maintaining state.
• Space Complexity: $$$O(n)$$$, for storing the dp array and state.

Implementation:

The solution calculates the minimal operations by: 1. Iterating through the array and updating dp. 2. Using a set to manage adjustments dynamically. 3. Combining results from dp and the tracked state to compute the answer.

Conclusion:

This approach ensures efficient computation for large arrays (( n \leq 5 \times 10^6 )) while maintaining correctness. The use of dynamic programming and an optimized state-tracking mechanism makes the solution both elegant and scalable.

Key Observations

1. The sum of cubes can be expressed as:

For large values of $$$r$$$, the sum can quickly grow to exceed $$$n$$$, so efficient calculations are critical.

1. The sum of cubes from $$$1$$$ to $$$x$$$ is given by:

Using this formula, the sum $$$S(l, r)$$$ can be computed as:

1. For each $$$l$$$, the problem is reduced to finding the smallest $$$r$$$ such that $$$S(l, r) = n$$$.

Algorithm

Step 1: Precomputation of Sum Function

• Define a helper function $$$s(x)$$$ that computes:

This function avoids overflow checks by using conditionals to cap results at $$$2 \times 10^{18}$$$.

Step 2: Binary Search for $$$r$$$

• For a fixed $$$l$$$, use binary search to find $$$r$$$ such that $$$S(l, r) = n$$$:
• Calculate $$$S(l, r)$$$ directly using $$$s(r)$$$ and $$$s(l-1)$$$.
• If $$$S(l, r) < n$$$, increase $$$r$$$.
• If $$$S(l, r) > n$$$, decrease $$$r$$$.

Step 3: Iterate Over $$$l$$$

• Iterate over $$$l$$$ from $$$1$$$ to $$$\sqrt[3]{n}$$$ (approximation of possible lower bounds).
• Use binary search for each $$$l$$$ to find a valid $$$r$$$.

Implementation

Key Functions

1. s(x): Calculates $$$\left( \frac{x(x+1)}{2} \right)^2$$$ with overflow checks.

2. f(l, r): Calculates $$$S(l, r)$$$ using the formula:

Includes overflow checks.

1. Binary Search: Finds $$$r$$$ for a given $$$l$$$ such that $$$S(l, r) = n$$$.

Key Observations

1. Cubic Sum Representation: The sum of cubes can be expressed as:

Here, $$$\text{prefix_sum}[i]$$$ stores the sum of cubes from $$$1$$$ to $$$i$$$.

1. Two-Pointer Technique: By maintaining two pointers, $$$l$$$ and $$$r$$$, and using the prefix sums:

• Increase $$$r$$$ when the sum is less than $$$n$$$.
• Increase $$$l$$$ when the sum exceeds $$$n$$$.
• Stop when the sum equals $$$n$$$ or when $$$r$$$ exceeds the maximum limit.

1. Precomputation: Precompute the cubes and their prefix sums for all integers up to $$$10^6$$$:

This allows efficient calculation of the sum of cubes for any segment $$$[l, r]$$$.

Algorithm

1. Precomputation: Precompute the cubes and their prefix sums up to $$$10^6$$$ in $$$\mathcal{O}(n)$$$ time.

2. Two-Pointer Search: For each query:

• Initialize $$$l = 1$$$, $$$r = 1$$$.
• Use the two-pointer approach to find a range $$$[l, r]$$$ such that the sum equals $$$n$$$:
  • If the sum of cubes from $$$l$$$ to $$$r$$$ equals $$$n$$$, output $$$l$$$ and $$$r$$$.
  • If the sum is less than $$$n$$$, increment $$$r$$$.
  • If the sum is greater than $$$n$$$, increment $$$l$$$.
• If no such range is found, output $$$-1$$$.

1. Complexity:

• Precomputation takes $$$\mathcal{O}(n)$$$ for $$$n = 10^6$$$.
• Each query runs in approximately $$$\mathcal{O}(n)$$$ due to the two-pointer traversal.
• Total complexity: $$$\mathcal{O}(n + qt)$$$, where $$$q$$$ is the number of queries.

GCD Problem Explanation

Key Idea:

1. Fix the leftmost element of the segment.

• The greatest common divisor (GCD) of (A[l, r)) changes at most (\log A_i) times.
• This is because the GCD value decreases as (r) increases and can only take specific values that divide the elements in the segment.

Example:

Consider the array:

Fix the leftmost element of the segment, (A[l] = 12).
Now, calculate the GCD for increasing values of (r):

• $$$ \text{GCD}(12) = 12 ) for \; r = 1, 2 $$$
• $$$ \text{GCD}(12, 18, 30) = 6 ) for \; r = 3, 4, 5 $$$
• $$$ \text{GCD}(12, 18, 30, 36, 16) = 2 ) for \; r = 6, 7 $$$
• $$$ \text{GCD}(12, 18, 30, 36, 16, 6, 9) = 1 ) for \; r = 8 $$$

Marked positions where the GCD changes: ( r = 2, 5, 7, 8 ).

Observations:

• If $$$[1, \text{marked}]$$$ is completely contained within a query, we can ignore the unmarked rightmost elements.
• This allows us to reduce the number of checks.

Efficient Implementation:

• Use a GCD segment tree to manage the structure of the array and determine the GCD for any segment efficiently.
• Sort the queries by $$$l_i$$$ in descending order $$$( l = N, N-1, \ldots, 1 )$$$.
• Utilize a max segment tree to answer the queries:
• First, check the query range $$$[l_i, r_i]$$$.
• Then, verify the GCD change points (marked positions) for fixed $$$r_i$$$.

Complexity:

The time complexity is:

where: - $$$N$$$: Length of the array. - $$$Q$$$: Number of queries. - $$$A_i$$$: Maximum value in the array.