HDU 4990 Reading comprehension 矩阵快速幂

题意：

给出一个序列，
\(f_n=\left\{\begin{matrix} 2f_{n-1}+1, n \, mod \, 2=1\\ 2f_{n-1}, n \, mod \, 2=0 \end{matrix}\right.\)
求\(f_n \, mod \, m\)的值。

分析：

我们可以两个两个的递推，这样就避免了奇偶讨论了。
\(\begin{bmatrix} 0 & 2 & 1 \\ 0 & 4 & 2\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} f_1\\ f_2\\ 1 \end{bmatrix} = \begin{bmatrix} f_3\\ f_4\\ 1 \end{bmatrix}\)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long LL;

LL n, MOD;

LL mul_mod(LL a, LL b) { return a * b % MOD; }

LL add_mod(LL& a, LL b) { a += b; if(a >= MOD) a -= MOD; }

struct Matrix
{
    LL a[3][3];
    Matrix() { memset(a, 0, sizeof(a)); }
    Matrix operator * (const Matrix& t) const {
        Matrix ans;
        for(int i = 0; i < 3; i++)
            for(int j = 0; j < 3; j++)
                for(int k = 0; k < 3; k++)
                    add_mod(ans.a[i][j], mul_mod(a[i][k], t.a[k][j]));
        return ans;
    }
};

Matrix pow_mod(Matrix a, LL n) {
    Matrix ans;
    for(int i = 0; i < 3; i++) ans.a[i][i] = 1;
    while(n) {
        if(n & 1) ans = ans * a;
        a = a * a;
        n >>= 1;
    }
    return ans;
}

int main()
{
    LL a0[3], a[3];
    a0[0] = a0[2] = 1; a0[1] = 2;
    Matrix M0;
    M0.a[0][1] = 2; M0.a[1][1] = 4;
    M0.a[0][2] = 1; M0.a[2][2] = 1;
    M0.a[1][2] = 2;

    while(scanf("%lld%lld", &n, &MOD) == 2) {
        Matrix M;
        for(int i = 0; i < 3; i++)
            for(int j = 0; j < 3; j++)
                M.a[i][j] = M0.a[i][j] % MOD;
        for(int i = 0; i < 3; i++)
            a[i] = a0[i] % MOD;

        M = pow_mod(M, (n - 1) / 2);
        int x = ((n & 1) ^ 1);
        LL ans = 0;
        for(int i = 0; i < 3; i++)
            add_mod(ans, mul_mod(M.a[x][i], a[i]));
        printf("%lld\n", ans);
    }

    return 0;
}