Yarra Gnisrever

Let $N$ and $K$ be two positive integers.
$F_n$ is the $n$-th Fibonacci number: $F_1 = F_2 = 1$, $F_n = F_{n – 1} + F_{n – 2}$ for all $n \geq 3$.
Let $s_n = F_{2n – 1} \bmod N$ and let $t_n = F_{2n} \bmod N$.
Start with an array of integers $A = (A[0], \cdots, A[N – 1])$ where initially every $A\text{[}i]$ is equal to $i$.
Now perform $K$ successive operations on $A$, where the $j$-th operation consists of reversing the order of those elements in $A$ with indices between $s_j$ and $t_j$ (both ends inclusive).
Define $R(N,K)$ to be $\sum_{i = 0}^{N – 1}i \times A\text {[}i]$ after $K$ operations.
For example, $R(5, 4) = 27$, as can be seen from the following procedure:
Initial position: $(0, 1, 2, 3, 4)$
Step 1 – Reverse $A[1]$ to $A[1]$: $(0, 1, 2, 3, 4)$
Step 2 – Reverse $A[2]$ to $A[3]$: $(0, 1, 3, 2, 4)$
Step 3 – Reverse $A[0]$ to $A[3]$: $(2, 3, 1, 0, 4)$
Step 4 – Reverse $A[3]$ to $A[1]$: $(2, 0, 1, 3, 4)$
$R(5, 4) = 0 \times 2 + 1 \times 0 + 2 \times 1 + 3 \times 3 + 4 \times 4 = 27$
Also, $R(10^2, 10^2) = 246597$ and $R(10^4, 10^4) = 249275481640$.
Find $R(10^{18}, 10^6)$ giving your answer modulo $10^9$.

To solve this problem, we need a clear understanding of Fibonacci numbers, iteration operations on arrays and modular arithmetic.

Let’s first observe the Fibonacci sequence modulo $N$. Since the Fibonacci sequence is defined recursively, and we apply the modulo operator to each term, the sequence will eventually start repeating. The sequence starts with $F_1 = 1,NF_2=1,N$.:

.. . . $F_{N+1}=F_N+F_{N-1}\bmod N$. . . .,

This will continue until repeating starting from the pair (1, 1). Therefore, the repeating sequence is the sequence for $F_1$, $F_2$,…$F_{d+1}$, where d is the period of the sequence modulo N.

Notably, since the sequence starts with 1 and 1 and each subsequent term is the sum of the preceding two terms modulo N, no term can be equal to N, meaning the period of the sequence will not exceed $N^2$.

To perform $K$ successive rotations to an initial array $A$, we’ll find the indices $s_j$ and $t_j$ for $2j-1$th and $2j$th terms of the Fibonacci sequence modulo $N$. However, these sequences may have shorter periods as well. We don’t need to generate more than $N^2$ Fibonacci numbers modulo $N$.

So, in practice, we calculate Fibonacci numbers modulo $N$, for $n=1$ up to $2*N^2$. Save each $F_{2n – 1}$ in the list $S$ and each $F_{2n}$ in the list $T$. At the same time, check the last $N^2$ terms for the repetition of initial $N^2$ elements in $S$ and $T$. If found, truncate the lists to the length of repeating sequence.

If $K$ is larger than the length of the repeating sequence, we only perform the operation for the first $K$ % repeat_length times.

The reversal operation on the array A is performed according to $s_i$ and $t_i$ indices for every $i$ from 1 to min(K, length of repetition). Perform the operation on a copy of the initial array if $K$ is more than the length of repetition.

Lastly, $R(N,K)$ can be calculated as described for $R(5,4)$. However to save time, we should also calculate the partial sums for arrays $A$ and $i*A[i]$, and adjust the sum for each reversal operation instead of re-calculating it from scratch.

Please note that the calculation of $R(10^{18},10^6)$ modulo $10^9$ is beyond ordinary computational capacity and may require an optimized code or algorithm for efficient computation.

Programming languages could be utilized to solve the problem. Concepts like dynamic programming, recurrence relation, Euclidean algorithm, Extended Euclidean algorithm and Lucas’s theorem will be very useful in writing and optimizing the algorithm.

In conclusion, the specific process to find $R(10^{18},10^6)$ modulo $10^9$ requires a combination of mathematical and computational strategies, with proper optimization considerations.

Due to the complexity of this problem, a direct solution or simplified form likely doesn’t exist. Computational resources are essential here.

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