1. Abridged Problem Statement
Given n sorted distinct points \(x_0,\dots,x_{n-1}\) on the real line, you want to reposition them (preserving their order) so that they become an arithmetic progression
\[
y_i = L + d\,i,\quad i=0,1,\dots,n-1,
\]
minimizing the total movement
\(\sum_{i=0}^{n-1}\bigl|x_i - y_i\bigr|\). Output this minimum sum and one optimal sequence \(y_i\).

2. Detailed Editorial

Let
\[
f(L,d) \;=\;\sum_{i=0}^{n-1}\bigl|\,x_i - (L + d\,i)\bigr|.
\]
We seek \(\min_{L\in\mathbb R,\;d\ge 0} f(L,d)\). The function \(f\) is convex in both \(L\) and \(d\), so we can minimize it by nesting two ternary searches.

(A) For a fixed \(d\), the function
\[
g(L)=f(L,d)=\sum_{i=0}^{n-1}\bigl|\, (x_i - d\,i) - L\bigr|
\]
is convex in \(L\). We can find its minimum by ternary search on \(L\) over a large interval (e.g.\ \([-10^{12}, 10^{12}]\)).

(B) The function \(F(d)=\min_L f(L,d)\) is also convex in \(d\). Hence we can perform a ternary search on \(d\) over \([0, 10^7]\). Each evaluation of \(F(d)\) itself uses an inner ternary search on \(L\) with 100 iterations.

Overall complexity: \(O(n \cdot \text{(outer iterations)} \cdot \text{(inner iterations)}) \approx O(400 \times 100 \times 100)\), well within limits.

Implementation Steps
1. Read \(n\) and sort array \(x\).
2. Define inner function \(f(L, d)\) = \(\sum_i |x_i - (L + d \cdot i)|\).
3. Define \(f(d)\) = ternary search on \(L \in [-10^{12}, 10^{12}]\) to minimize \(f(L, d)\); return the best cost and optimal \(L\).
4. Ternary-search \(d\in[0,10^7]\), calling \(f(d)\) at each evaluation.
5. Output \(f(\text{best\_d}).\text{cost}\) and the positions \(y_i = \text{best\_L} + \text{best\_d} \cdot i\).

3. 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;
}
```

4. Python Solution 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()
```

5. Compressed Editorial
- Parametrize new positions as \(y_i = L + d\,i\).
- The total shift \(f(L,d) = \sum_i |x_i - (L + d \cdot i)|\) is convex in both \(L\) and \(d\).
- Nest two ternary searches: the outer one over \(d \in [0, 10^7]\), and for each \(d\) an inner ternary search over \(L \in [-10^{12}, 10^{12}]\) to minimize \(f(L, d)\).
- The inner search reports both the optimal cost and optimal \(L\), which is then used to reconstruct all coordinates.
- Recover the best \(L, d\) and output the result.
