I think it is hard because how could a team perfomrance an indication for every member's level. But I believe that it will be interesting if codeforces add a new feature of rating for teams and then introduce this new type of contests.

Sorry for the long wait, I was unable to reply faster.

This contest is most suitable for solving it using only one computer per team (same as during the onsite event). However, we cannot enforce this rule, so it is not forbidden to use multiple computers.

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true

It's direct greedy. You won't need dp or anything complicated.

FFT isn't that hard

everytime a one occurs, the element next to it can be 1 or 0. in a string where there is 0 at last, no of 1s = no of 10 + no of 11 follows. in a string where there is 1 at last, the sum will be no of ones -1

I referred to Jiangly's submission 292165413. He made all the elements of array equal and less than 3 (if possible) while storing the count of number of operations required on each index and if it's possible the minimum operations can be achieved by simply subtracting minimum value of an operation on any index times n from total number of operations......

// this is solution ? 

I solved it a bit differently. Let $$$f[i]$$$ be the number of operations you apply at $$$i$$$. Note that $$$f[i]$$$ is cyclic, meaning $$$f[i] = f[i + N]$$$. Let the final element be $$$x$$$. Then, we must have:

If you solve this equation for $$$f[i]$$$ (I took $$$N = 5$$$ as a reference), you find:

Let the second term be $$$q[i]$$$, where:

Observe that $$$q[i]$$$ satisfies the recurrence:

Thus, we can compute $$$q[N - 1]$$$ first and then evaluate all $$$q[i]$$$. However, if $$$q[N - 1]$$$ is not an integer, we immediately print

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Additionally, since

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Hence,

After this, we check whether $$$x$$$ is non-negative and then print the final answer.

solution: First enumerate offline all distinct OR-value intervals ending at each i → sort those intervals in ascending order of their OR value → perform an online DP using a Fenwick tree. When dealing with large data scales such as 2e5 or even 1e6, it’s critical to avoid small objects and wrapper classes like Integer, int[], record Segment(...), or data structures like PriorityQueue and PriorityQueue<int[]>—especially when performance matters. These objects carry significant overhead due to Java’s object model: every object includes a header, introduces pointer indirection, and scatters data across the heap, leading to poor cache locality. Unlike Rust and c++ where structs are often plain, tightly packed value types, Java’s heap-allocated objects are far less memory-efficient and CPU-friendly. In performance-critical cases like this problem, replacing Segment[] or PriorityQueue<int[]> sorted via lambda comparators with a custom radix sort on 4 flat int[] avoids comparisons altogether, works directly on raw bits, and accesses memory sequentially. This results in branchless, linear-time sorting with excellent cache usage and significantly lower overhead—often 3–5× faster than Arrays.sort. Valhalla aims to bring value class, value record to Java, allowing objects without identity and header overhead. This would enable Rust-like performance for dense data structures. So, after working for over ten years, when can Valhalla be previewed??????(considering the poor Vector api incubated for over ten times lol).

(I misread, so ignore, sorry) the path forms a sort of an reversed L(goes right and then down)

I am going to part from the fact that gcd(x,a) = 1 and gcd(y,b) = 1, and that the lower bound is

Then the cost of the path (goes right and then down) is

going from $$$(1, 1)$$$ to $$$(1, y)$$$ costs $$$\sum_{i=1}^{y} gcd(1,a)+ gcd(i, b)$$$, and

and going from $$$(1, y)$$$ to $$$(x, y)$$$ costs $$$\sum_{i=2}^{x} gcd(i, a) + gcd(y, b)$$$, and recall that $$$gcd(y, b) = 1$$$, so we have

So the cost along all the path can be expressed as

Which is the lower bound, hence it is optimal

Let's look at "border" $$$(n, y) - (x, y) - (x, n)$$$. Any path from $$$(1, 1)$$$ to $$$(n, n)$$$ crosses it.

WLOG, let's say that the path crosses the border at cell $$$(x, y')$$$ where $$$y \le y' \le n$$$.

Let's find the minimum path from $$$(1, 1)$$$ to $$$(x, y')$$$. It's easy to see that we can go from $$$(1, 1)$$$ to $$$(x, 1)$$$ and then to $$$(x, y')$$$ since $$$\gcd(x, a) = 1$$$.

So, the path $$$(1, 1) - (x, y')$$$ goes through cell $$$(x, y)$$$ for any $$$(x, y')$$$.