注意是n位数！

思路：

对于给的n都会包含有四种状态

0、7和9的个数都是奇数

1、7是奇数，9是偶数

2、7是偶数，9是奇数

3、7是偶数，9是偶数

显然状态3是我们要状态，但是他们之间是可以互相转移的

所以对于每次添加一个空位放数字，建立转移矩阵

| 3 1 1 0 |

| 1 3 0 1 |

| 1 0 3 1 |

| 0 1 1 3 |

初始状态为 （0，0，0，1）

然后就是n次方了~利用矩阵快速幂

最后ans.mat[0][3]就是答案了-Note that n is the number of bits!

Ideas:

For given n will contain four states

Number 0,7 and 9 are odd

1,7 is an odd number, an even number 9

2,7 is an even number, an odd number 9

3,7 is an even number, an even number 9

Obviously we want to state 3 is a state, but they can be transferred between each other

So for every time you add a space to put numbers to establish transfer matrix

| 3 1 1 0 |

| 1 3 0 1 |

| 1 0 3 1 |

| 0 1 1 3 |

The initial state is (0,0,0,1)

And then the n-th power of matrix fast power ~

Finally ans.mat [0] [3] is the answer