1. Abridged Problem Statement  
Given a starting voltage of 1 and two amplifier types—  
  • Type 1: X → 2·X − 1  
  • Type 2: X → 2·X + 1  
you need to reach exactly N volts (1≤N≤2·10⁹) using the fewest amplifiers. Output the minimum number of amplifiers and a sequence (1s and 2s) describing the types in order. If it’s impossible, print “No solution.”

2. Detailed Editorial  
Goal: build N from 1 with operations X↦2X−1 or X↦2X+1. Observe:  
  – Starting from 1, every result is odd. Thus N must be odd; if N is even, no solution.  
  – Brute‐force BFS over all reachable voltages is infeasible (N up to 2·10⁹).  
Idea: work backwards from N to 1 by inverting operations:  
  – Inverse of X→2X−1 is N→(N+1)/2, corresponding to a type-1 amplifier just before N.  
  – Inverse of X→2X+1 is N→(N−1)/2, corresponding to a type-2 amplifier.  
Since N is odd, both (N±1)/2 are integers. We need the shortest reverse path to 1. A greedy rule works in O(log N) steps:  
  1. While N>1, compute a=(N−1)/2 and b=(N+1)/2.  
  2. Exactly one of a,b is odd (the other even). To stay integer in the next step, pick the odd predecessor:  
     – If a is odd, record “2” and set N←a.  
     – Otherwise b is odd, record “1” and set N←b.  
  3. Repeat until N=1.  
  4. Reverse the recorded operations: this is the forward sequence from 1 to the original N.  
Correctness: always keeps the chain valid (next reverse step has an odd N), and it yields the minimal number of steps because any deviation would force an even intermediate and dead-end.  
Complexity: each step halves N roughly, so O(log N) time and memory.

3. Provided C++ Solution with Detailed Comments  
#include <bits/stdc++.h>  
using namespace std;

// Global variable to hold the target voltage
int n;

// Read input
void read() {
    cin >> n;
}

// Solve the problem
void solve() {
    // If N is even, we can never reach it (we only generate odd voltages)
    if (n % 2 == 0) {
        cout << "No solution\n";
        return;
    }

    // Vector to store the reverse sequence of amplifier types
    // We'll build it from target back to 1, then reverse it.
    vector<int> ans;

    // Work backwards until we reduce N to 1
    while (n > 1) {
        // Candidate previous voltages for each inverse operation
        int minus_op = (n - 1) / 2; // inverse of type-2 forward: X->2X+1
        int plus_op  = (n + 1) / 2; // inverse of type-1 forward: X->2X-1

        // Exactly one of minus_op or plus_op is odd. We pick the odd one
        // so that the next iteration n remains odd and valid.
        if (minus_op % 2 == 1) {
            // minus_op is odd -> we used type-2 forward to reach n
            ans.push_back(2);
            n = minus_op;
        } else {
            // plus_op must be odd -> we used type-1 forward to reach n
            ans.push_back(1);
            n = plus_op;
        }
    }

    // Reverse to get the forward sequence from 1 to original target
    reverse(ans.begin(), ans.end());

    // Output the number of amplifiers and the sequence
    cout << ans.size() << "\n";
    for (int t : ans) cout << t << " ";
    cout << "\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 main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    n = int(data[0])

    # If n is even, no sequence of 2X±1 from 1 can reach it
    if n % 2 == 0:
        print("No solution")
        return

    ops = []  # will store reverse operations
    while n > 1:
        # Compute both possible predecessors
        a = (n - 1) // 2  # from a with type-2 forward gives n = 2a+1
        b = (n + 1) // 2  # from b with type-1 forward gives n = 2b-1

        # Choose the one that remains odd (so next iteration is valid)
        if a % 2 == 1:
            ops.append(2)
            n = a
        else:
            ops.append(1)
            n = b

    # We built ops from target back to 1; reverse to get forward sequence
    ops.reverse()

    # Output result
    print(len(ops))
    if ops:
        print(" ".join(map(str, ops)))

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

5. Compressed Editorial  
Only odd targets are reachable because every operation yields an odd. Reverse greedily: from N repeatedly pick the inverse that produces an odd predecessor—(N−1)/2 if it’s odd implies forward type 2, otherwise (N+1)/2 implies type 1—until reaching 1. Reverse that list for the forward amplifier sequence. This takes O(log N) time and is optimal.