乘法分配率

 A^(N*N) * B^(N*N) = A*B*A*B*A*B*A··· = A*(B*A)*(B*A)···

然后里面的结果就是6*6的格子，然后快速幂一下。

#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

const int MAX_N = 1007;
const int MAX_K = 7;
const int MOD = 6;

mat mul(mat &A, mat &B) {
    mat C(A.size(), vec(B[0].size()));
    for (int i = 0; i < A.size(); ++i) {
        for (int k = 0; k < B.size(); ++k) {
            for (int j = 0; j < B[0].size(); ++j) {
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
            }
        }
    }
    return C;
}

mat Pow(mat A, ll n) {
    mat B(A.size(), vec(A.size()));
    for (int i = 0; i < A.size(); ++i) {
        B[i][i] = 1;
    }
    while (n > 0) {
        if (n & 1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}

inline bool rd(ll &n){  
    ll x = 0, tmp = 1;  
    char c = getchar();  
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();  
    if(c == EOF) return false;  
    if(c == '-') c = getchar(), tmp = -1;  
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();  
    n = x*tmp;  
    return true;
} 


int main() {
    int n, k;
    while (2 == scanf("%d%d", &n, &k)) {
        if (0 == n && k == 0) break;
        mat A(n, vec(k)), B(k, vec(n));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < k; ++j) {
                rd(A[i][j]);
            }
        }
        for (int i = 0; i < k; ++i) {
            for (int j = 0; j < n; ++j) {
                rd(B[i][j]);
            }
        }
        long long nn = n * n;
        if (nn == 1) {
            int ans = 0;
            for (int i = 0; i < k; ++i) {
                ans = (ans + A[0][i] * B[i][0]) % MOD;
            }
            printf("%d\n", ans);
        } else {
            mat C(k, vec(k));
            for (int i = 0; i < k; ++i) {
                for (int j = 0; j < k; ++j) {
                    for (int l = 0; l < n; ++l) {
                        C[i][j] = C[i][j] + B[i][l] * A[l][j];
                    }
                }
            }
            C = Pow(C, nn - 1);
            mat D(n, vec(n));
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < k; ++j) {
                    for (int l = 0; l < k; ++l) {
                        D[i][j] = (D[i][j] + A[i][l] * C[l][j]) % MOD;
                    }
                }
            }
            mat E(n, vec(n));
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    for (int l = 0; l < k; ++l) {
                        E[i][j] = (E[i][j] + D[i][l] * B[l][j]) % MOD;
                    }
                }
            }
            long long sum = 0;
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    sum += E[i][j];
                }
            }
            printf("%I64d\n", sum);
        }
    }
    return 0;
}