logn的斐波那契

This another O(n) which relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} to itself (in other words calculate power(M, n )), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix.

The matrix representation gives the following closed expression for the Fibonacci numbers:
\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}. 

 1 #include <stdio.h>
 2 
 3 void multiply(int F[2][2], int M[2][2]);
 4 
 5 void power(int F[2][2], int n);
 6 
 7 /* function that returns nth Fibonacci number */
 8 int fib(int n)
 9 {
10   int F[2][2] = {{1,1},{1,0}};
11   if (n == 0)
12     return 0;
13   power(F, n-1);
14   return F[0][0];
15 }
16 
17 /* Optimized version of power() in method 4 */
18 void power(int F[2][2], int n)
19 {
20   if( n == 0 || n == 1)
21       return;
22   int M[2][2] = {{1,1},{1,0}};
23 
24   power(F, n/2);
25   multiply(F, F);
26 
27   if (n%2 != 0)
28      multiply(F, M);
29 }
30 
31 void multiply(int F[2][2], int M[2][2])
32 {
33   int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
34   int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
35   int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
36   int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];
37 
38   F[0][0] = x;
39   F[0][1] = y;
40   F[1][0] = z;
41   F[1][1] = w;
42 }
43 
44 /* Driver program to test above function */
45 int main()
46 {
47   int n = 9;
48   printf("%d", fib(9));
49   getchar();
50   return 0;
51 }

 

posted on 2014-08-15 15:13  雨歌_sky  阅读(227)  评论(0)  编辑  收藏  举报

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