斐波那契数列问题的解

解法一：递归

1 int Fib(int n)

2 {

3     if (n <= 0)

4         return 0;

5     else if (n == 1)

6         return 1;

7     else

8         return Fib(n - 1) + Fib(n - 2);

9 }

解法二：递推

 1 int Fib(int n)

 2 {

 3     if (n <= 0)

 4         return 0;

 5     else if (n == 1)

 6         return 1;

 7 

 8     int first = 0, second = 1;

 9     int res = 0;

10     for (int i = 1; i < n; ++i) {

11         res = first + second;

12         first = second;

13         second = res;

14     }

15     return res;

16 }

解法三：分治

通项之间有如下关系：

其中矩阵A为：

根据以下公式，可以log(n)次乘法计算出Aⁿ。

 1 class Matrix {

 2 public:

 3     Matrix() : m00(0), m01(0), m10(0), m11(0) {}

 4     Matrix(int m00, int m01, int m10, int m11) : 

 5             m00(m00), m01(m01), m10(m10), m11(m11) {}

 6     Matrix& operator=(const Matrix& m) {

 7         this->m00 = m.m00;

 8         this->m01 = m.m01;

 9         this->m10 = m.m10;

10         this->m11 = m.m11;

11         return *this;

12     }

13 

14     int m00;

15     int m01;

16     int m10;

17     int m11;

18 };

19 Matrix operator*(const Matrix& lhs, const Matrix& rhs)

20 {

21     Matrix res;

22     res.m00 = lhs.m00 * rhs.m00 + lhs.m01 * rhs.m10;

23     res.m01 = lhs.m00 * rhs.m01 + lhs.m01 * rhs.m11;

24     res.m10 = lhs.m10 * rhs.m00 + lhs.m11 * rhs.m10;

25     res.m11 = lhs.m10 * rhs.m01 + lhs.m11 * rhs.m11;

26     return res;

27 }

28 

29 Matrix MatrixPow(const Matrix& m, int n)

30 {

31     Matrix res(1, 0, 0, 1);

32     Matrix base = m;

33     for (; n; n >>= 1) {

34         if (n & 1)

35             res = res * base;

36         base = base * base;

37     }

38     return res;

39 }

40 

41 int Fib(int n)

42 {

43     if (n <= 0)

44         return 0;

45     else if (n == 1)

46         return 1;

47 

48     Matrix A(1, 1, 1, 0);

49     Matrix m = MatrixPow(A, n - 1);

50 

51     return m.m00;

52 }

 函数MatrixPow的写法似乎不像分治，实际上我们把它写成下面这样就比较明显了，时间复杂度O(logn)。

1 Matrix MatrixPow(const Matrix& m, int n)

2 {

3     if (n == 1)

4         return m;

5     if (n % 2 == 0)

6         return MatrixPow(m * m, n / 2);

7     else

8         return MatrixPow(m * m, n / 2) * m;

9 }