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。

 

 1 #include <cassert>
 2 
 3 struct Matrix2By2
 4 {
 5     Matrix2By2
 6     (
 7         long long m00 = 0,
 8         long long m01 = 0,
 9         long long m10 = 0,
10         long long m11 = 0
11     )
12         :m_00(m00), m_01(m01), m_10(m10), m_11(m11)
13     {
14     }
15 
16     long long m_00;
17     long long m_01;
18     long long m_10;
19     long long m_11;
20 };
21 
22 Matrix2By2 MatrixMultiply
23 (
24     const Matrix2By2& matrix1,
25     const Matrix2By2& matrix2
26 )
27 {
28     return Matrix2By2(
29         matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,
30         matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,
31         matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,
32         matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);
33 }
34 
35 Matrix2By2 MatrixPower(unsigned int n)
36 {
37     assert(n > 0);
38 
39     Matrix2By2 matrix;
40     if (n == 1)
41     {
42         matrix = Matrix2By2(1, 1, 1, 0);
43     }
44     else if (n % 2 == 0)
45     {
46         matrix = MatrixPower(n / 2);
47         matrix = MatrixMultiply(matrix, matrix);
48     }
49     else if (n % 2 == 1)
50     {
51         matrix = MatrixPower((n - 1) / 2);
52         matrix = MatrixMultiply(matrix, matrix);
53         matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
54     }
55 
56     return matrix;
57 }
58 
59 long long Fibonacci_Solution3(unsigned int n)
60 {
61     int result[2] = { 0, 1 };
62     if (n < 2)
63         return result[n];
64 
65     Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
66     return PowerNMinus2.m_00;
67 }

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标签：00,数列,long,斐波,Matrix2By2,那契,matrix1,matrix2,matrix
来源： https://www.cnblogs.com/larry-xia/p/10659613.html