Domino Cover Problem • Two

Time limit: 10000ms single point time limit: 1000ms memory limit: 256MB description

Last week we studied the 2xN Domino problem, this week we might as well increase the difficulty, study the 3xN Domino problem?
So our question is: for the 3xN chessboard, how many different coverage methods are used to cover a 1x2 Domino?
First of all we can be sure that the odd length must not be covered, for even lengths, such as 2, 4, we have the following types of coverage:

Input

Line 1th: an integer n. Represents the board length. 1≤n≤100,000,000

Output

Line 1th: An integer representing the number of coverage scenarios MOD 12357

Sample input

62247088

Sample output

4037
Tips:

First, when n is odd, there is no solution, and the output is 0;

When n is even, from a small view:

There are 3 kinds of n=2. Each additional two columns can be considered as a situation:

1. These two columns are not connected with the previous one, and are placed separately, with 3 kinds, namely 3*f (n-2);

2. The two columns are connected to the middle of the previous two columns, that is, the midline is horizontally placed (n-2 and n-1 columns), there is this, you can also put the bottom two transverse to the top, so there are two cases, the latter 4 columns as a whole, namely 2*f (N-4);

And there may be n-3 and n-4 columns, that is, the post 6 columns as a whole, there are two kinds, namely, 2*f (n-6), recursion.

The end of the recursion is F (0) =1;f (2) = 3;

#include <iostream>using namespace Std;typedef unsigned long long ll;const ll MOD = 12357;ll n;ll a[5];void Solve () {    a[0] = 0;    A[1] = 2;    A[2] = 3;    for (int i = 3; I <= N; ++i) {        cout<< (i&1) <<endl;        if (i&1) {                                                        //i is odd            a[i%5] = (2*a[(i-1+5)%5] + a[(i-2+5)%5])% MOD;        } else {                                                            //i is even            a[i%5] = (3*a[( I-2+5)%5] + a[(i-3+5)%5])% MOD;        }    }    cout << a[n%5] << Endl;} int main () {    while (CIN >> N) {        if (N & 1) {            cout << "0" << Endl;        } else {            Solve ();        }    }    return 0;}

Note: The code goes

Domino Cover Problem • Two