https://blog.csdn.net/zhangxiaoduoduo/article/details/81807338
这个的矩阵快速幂讲的蛮好的

这题简单的说就是你要在一个骰子上垒n个骰子，
就要在{4，4，4，4，4，4}(最底下的骰子每个面朝上有四种可能)
上乘n个矩阵A(A[i][j]表示如果下面骰子i朝上，上面骰子j朝上有多少种可能)

因为数据太大需要用到快速幂，模板如下

long long fastPower(long long base, long long power) {
    long long result = 1;
    while (power > 0) {
        if (power & 1) {
            result = result * base % mod;
        }
        power >>= 1;
        base = (base * base) % mod;
    }
    return result;
}

将底数换成矩阵就行了

import java.util.*;
import java.io.*;

public class Main {
    static class Input {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        public int nextInt() throws IOException {
            in.nextToken();
            return (int) in.nval;
        }
    }

    public static void main(String[] args) throws IOException {
        Input in = new Input();
        int n = in.nextInt() - 1,m = in.nextInt(),mod = 1000000007;
        long[][] A = new long[7][7];
        for (int i = 1; i <= 6; i++) Arrays.fill(A[i],4);
        for (int i = 0; i < m; i++) {
            int a = in.nextInt(),b = in.nextInt();
            A[a][(b + 2) % 6 + 1] = 0;
            A[b][(a + 2) % 6 + 1] = 0;
        }
        long[][] Res = new long[7][7];
        for (int i = 1; i <= 6; i++) Res[i][i] = 1;
        long[][] T;
        while (n > 0) { // 矩阵快速幂
            T = new long[7][7];
            if(n % 2 == 1) {
                for (int i = 1; i <= 6; i++) {
                    for (int j = 1; j <= 6; j++) {
                        for (int k = 1; k <= 6; k++) {
                            T[i][j] = (T[i][j] + Res[i][k] * A[k][j]) % mod;
                        }
                    }
                }
                Res = T;
            }
            n >>= 1;
            T = new long[7][7];
            for (int i = 1; i <= 6; i++) {
                for (int j = 1; j <= 6; j++) {
                    for (int k = 1; k <= 6; k++) {
                        T[i][j] = (T[i][j] + A[i][k] * A[k][j]) % mod;
                    }
                }
            }
            A = T;
        }
        long result = 0;
        for (int i = 1; i <= 6; i++) {
            for (int j = 1; j <= 6; j++) {
                result = (result + Res[i][j] * 4) % mod;
            }
        }
        System.out.println(result);
    }
}