【算法学习】POJ3070——利用分治法来计算Fibonacci数列的值

Description

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.

Input

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.

Output

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).

Sample Input

099999999991000000000-1

Sample Output

0346266875

Hint

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（lgn），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;}