The objective is to determine the highest possible integer achievable by merging adjacent equal values in a sequence. When two adjacent numbers share the same value $v$, they can be combined into a single number with value $v+1$. This process repeats until no further merges are possbile or the desired maximum is reached.

To solve this efficiently, an interval dynamic programming approach is suitable given the constraint $N \le 248$. Define a state table where dp[i][j] represents the single value obtained if the subarray from index $i$ to $j$ is completely merged. If the range cannot be reduced to a single number, the value remains 0.

Initialization sets dp[i][i] to the input sequence values. The algorithm iterates through all posssible subarray lengths. For each range [i, j], it checks every possible split point k. If the left segment [i, k] and the right segment [k+1, j] both merge into the same non-zero value, the current range [i, j] can merge into that value plus one. Throughout this process, track the global maximum value encountered.

The time complexity is $O(N^3)$ due to the three nested loops (length, start index, split point), which fits with in the limits for $N \approx 250$. Space complexity is $O(N^2)$ for the DP table.

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int size;
    if (!(cin >> size)) return 0;

    vector<int> initial_values(size);
    for (int i = 0; i < size; ++i) {
        cin >> initial_values[i];
    }

    // dp[i][j] stores the merged value of range [i, j], or 0 if impossible
    vector<vector<int>> dp(size, vector<int>(size, 0));
    int max_val = 0;

    // Base case: single elements
    for (int i = 0; i < size; ++i) {
        dp[i][i] = initial_values[i];
        max_val = max(max_val, dp[i][i]);
    }

    // Iterate over subarray lengths
    for (int span = 2; span <= size; ++span) {
        // Iterate over start index
        for (int start = 0; start <= size - span; ++start) {
            int end = start + span - 1;
            // Iterate over split point
            for (int pivot = start; pivot < end; ++pivot) {
                int left_val = dp[start][pivot];
                int right_val = dp[pivot + 1][end];

                if (left_val > 0 && left_val == right_val) {
                    int merged = left_val + 1;
                    dp[start][end] = max(dp[start][end], merged);
                    max_val = max(max_val, merged);
                }
            }
        }
    }

    cout << max_val << endl;

    return 0;
}