<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

524. Buoys
Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



The swimming area of Berhattan's city beach is marked out with n buoys. The buoys form a straight line. When the buoys were being put into the water, nobody cared to observe the same distance between each pair of adjacent buoys.

Now the beach keeper wants the distance between any two adjacent buoys to be the same. He plans to shift some or all of the buoys without changing their respective order. To facilitate the task, he wants the total length of all shifts to be as small as possible.

Given coordinates of the buoys, you should find the minimum possible length of all shifts, as well as new coordinates of the buoys.

Input
The first line of input contains a single integer n (2 ≤ n ≤ 400), n — the number of buoys. The second line contains buoys' integer coordinates x1, x2,..., xn (-10000 ≤ xi ≤ 10000). No two given buoys will share the same place. The coordinates are given in strictly increasing order.

Output
To the first line print a real number t — the minimum possible total length of required shifts. Output this value with at least 4 digits after the decimal point.

To the second line print n numbers — new coordinates of the buoys. The new coordinates should be printed in strictly increasing order with at least 7 digits after the decimal point. If there are several optimal ways to shift the buoys, you may output any of them.

Example(s)
sample input
sample output
4 -2 2 6 9
1.0000
-2.0000000000 1.6666666667 5.3333333333 9.0000000000



Note
All buoys are located on the Ox axis. You may move buoys only along the Ox axis.

<|response|>
1. Abridged Problem Statement

You are given n distinct points x[0…n−1] on the real line in strictly increasing order. You want to move them (without changing their order) so that they form an arithmetic progression

  y[i] = L + d·i,  for i = 0…n−1,  with d ≥ 0,

minimizing the total movement

  T = ∑ᵢ |x[i] − y[i]|.

Output the minimum T and one choice of (L, d) (hence the y[i]) achieving it.

2. Key Observations

- The total shift is f(L, d) = ∑ᵢ |x[i] − (L + d·i)|, which is convex in both L and d.

- For a fixed d, define g(L) = f(L, d) = ∑ᵢ |(x[i] − d·i) − L|. As a function of L alone, this is convex and piecewise-linear, minimized by ternary search.

- The outer function F(d) = min_L f(L, d) is also convex in d, so we can apply a ternary search on d over [0, D_max] (e.g. D_max = 10⁷).

- Nesting two ternary searches, each with ~100 iterations, gives sufficient precision.

3. Full Solution Approach

1. Read n and array x[0…n−1].
2. Define f(l, d) = ∑ᵢ |x[i] − (l + d·i)| (the total shift cost).
3. Define f(d): ternary-search l ∈ [−10¹², 10¹²] to minimize f(l, d). Return {best cost, best l}.
4. Outer ternary-search d ∈ [0, 10⁷], using f(d).first as the objective. Track the best d found.
5. After the outer search, call f(best_d) again to recover the optimal L and cost precisely.
6. Output cost (with ≥4 decimals) and the sequence y[i] = L + d·i (with ≥7 decimals).

Complexity: each inner ternary search is O(n × 100), and the outer search is also 100 iterations → O(n × 10000), which is fine for n ≤ 400.

4. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<int> a;

void read() {
    cin >> n;
    a.resize(n);
    cin >> a;
}

long double f(long double l, long double d) {
    long double ans = 0;
    for(int i = 0; i < n; i++) {
        ans += abs(a[i] - (l + d * i));
    }
    return ans;
}

pair<long double, long double> f(long double d) {
    long double l = -1e12, r = 1e12, m1, m2, ans_l = 0;
    for(int steps = 0; steps < 100; steps++) {
        m1 = l + (r - l) / 3;
        m2 = r - (r - l) / 3;
        if(f(m1, d) < f(m2, d)) {
            r = m2;
            ans_l = m1;
        } else {
            l = m1;
            ans_l = m2;
        }
    }

    return {f(ans_l, d), ans_l};
}

void solve() {
    // The final buoys form an arithmetic progression coord(i) = l + d * i, so
    // the total shift is g(l, d) = sum_i |a[i] - (l + d * i)|. For a fixed
    // spacing d this is convex in the starting offset l, and the optimum cost
    // over l is in turn convex in d. We therefore nest two ternary searches:
    // the outer one over d in [0, 1e7] and the inner one (function f(d)) over
    // l, each returning the best cost. The inner search reports both the cost
    // and the optimal l, which then reconstructs all coordinates.

    sort(a.begin(), a.end());

    long double l = 0, r = 1e7, m1, m2, ans_d = 1;
    for(int steps = 0; steps < 100; steps++) {
        m1 = l + (r - l) / 3;
        m2 = r - (r - l) / 3;
        if(f(m1).first < f(m2).first) {
            r = m2;
            ans_d = m1;
        } else {
            l = m1;
            ans_d = m2;
        }
    }

    pair<long double, long double> ans = f(ans_d);
    cout << setprecision(9) << fixed << ans.first << '\n';
    for(int i = 0; i < n; i++) {
        cout << setprecision(9) << fixed << ans.second + ans_d * i << ' ';
    }
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python Implementation with Detailed Comments

```python
import sys

def read_ints():
    return list(map(int, sys.stdin.readline().split()))

def compute_cost_for_d(x, d):
    # Build b_i = x_i - d*i
    b = [xi - d * i for i, xi in enumerate(x)]
    # Find median of b
    b.sort()
    n = len(b)
    L = b[n//2]
    # Compute sum of absolute deviations from the median
    cost = sum(abs(bi - L) for bi in b)
    return cost, L

def main():
    data = sys.stdin.read().split()
    n = int(data[0])
    x = list(map(int, data[1:]))

    # Ternary search over d in [0, 1e7]
    lo, hi = 0.0, 1e7
    best_d = 0.0
    best_cost = float('inf')
    best_L = 0.0

    # Do ~100 iterations for high precision
    for _ in range(100):
        m1 = lo + (hi - lo) / 3
        m2 = hi - (hi - lo) / 3
        cost1, L1 = compute_cost_for_d(x, m1)
        cost2, L2 = compute_cost_for_d(x, m2)

        # Narrow the search range
        if cost1 < cost2:
            hi = m2
            if cost1 < best_cost:
                best_cost, best_d, best_L = cost1, m1, L1
        else:
            lo = m1
            if cost2 < best_cost:
                best_cost, best_d, best_L = cost2, m2, L2

    # Recompute L at best_d for accuracy
    best_cost, best_L = compute_cost_for_d(x, best_d)
    # Output the results
    # Total minimal shift (at least 4 decimal digits)
    print(f"{best_cost:.4f}")
    # New buoy positions, each with 7 decimal digits
    y = [best_L + best_d * i for i in range(n)]
    print(" ".join(f"{yi:.7f}" for yi in y))

if __name__ == "__main__":
    main()
```

Explanation of Key Steps:
- We reduce the two-parameter minimization over (L, d) to nested ternary searches exploiting the convexity of the total shift in both variables.
- The inner ternary search finds optimal L for fixed d; the outer finds optimal d.
- Finally, we reconstruct the optimal L and output the aligned arithmetic progression.
