1. Abridged Problem Statement  
Given integers m,n (1≤m,n≤1000), count the exact number of ways to tile an m×n grid with 1×2 and 2×1 dominoes so that no four dominoes meet at a single grid corner. If mn is odd, the answer is 0. Print the total count (no modulo).

2. Detailed Editorial  

Overview  
We must count domino‐tilings of an m×n board avoiding any point where four tiles meet. Key idea: choose the smaller dimension as n (swap if needed) and handle special cases n=1,2, then derive recurrences for odd n>2 and even n>2.

A. Impossible when mn is odd  
A domino covers two cells; if the board has odd area, return 0.

B. Case n=1  
A 1×m strip can be covered only by horizontal dominoes, one way (assuming m even; odd m is already excluded).

C. Case n=2  
Let dp[i] = number of valid tilings of 2×i strip.  
- Base: dp[0]=1 (empty), dp[1]=1 (one vertical domino), dp[2]=2.  
- Recurrence: to extend to width i, either place a vertical domino at column i (dp[i−1] ways), or place two horizontal dominoes covering columns i−2,i−1 and another horizontal pair covering i,i+1 etc. Detailed combinatorics yield dp[i] = dp[i−1] + dp[i−3].

D. Case n>2  
We observe that large blocks of size n×n can only be tiled in two “cornerless” patterns (one rotated version of the other). This lets us form a one‐dimensional DP on columns.

D1. Odd n  
Every valid tiling consists of repeating an (n×(n+1)) block or its transpose (n×(n−1)) block, up to reflection. Let f[i] count ways for n×i strip (i columns):  
f[i] = f[i−(n+1)] + f[i−(n−1)]  
Multiply final result by 2 for the two orientations.

D2. Even n  
We keep two DP states: for the first i columns,  
  dp[i][0] = ways ending with a fully vertical column at i,  
  dp[i][1] = ways ending with a “mixed” column (part horizontal).  
Transitions:  
  dp[i][0] = dp[i−1][1]  
  dp[i][1] = dp[i−(n−2)][0] + dp[i−n][0]  
Answer is dp[m][0]+dp[m][1].

Implementation details  
Use big integers (e.g., boost::multiprecision::cpp_int in C++) because counts grow exponentially.

3. C++ Solution with Detailed Comments

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using bigint = boost::multiprecision::cpp_int;

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

    int m, n;
    cin >> m >> n;
    // If area is odd, no tiling possible
    if ((long long)m * n % 2 == 1) {
        cout << 0 << "\n";
        return 0;
    }
    // Ensure n <= m for convenience
    if (n > m) swap(n, m);

    // Case n=1: only one way to place horizontal dominoes
    if (n == 1) {
        cout << 1 << "\n";
        return 0;
    }

    // Case n=2: use 1D DP with recurrence dp[i]=dp[i-1]+dp[i-3]
    if (n == 2) {
        vector<bigint> dp(m+1);
        dp[0] = 1;  // empty strip
        dp[1] = 1;  // one vertical domino
        dp[2] = 2;  // two verticals or two horizontals
        for (int i = 3; i <= m; ++i) {
            dp[i] = dp[i-1] + dp[i-3];
        }
        cout << dp[m] << "\n";
        return 0;
    }

    // For n > 2 and odd n: 1D DP with two possible column-block widths
    if (n % 2 == 1) {
        vector<bigint> dp(m+1);
        dp[0] = 1;
        // transitions by adding an (n-1)-wide or (n+1)-wide block
        for (int i = 1; i <= m; ++i) {
            if (i - (n-1) >= 0) dp[i] += dp[i - (n-1)];
            if (i - (n+1) >= 0) dp[i] += dp[i - (n+1)];
        }
        // two orientations of the global pattern
        cout << dp[m] * 2 << "\n";
        return 0;
    }

    // For n > 2 and even n: two-state DP
    // dp[i][0]: ways ending with fully vertical column at i
    // dp[i][1]: ways ending with a mixed column at i
    vector<array<bigint,2>> dp(m+1);
    dp[0][0] = 1;  // empty prefix counted as vertical
    dp[0][1] = 1;  // and as mixed (dummy)
    for (int i = 1; i <= m; ++i) {
        // To end with vertical column at i, previous must be mixed
        dp[i][0] = dp[i-1][1];
        // To end mixed: we can place a horizontal block of width n
        // or of width n-2, both require previous vertical boundary
        if (i - (n-2) >= 0) dp[i][1] += dp[i - (n-2)][0];
        if (i - n >= 0)         dp[i][1] += dp[i - n][0];
    }
    // total ways is sum of both ending types
    cout << dp[m][0] + dp[m][1] << "\n";
    return 0;
}

4. Python Solution with Detailed Comments

# Uses Python's built-in big integers automatically.
import sys

def main():
    data = sys.stdin.read().strip().split()
    m, n = map(int, data)
    # If area is odd, no tiling possible
    if (m * n) & 1:
        print(0)
        return

    # Work with n <= m
    if n > m:
        n, m = m, n

    # Case n = 1: only one horizontal packing
    if n == 1:
        print(1)
        return

    # Case n = 2: dp[i] = dp[i-1] + dp[i-3]
    if n == 2:
        dp = [0] * (m + 1)
        dp[0], dp[1], dp[2] = 1, 1, 2
        for i in range(3, m + 1):
            dp[i] = dp[i - 1] + dp[i - 3]
        print(dp[m])
        return

    # Case n > 2 and odd: f[i] = f[i-(n-1)] + f[i-(n+1)], answer = 2*f[m]
    if n % 2 == 1:
        dp = [0] * (m + 1)
        dp[0] = 1
        for i in range(1, m + 1):
            if i - (n - 1) >= 0:
                dp[i] += dp[i - (n - 1)]
            if i - (n + 1) >= 0:
                dp[i] += dp[i - (n + 1)]
        print(dp[m] * 2)
        return

    # Case n > 2 and even: two-state DP
    # dp[i][0]: ends with vertical column; dp[i][1]: ends mixed
    dp = [[0, 0] for _ in range(m + 1)]
    dp[0][0] = dp[0][1] = 1
    for i in range(1, m + 1):
        # vertical at i => previous must be mixed
        dp[i][0] = dp[i - 1][1]
        # mixed at i => add a block of width n-2 or n from previous vertical
        if i - (n - 2) >= 0:
            dp[i][1] += dp[i - (n - 2)][0]
        if i - n >= 0:
            dp[i][1] += dp[i - n][0]
    # total ways is sum of both ending states
    print(dp[m][0] + dp[m][1])

if __name__ == "__main__":
    main()

5. Compressed Editorial  

- If m·n is odd ⇒ answer=0.  
- Let n≤m.  
- n=1 ⇒ answer=1.  
- n=2 ⇒ dp[i]=dp[i−1]+dp[i−3], base dp[0]=1,dp[1]=1,dp[2]=2.  
- odd n>2 ⇒ dp[i]=dp[i−(n−1)]+dp[i−(n+1)], answer=2·dp[m].  
- even n>2 ⇒ two states dp[i][0/1]:  
    dp[i][0]=dp[i−1][1],  
    dp[i][1]=dp[i−(n−2)][0]+dp[i−n][0],  
  answer=dp[m][0]+dp[m][1]. Use big integers.