【算法学习】POJ3070——利用分治法来计算Fibonacci数列的值

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

我的解决办法是利用了类似于求x^n的分治法来加快矩阵运算，使得计算Fibonacci数列第n项的时间复杂度变成了O（logn），比普通算法的O（n）更快

这题有一个比较坑的点就是每一步矩阵运算都要进行一次mod1000

代码如下：

#include <iostream>using namespace std;struct matrix {long x00;long x01;long x10;long x11;
};matrix matrix_mul ( matrix a, matrix b ) {matrix c = {0,0,0,0};c.x00 = (a.x00*b.x00 + a.x01*b.x10)%10000;c.x01 = (a.x00*b.x01 + a.x01*b.x11)%10000;c.x10 = (a.x10*b.x00 + a.x11*b.x10)%10000;c.x11 = (a.x10*b.x01 + a.x11*b.x11)%10000;return c;
}matrix matrix_pow ( matrix a, long n ) {if ( n == 1 ) {return a;}else if ( n%2 == 0 ) {matrix temp = matrix_pow(a,n/2);return matrix_mul(temp,temp);}else {matrix temp = matrix_pow(a,(n-1)/2);matrix temp1 = matrix_mul(temp,temp);return matrix_mul(temp1,a);}}long fibonacci_value ( long n ) {if( n == 0 ) {return 0;}else {matrix root = {1,1,1,0};root = matrix_pow(root,n);return root.x01 % 10000;}
}int main()
{long input[100] = {0};int num = 0;long temp;while (true) {cin >> temp;if( temp != -1 ) {input[num++] = temp;}else {break;}}for( int i = 0; i < num; ++i ) {cout << fibonacci_value(input[i]) << endl;}return 0;
}