HDU 4990 Reading comprehension

题目链接：HDU 4990 Reading comprehension

题目大意：
\(f_0 = 0\)，\(n\)为奇数时：\(f_n = f_{n-1} \times 2 + 1\)，\(n\)为偶数时：\(f_n = f_{n-1} \times 2\)。
已知\(n\)和\(m\)，求\(f_n\%m。\)

题解：
假设\(n\)为奇数，那么\(f_n=f_{n-1}\times 2+1\)，而\(n-1\)必定是偶数，所以有：\(f_n=f_{n-1}+f_{n-2}\times 2+1\)；
同理，假设\(n\)为偶数，那么\(f_n=f_{n-1}\times 2\)，而\(n-1\)必定是奇数，所以有：\(f_n=f_{n-1}+f_{n-2}\times 2+1\)。
那么，\(f_n=f_{n-1}+f_{n-2}\times 2+1\)对于任意\(n(n>0)\)恒成立。
构造矩阵：

\[\left(\begin{matrix} f_i \\ f_{i-1} \\ 1 \end{matrix}\right) = \left(\begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right) \times \left(\begin{matrix} f_{i-1} \\ f_{i-2} \\ 1 \end{matrix}\right) \]

所以：

\[\left(\begin{matrix} f_n \\ f_{n-1} \\ 1 \end{matrix}\right) = \left(\begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right)^{n-1} \times \left(\begin{matrix} f_1 \\ f_0 \\ 1 \end{matrix}\right) \]

由\(f_1 = 1,f_0 = 0\)得：

\[\left(\begin{matrix} f_n \\ f_{n-1} \\ 1 \end{matrix}\right) = \left(\begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right)^{n-1} \times \left(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}\right) \]

#include <iostream>
#include <cstring>
using namespace std;

struct Matrix { // 矩阵
    int row, col;
    long long num[3][3];
};
long long n, m;

Matrix multiply(Matrix a, Matrix b) { // 矩阵乘法
    Matrix temp;
    temp.row = a.row, temp.col = b.col;
    memset(temp.num, 0, sizeof(temp.num));
    for (int i = 0; i < a.row; ++i)
        for (int j = 0; j < b.col; ++j)
            for (int k = 0; k < a.col; ++k)
                temp.num[i][j] = (temp.num[i][j] + a.num[i][k] * b.num[k][j] % m) % m;
    return temp;
}

Matrix fastPow(Matrix base, long long k) { // 矩阵快速幂
    Matrix ans;
    ans.row = ans.col = 3;
    memset(ans.num, 0, sizeof(ans.num));
    ans.num[0][0] = ans.num[1][1] = ans.num[2][2] = 1;
    while (k) {
        if (k & 1) {
            ans = multiply(ans, base);
        }
        base = multiply(base, base);
        k >>= 1;
    }
    return ans;
}

int main() {
    Matrix base;
    base.row = base.col = 3;
    base.num[0][0] = base.num[0][2] = base.num[1][0] = base.num[2][2] = 1;
    base.num[1][1] = base.num[1][2] = base.num[2][0] = base.num[2][1] = 0;
    base.num[0][1] = 2;
    while (cin >> n >> m) {
        Matrix ans = fastPow(base, n - 1);
        cout << (ans.num[0][0] + ans.num[0][2]) % m << endl;
    }
    return 0;
}