51nod 1033 Domino Cover V2

Matrix multiplication problem

It's just a matter of pressing the state of a column and then preprocessing the number of transitions for a given two states.

Then construct a matrix A[i][j] representing the number of States of I to the state of J

And then the direct matrix multiplication is possible.

#include <algorithm> #include <iostream> #include <cmath> #include <vector> using namespace std;
const int mod=1000000007;
typedef vector<long Long> VEC;
typedef vector<vec> MAT;
    A*b Mat Mul (mat& A, mat& B) {Mat C (A.size (), VEC (B[0].size ())); for (int i=0;i< (int) a.size (), ++i) {for (int j=0;j< (int) b[0].size (), ++j) {for (int k=0;k< (int) b.size (); ++k
			) {c[i][j]= (c[i][j]+a[i][k]*b[k][j])%mod;
}}} return C;
    }//A^n Mat POW (Mat a,int N) {Mat B (A.size (), VEC (A.size ()));
	for (int i=0;i< (int) a.size (); ++i) {b[i][i]=1;
		} while (n) {if (n&1) {B=mul (b,a); } n>>=1;
    A=mul (A,a);
} return B;
} int n,m;
Long Long dp[1<<5][1<<5];
	void Dfs (int c,int pre,int now) {if (c>n) {return;
        } if (c==n) {dp[pre][now]++;
    Return
    } dfs (C+1,PRE&LT;&LT;1,NOW&LT;&LT;1|1);
    DFS (C+1,PRE&LT;&LT;1|1,NOW&LT;&LT;1); DFS (c+2,pRE&LT;&LT;2,NOW&LT;&LT;2);
    } int main () {cin>>m>>n;
    Mat A (1<<n,vec (1<<n));
    DFS (0,0,0);
        for (int i=0;i< (1<<n), i++) {for (int j=0;j< (1<<n); j + +) {A[i][j]=dp[i][j];
    }} a=pow (a,m+1);
    printf ("%lld\n", a[0][(1<<n)-1]);
return 0; }/* In:2 3 Out:3 */