1220. 统计元音字母序列的数目

 

 

 

 

 矩阵快速幂

 

 

#include <iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
class Solution {
public:
    typedef long long ll;
    const ll mod = 1000000007;
    struct mat//定义矩阵结构体
    {
        ll m[5][5];
        mat()
        {
            memset(m, 0, sizeof(m));
        }
    };
    mat mul(mat &A, mat &B)
    {
        mat C;
        for (int i = 0; i < 5; i++)
        {
            for (int j = 0; j < 5; j++)
            {
                for (int k = 0; k < 5; k++)
                {
                    C.m[i][j] = (C.m[i][j] + A.m[i][k] * B.m[k][j]) % mod;
                }
            }
        }
        return C;
    }
    mat pow(mat A, ll n)
    {
        mat B;
        for (int i = 0; i < 5; i++)//初始化方阵
            B.m[i][i] = 0;

        //初始被乘矩阵的初值
        B.m[0][0] = 1;
        B.m[1][0] = 1;
        B.m[2][0] = 1;
        B.m[3][0] = 1;
        B.m[4][0] = 1;

        while (n)
        {
            if (n & 1)
                B = mul(A, B);//注意这里，矩阵的左乘和右乘是不一样的，对应的系数矩阵也不一样
            A = mul(A, A);
            n >>= 1;
        }
        return B;
    }
    int countVowelPermutation(int n) {
        mat A;//矩阵A是系数矩阵（转移矩阵）
        A.m[0][1] = 1;
        A.m[1][0] = 1, A.m[1][2] = 1;
        A.m[2][0] = 1, A.m[2][1] = 1, A.m[2][3] = 1, A.m[2][4] = 1;
        A.m[3][2] = 1, A.m[3][4] = 1;
        A.m[4][0] = 1;


        if (n == 1)
        {
            printf("5\n");
        }
        // else if(n==2)
        // {
        //   printf("45\n");
        // }
        else
        {
            mat B = pow(A, n - 1);
            ll ans = B.m[0][0] + B.m[1][0] + B.m[2][0] + B.m[3][0] + B.m[4][0];
            return ans % mod;
        }
    }

};


int main()
{
    Solution A;
    cout << A.countVowelPermutation(2) << endl;
    system("pause");
    return 0;
}