1) Abridged problem statement
- You are given a 4x4 sliding puzzle (numbers 0..15), where 0 is the blank. You may swap 0 with an adjacent tile (up/down/left/right). Determine whether the given configuration can be transformed into the target:
  1  2  3  4
  5  6  7  8
  9 10 11 12
 13 14 15  0
- Input: 4 lines of 4 integers each.
- Output: YES if the target is reachable, otherwise NO.

2) Detailed editorial
Key facts:
- For the 15-puzzle on a 4x4 board, configurations split into two equivalence classes (reachable vs unreachable). A simple invariant decides solvability.
- Flatten the board row-wise into an array. Replace 0 with 16 (treat the blank as the largest tile).
- Let I be the inversion count of this array: the number of pairs (i < j) with a[i] > a[j].
- Let d be the Manhattan distance of the blank to the target position (bottom-right), i.e., if 0 is at row r, col c (0-indexed), then d = (3 - r) + (3 - c).

Crucial invariant:
- Each legal move (swapping the blank with a neighbor) changes the inversion count I by an odd number (i.e., toggles I mod 2) when 0 is treated as 16.
  - Horizontal swap: the 16 crosses 1 element ⇒ I changes by ±1 (odd).
  - Vertical swap: although the swap crosses several elements in the linearization, one can show the parity change is still 1 (odd).
- Each move changes the blank’s Manhattan distance by exactly 1 (also toggles parity).
- Therefore, along any sequence of moves, I mod 2 and d mod 2 always stay equal. In the target state, I = 0 and d = 0, so the solvability condition is:
  (I + d) % 2 == 0.

Equivalence with the classical rule:
- The standard 4x4 test says: (inversions ignoring 0) + (row index of blank counted from bottom, 1-based) must be odd for solvability with the given target.
- The test in this solution (I with 0→16 plus Manhattan distance to bottom-right) is equivalent modulo 2.

Algorithm:
- Read the 16 numbers into an array, locate the zero at index p (row r = p//4, col c = p%4).
- Compute d = (3 - r) + (3 - c).
- Replace 0 with 16 and compute inversion count I over all 16 values.
- If (I + d) % 2 == 0, print YES; else print NO.

Complexity:
- Inversions over 16 elements is O(16^2) = 256 operations. Overall O(1) for this fixed-size puzzle.

3) Provided C++ solution with detailed comments
#include <bits/stdc++.h>              // Pulls in standard C++ headers (IO, containers, etc.)

using namespace std;

// Overload output operator for pair<T1,T2> (not used here, but present as utility).
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload input operator for pair<T1,T2> (not used here).
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload input operator for vector<T> to read all elements from stream.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;                      // Read each element into the vector.
    }
    return in;
};

// Overload output operator for vector<T> to print elements separated by spaces (unused).
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

vector<int> a;                        // Global array to hold the 16 puzzle values.

void read() {
    a.resize(16);                     // Ensure vector has 16 slots.
    cin >> a;                         // Read all 16 integers using the overloaded operator>>.
}

// Compute inversion count of the given array (number of pairs i<j with arr[i] > arr[j]).
int inversions(vector<int>& arr) {
    int count = 0;
    for(int i = 0; i < arr.size(); i++) {
        for(int j = i + 1; j < arr.size(); j++) {
            if(arr[i] > arr[j]) {
                count++;              // Found an inversion.
            }
        }
    }
    return count;
}

void solve() {
    // We use the well-known parity invariant of the 15-puzzle.
    // Replace 0 with 16 and compute:
    //   sum = (Manhattan distance of blank to bottom-right) + (inversion count of the array)
    // The configuration is solvable iff sum is even.

    int sum = 0;
    for(int i = 0; i < 16; i++) {
        if(a[i] == 0) {               // Find the blank (0) position.
            // Compute Manhattan distance from current blank (row=i/4, col=i%4)
            // to the target blank position (3,3).
            sum = (3 - i / 4) + (3 - i % 4);
            a[i] = 16;                // Treat blank as the largest tile for inversion counting.
            break;                    // Only one blank exists; stop.
        }
    }

    sum += inversions(a);             // Add inversion count to the Manhattan distance.

    // If the total parity is even, the state is reachable.
    cout << (sum % 2 == 0 ? "YES" : "NO") << '\n';
}

int main() {
    ios_base::sync_with_stdio(false); // Fast IO.
    cin.tie(nullptr);                 // Untie cin from cout.

    int T = 1;                        // The problem has a single test case in this setup.
    // cin >> T;                      // (Left here if you ever need multiple test cases.)
    for(int test = 1; test <= T; test++) {
        read();                       // Read the 4x4 puzzle values.
        // cout << "Case #" << test << ": ";
        solve();                      // Decide solvability and print YES/NO.
    }

    return 0;                         // End of program.
}

4) Python solution (well-commented)
import sys

def main():
    # Read all tokens (integers) from stdin.
    data = sys.stdin.read().strip().split()
    if not data:
        return

    # Parse the first 16 integers as the puzzle (row-major order).
    vals = list(map(int, data[:16]))

    # Locate the blank (0).
    zero_index = vals.index(0)           # Index in [0..15]
    r, c = divmod(zero_index, 4)         # Row and column (0-based)

    # Manhattan distance from blank to the target position (3,3).
    d = (3 - r) + (3 - c)

    # Replace 0 with 16 (treat blank as largest tile) for inversion counting.
    vals[zero_index] = 16

    # Compute inversion count over all 16 values.
    inv = 0
    for i in range(16):
        for j in range(i + 1, 16):
            if vals[i] > vals[j]:
                inv += 1

    # Solvable iff (inv + d) is even.
    print("YES" if (inv + d) % 2 == 0 else "NO")

if __name__ == "__main__":
    main()

5) Compressed editorial
- Flatten the board row-wise, replace 0 with 16, and count inversions I over all 16 elements.
- Let d be the Manhattan distance from the blank to the target corner (3,3).
- Any legal move swaps the blank with a neighbor; it toggles both the inversion parity (with 0→16) and the blank distance parity. Hence I ≡ d (mod 2) for all reachable states; in the target, both are 0.
- Therefore the given configuration is solvable iff (I + d) is even. This is equivalent to the classic 4x4 criterion: (inversions ignoring 0) + (row of blank from bottom, 1-based) is odd.
- Time complexity is O(16^2) for counting inversions, effectively constant.