【算法学习】POJ3070——利用分治法来计算Fibonacci数列的值

DescriptionIn the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.
InputThe 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.
OutputFor each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input0
9
999999999
1000000000
-1Sample Output0
34
626
6875HintAs 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;
}