题目链接

给一个长度为n的字符串， 每个字符可以使f或m。 问你不包含子串fmf以及fff的字符串数量有多少。

令0表示mm结尾， 1表示mf， 2表示ff， 3表示fm。

那么

f(n+1, 0) = f(n, 0) + f(n, 3)

f(n+1, 1) = f(n, 0)

f(n+1, 2) = f(n, 1)

f(n+1, 3) = f(n, 1) + f(n, 2)

所以构造出矩阵

{1, 0, 0, 1}

{1, 0, 0, 0}

{0, 1, 0, 0}

{0, 1, 1, 0}

然后快速幂， 最后的答案就是矩阵第一列的值相加。

#include <iostream>

#include <vector>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <complex>

#include <cmath>

#include <map>

#include <set>

#include <string>

#include <queue>

#include <stack>

#include <bitset>

using namespace std;

#define pb(x) push_back(x)

#define ll long long

#define mk(x, y) make_pair(x, y)

#define lson l, m, rt<<1

#define mem(a) memset(a, 0, sizeof(a))

#define rson m+1, r, rt<<1|1

#define mem1(a) memset(a, -1, sizeof(a))

#define mem2(a) memset(a, 0x3f, sizeof(a))

#define rep(i, n, a) for(int i = a; i<n; i++)

#define fi first

#define se second

typedef complex <double> cmx;

typedef pair<int, int> pll;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int inf = 1061109567;

const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };

int a[4][4] = {

    {1, 0, 0, 1},

    {1, 0, 0, 0},

    {0, 1, 0, 0},

    {0, 1, 1, 0},

};

int mod;

struct Matrix

{

    int f[4][4];

    Matrix() {

        mem(f);

    }

}m;

Matrix operator * (Matrix a, Matrix b) {

    Matrix c;

    for(int k = 0; k<4; k++) {

        for(int j = 0; j<4; j++) {

            for(int i = 0; i<4; i++) {

                c.f[i][j] += a.f[i][k]*b.f[k][j];

                c.f[i][j] %= mod;

            }

        }

    }

    return c;

}

Matrix operator ^(Matrix a, ll b) {

    Matrix tmp;

    for(int i = 0; i<4; i++)

        tmp.f[i][i] = 1;

    while(b) {

        if(b&1)

            tmp = tmp*a;

        a = a*a;

        b>>=1;

    }

    return tmp;

}

int main()

{

    int n;

    while(cin>>n>>mod) {

        memcpy(m.f, a, sizeof(a));

        m = m^n;

        int ans = 0;

        for(int i = 0; i < 4; i++) {

            ans = (ans + m.f[i][0])%mod;

        }

        cout<<ans<<endl;

    }

    return 0;

}