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 \(\min_{L\in\mathbb R,\;d>0} f(L,d)\).  Observe two facts:

(A) For a fixed \(d\),  
\[
g(L)=f(L,d)=\sum_{i=0}^{n-1}\bigl|\, (x_i - d\,i) - L\bigr|
\]  
is a convex, piecewise-linear function in \(L\).  Its minimizer is any median of the multiset  
\[
\{\,x_i - d\,i\mid i=0,1,\dots,n-1\}.
\]  
Thus one can compute  
\[
L^*(d) = \mathrm{median}\bigl\{x_i - d\,i\bigr\}
\quad\text{and}\quad
F(d)=f\bigl(L^*(d),d\bigr)
\]  
in \(O(n\log n)\) (to find the median and sum absolute values).

(B) The function \(F(d)\) is convex in \(d\).  Hence we can perform a ternary search on \(d\) over some large interval \([0,D_{\max}]\) (e.g.\ \(D_{\max}=10^7\)).  Each evaluation of \(F(d)\) costs \(O(n\log n)\).  With \(O(100)\) iterations we get high precision.

Overall complexity: \(O(n\log n\cdot\text{(iterations)})\approx O(400\log 400\times100)\), well within limits.

Implementation Steps  
1. Read \(n\) and array \(x\).  
2. Ternary-search \(d\in[0,10^7]\): at each candidate \(d\),
   - Form the array \(b_i = x_i - d\,i\).
   - Find the median of \(b\).
   - Compute \(\sum_i |b_i - \mathrm{median}|\).  
3. Keep the best \(d\).  Once found, recompute \(L=\) median of \(\{x_i - d\,i\}\).  
4. Output \(\sum_i |x_i - (L+d i)|\) and the new positions \(y_i=L+d i\).  

3. C++ Solution with Line-by-Line Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overloads to read/write pairs and vectors conveniently
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;

// Read input: n and the array a[]
void read() {
    cin >> n;
    a.resize(n);
    cin >> a;
}

// For a fixed (L,d), compute sum |a[i] - (L + d*i)|
long double cost_Ld(long double L, long double d) {
    long double s = 0;
    for(int i = 0; i < n; i++)
        s += fabsl(a[i] - (L + d * i));
    return s;
}

// For a fixed d, find the best L and return { cost, L }
pair<long double,long double> eval_d(long double d) {
    // Build transformed values b[i] = a[i] - d*i
    vector<long double> b(n);
    for(int i = 0; i < n; i++)
        b[i] = a[i] - d * i;
    // Find median of b[]
    sort(b.begin(), b.end());
    long double L = b[n/2];
    // Compute cost at this L
    long double total = 0;
    for(int i = 0; i < n; i++)
        total += fabsl(b[i] - L);
    return { total, L };
}

void solve() {
    // Sort input array just in case (it is already sorted by problem statement)
    sort(a.begin(), a.end());

    // Ternary search on d in [0, 1e7]
    long double lo = 0, hi = 1e7;
    long double best_d = 0, best_cost = 1e300, best_L = 0;
    for(int it = 0; it < 100; it++) {
        long double m1 = lo + (hi - lo) / 3;
        long double m2 = hi - (hi - lo) / 3;
        auto r1 = eval_d(m1);
        auto r2 = eval_d(m2);
        if(r1.first < r2.first) {
            hi = m2;
            if(r1.first < best_cost) {
                best_cost = r1.first;
                best_d = m1;
                best_L = r1.second;
            }
        } else {
            lo = m1;
            if(r2.first < best_cost) {
                best_cost = r2.first;
                best_d = m2;
                best_L = r2.second;
            }
        }
    }

    // We have best_d and best_L (approx). Re-evaluate best_L precisely:
    auto final_pair = eval_d(best_d);
    best_cost = final_pair.first;
    best_L    = final_pair.second;

    // Output total minimal shift
    cout << fixed << setprecision(9) << best_cost << "\n";
    // Output new positions y[i] = best_L + best_d * i
    for(int i = 0; i < n; i++) {
        long double y = best_L + best_d * i;
        cout << fixed << setprecision(9) << y << (i+1<n?' ':'\n');
    }
}

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

    read();
    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 \(y_i = L + d\,i\).  
- For fixed \(d\), optimal \(L\) is the median of \(\{x_i - d\,i\}\).  
- Let \(F(d)\) be the minimal total shift at that \(d\).  Then \(F(d)\) is convex.  
- Perform a ternary search on \(d\in[0,10^7]\), computing \(F(d)\) in \(O(n\log n)\) each time.  
- Recover \(L\) from the median and output the result.