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Sums of subarrays

Published on Sunday, 31st March 2019, 04:00 am; Solved by 165

Problem 663

Let $t_k$ be the tribonacci numbers defined as:
$\quad t_0 = t_1 = 0$;
$\quad t_2 = 1$;
$\quad t_k = t_{k-1} + t_{k-2} + t_{k-3} \quad \text{ for } k \ge 3$.

For a given integer $n$, let $A_n$ be an array of length $n$ (indexed from 0 to $n-1$), that is initially filled with zeros.
The array is changed iteratively by replacing $A_n[(t_{2 i-2} \text{ mod } n)]$ with $A_n[(t_{2 i-2} \text{ mod } n)]+2 (t_{2 i-1} \text{ mod } n)-n+1$ in each step $i$.
After each step $i$, define $M_n(i)$ to be $\displaystyle \max\{\sum_{j=p}^q A_n[j]: 0\le p\le q<n\}$, the maximal sum of any contiguous subarray of $A_n$.

The first 6 steps for $n=5$ are illustrated below:
Initial state: $\, A_5=\{0,0,0,0,0\}$
Step 1: $\quad \Rightarrow A_5=\{-4,0,0,0,0\}$ , $M_5(1)=0$
Step 2: $\quad \Rightarrow A_5=\{-4, -2, 0, 0, 0\}$ , $M_5(2)=0$
Step 3: $\quad \Rightarrow A_5=\{-4, -2, 4, 0, 0\}$ , $M_5(3)=4$
Step 4: $\quad \Rightarrow A_5=\{-4, -2, 6, 0, 0\}$ , $M_5(4)=6$
Step 5: $\quad \Rightarrow A_5=\{-4, -2, 6, 0, 4\}$ , $M_5(5)=10$
Step 6: $\quad \Rightarrow A_5=\{-4, 2, 6, 0, 4\}$ , $M_5(6)=12$

Let $\displaystyle S(n,l)=\sum_{i=1}^l M_n(i)$. Thus $S(5,6)=32$.
You are given $S(5,100)=2416$, $S(14,100)=3881$ and $S(107,1000)=1618572$.

Find $S(10\,000\,003,10\,200\,000)-S(10\,000\,003,10\,000\,000)$.