Let $$$k=\lfloor\frac{n}{m}\rfloor$$$. The answer is $$$(1+\cdots+k)\cdot m=\frac{k(k+1)}{2}\cdot m$$$.

Consider modifying each character from left to right.

To change character $$$s_i$$$ to $$$'a'$$$, it is not difficult to see that the minimum number of operations needed is $$$min(s_i-'a','z'+1-s_i)$$$.

If the remaining number of operations $$$r$$$ is less than $$$min(s_i-'a','z'+1-s_i)$$$, then shift the current character to the left $$$r$$$ times.

You need to be careful: if all characters become $$$'a'$$$ and the remaining number of operations is odd, you need to change $$$s_n$$$ to $$$'b'$$$.

Consider the following problem:

For a string $$$t$$$ of length $$$m$$$, if $$$t$$$ is palindrome and $$$t+t_m$$$ is also palindrome, what kind of string is $t$？

We can find that $$$t_1=t_2=\cdots=t_m$$$ must hold.

Then the solution is clear. We can do bruteforce until we obtain a string where all characters are equal. Afterwards, you only need to add the same character at the end every time.

There will be only $$$O(\log n)$$$ iterations because after every time we add character to the end we will divide size of string by $$$2$$$.

What's the minimum sum we can get? The answer is $$$2l+2$$$.

Consider starting from the minimum sum and only increasing the sum by $$$1$$$ at a time.

Construction: $$$l+2, l, l+3, l+1, l+4 \dots$$$

Let $$$dp_{i,fa,fb}$$$ be answer if we consider only first $$$i$$$ columns, where $$$fa$$$ and $$$fb$$$ respectively represent whether $$$a_i$$$ and $$$b_i$$$ are covered by dominoes or not. Transitions are obvious.

let's say that $$$l=r=n$$$.

It's obvious that if $$$a$$$ is not a divisor of $$$n$$$, then vertex $$$a$$$ cannot contribute to $$$flow(n)$$$.

Think about the divisors less than $$$\sqrt{n}$$$ or more than $$$\sqrt{n}$$$.

How about the divisors which is just equal to $$$\sqrt{n}$$$?

How can we calculate $$$\Sigma_{i=l} ^r flow(i)$$$ faster? How about count the multiples instead of divisors?

First notice that we only need to make first $$$m$$$ characters contain $$$k$$$ ones, and $$$s_i=s_{i-m}$$$ for all $$$i$$$ such that $$$m+1\leq i\leq n$$$.

Now let's say you fixed your answer string and you want to find the minimum number of operations to get it. Obviously the answer is the number of segments of consecutive ones after xoring the answer string with $$$s$$$.

Number of segments of consecutive ones can be calculated using amount of pairs of distinct adjacent numbers, so our answer is $$$\frac{\sum{x_i\oplus x_{i-1}}}{2}$$$ where $$$x$$$ is xor string. (you need to add zeros on left and right of string).

You can calculate it using dp.