hdu 2604Queuing dp+ 矩阵快速幂

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;
}