Summary

This problem asks for the minimum operation cost to withdraw guards such that the conditions are satisfied.
However, it turns out that doing nothing is always optimal.

Analysis

First, the crucial point is that the operation can be performed “\(0\) or more times.”
In other words, it is not necessarily required to withdraw any guards.

In the initial state, one guard is stationed at each lamppost.

Let’s check the conditions under this state.

Condition 1: At least \(2\) guards remain

From the constraints, \(N \geq 2\).
Since there are \(N\) guards in the initial state, there are always at least \(2\) guards.

Condition 2: All remaining guards satisfy the reassurance condition

Suppose there is a guard at lamppost \(i\).

The reassurance condition is that there is also a guard at least at one of:

• the lamppost \(\lfloor N/2 \rfloor\) positions ahead clockwise
• the lamppost \(\lfloor N/2 \rfloor\) positions ahead counter-clockwise

of lamppost \(i\).

In the initial state, there is a guard at every lamppost, so naturally, there are guards at these lampposts as well.

For example:

• When \(N\) is even
  The lampposts \(\lfloor N/2 \rfloor\) positions ahead clockwise and counter-clockwise are the same lamppost, and there is a guard stationed there.

• When \(N\) is odd
  The clockwise and counter-clockwise destinations are different lampposts, but there are guards at both.

Therefore, the reassurance condition is always satisfied in the initial state.

About the Cost

If we do not perform any operations, the total cost is \(0\).

Additionally, the cost of each operation is the maximum cost of the withdrawn guards, and all costs \(A_i\) are at least \(1\).
Therefore, the operation cost is always at least \(0\), and the total cost can never be less than \(0\).

Thus, the minimum cost is always \(0\).

If you naively try to search for “which guards to keep,” the number of choices becomes extremely large, which is impractical for \(N \leq 10^6\).
However, in this problem, the option of “doing nothing” always satisfies the conditions, so such a search is unnecessary.

Algorithm

1. Read the input.
2. If no operations are performed, the conditions are always satisfied.
3. Therefore, the answer is always \(0\).
4. Output \(0\).

Complexity

• Time Complexity: \(O(N)\)
  This is because we need to read \(A_1, A_2, \ldots, A_N\) as input.
• Space Complexity: \(O(N)\)
  This is because the provided code stores the input in an array \(A\).
  However, since the values are not used, we can discard them as we read them to achieve \(O(1)\) space complexity.

Implementation Details

In this problem, the values of the withdrawal costs \(A_i\) do not affect the answer.
You only need to read the input correctly and always output 0.

Source Code

#include <bits/stdc++.h>
using namespace std;

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

    int N;
    cin >> N;
    vector<long long> A(N);
    for (auto &x : A) cin >> x;

    cout << 0 << '\n';
    return 0;
}

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This editorial was generated by gpt-5.5-high.