计算斐波那契数列

• O(n)复杂度方法

• O(logn)复杂度方法

计算矩阵的n次方，可以先求矩阵的n/2次方，然后再平方即可。

#include <cassert>

struct Matrix2By2
{
    Matrix2By2
    (
        long long m00 = 0,
        long long m01 = 0,
        long long m10 = 0,
        long long m11 = 0
    )
        :m_00(m00), m_01(m01), m_10(m10), m_11(m11)
    {
    }

    long long m_00;
    long long m_01;
    long long m_10;
    long long m_11;
};

Matrix2By2 MatrixMultiply
(
    const Matrix2By2& matrix1,
    const Matrix2By2& matrix2
)
{
    return Matrix2By2(
        matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,
        matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,
        matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,
        matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);
}

Matrix2By2 MatrixPower(unsigned int n)
{
    assert(n > 0);

    Matrix2By2 matrix;
    if (n == 1)
    {
        matrix = Matrix2By2(1, 1, 1, 0);
    }
    else if (n % 2 == 0)
    {
        matrix = MatrixPower(n / 2);
        matrix = MatrixMultiply(matrix, matrix);
    }
    else if (n % 2 == 1)
    {
        matrix = MatrixPower((n - 1) / 2);
        matrix = MatrixMultiply(matrix, matrix);
        matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
    }

    return matrix;
}

long long Fibonacci_Solution3(unsigned int n)
{
    int result[2] = { 0, 1 };
    if (n < 2)
        return result[n];

    Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
    return PowerNMinus2.m_00;
}