C. Ultimate Value

Problem

Let’s define a function $f(a)$ for an array $a$ of length $n$ as $f(a) = cost + (a_1 - a_2 + a_3 - a_4 + \ldots + a_n)$ where cost is zero initially.

Now consider the scenario where Alice and Bob are given an array $a$ of length $n$. They play a game taking at most $10^{100}$ turns alternately with Alice going first.

In each turn, they must perform any one (only one) of the following operations:

End the game for both Alice and Bob. Choose two indices $l, r$ with $1 \leq l \leq r \leq n$ and swap $a_l$ and $a_r$; this adds $(r - l)$ to the cost.

Assume that Alice tries to maximize $f(a)$ and Bob tries to minimize it.

Your task is to determine the final value of $f(a)$ assuming both players play optimally.

Solution and Observations

The key insight to solve this problem comes from combining two factors: the index position and its associated swap cost.

1. Game Simplification:

   • The game effectively reduces to finding a single optimal swap, as Alice can always undo Bob’s moves

2. Key Strategy - Index-Cost Combination:

   • For each element at position $i$, we consider:
     • The cost of swapping (which depends on position $i$)
     • The change in value if we move this element
   • By combining these factors, we can evaluate each potential move’s total impact
   • We track the best achievable differences separately for odd and even positions

3. Final Value: The answer is the maximum of:

   • The initial alternating sum (without any swaps)
   • The best achievable value after considering all possible index-cost combinations

The solution combines dynamic programming with careful tracking of potential value changes from swaps at different positions.

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