题目：写一个函数，输入n求斐波那契数列的第n项。

暴力简单解法：

1. long long Fibonacci(unsigned int n)
2. {
3. if(n<=0)
4. return 0;
5. if(n==1)
6. return 1;
8. return Fibonacci(n-1)+Fabonacci(n-2);
9. }

利用循环改进之后，这个的时间复杂度是o（n)

1. // ====================方法2：循环====================
2. long long Fibonacci_Solution2(unsigned n)
3. {
4. int result[2] = {
   0, 1};
5. if(n < 2)
6. return result[n];
8. long long fibNMinusOne = 1;
9. long long fibNMinusTwo = 0;
10. long long fibN = 0;
11. for(unsigned int i = 2; i <= n; ++ i)
12. {
13. fibN = fibNMinusOne + fibNMinusTwo;
15. fibNMinusTwo = fibNMinusOne;
16. fibNMinusOne = fibN;
17. }
19. return fibN;
20. }

第三种方法：利用如下的一个公式

1. // ====================方法3：基于矩阵乘法====================
2. #include <cassert>
4. struct Matrix2By2
5. {
6. Matrix2By2
7. (
8. long long m00 = 0,
9. long long m01 = 0,
10. long long m10 = 0,
11. long long m11 = 0
12. )
13. :m_00(m00), m_01(m01), m_10(m10), m_11(m11)
14. {
15. }
17. long long m_00;
18. long long m_01;
19. long long m_10;
20. long long m_11;
21. };
23. Matrix2By2 MatrixMultiply
24. (
25. const Matrix2By2& matrix1,
26. const Matrix2By2& matrix2
27. )
28. {
29. return Matrix2By2(
30. matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,
31. matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,
32. matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,
33. matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);
34. }
36. Matrix2By2 MatrixPower(unsigned int n)
37. {
38. assert(n > 0);
40. Matrix2By2 matrix;
41. if(n == 1)
42. {
43. matrix = Matrix2By2(1, 1, 1, 0);
44. }
45. else if(n % 2 == 0)
46. {
47. matrix = MatrixPower(n / 2);
48. matrix = MatrixMultiply(matrix, matrix);
49. }
50. else if(n % 2 == 1)
51. {
52. matrix = MatrixPower((n - 1) / 2);
53. matrix = MatrixMultiply(matrix, matrix);
54. matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
55. }
57. return matrix;
58. }
60. long long Fibonacci_Solution3(unsigned int n)
61. {
62. int result[2] = {
    0, 1};
63. if(n < 2)
64. return result[n];
66. Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
67. return PowerNMinus2.m_00;
68. }