【原】 POJ 3070 Fibonacci 斐波那契数列 解题报告

http://poj.org/problem?id=3070

方法：
divide and conquer，类似于求幂
矩阵求幂
复杂度：O(logn)

| F(n+1) F(n)   | = | 1  1 |^n
| F(n)   F(n-1) |   | 1  0 |

注意：
矩阵的0次幂是单位矩阵

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

0

9

999999999

1000000000

-1

Sample Output

0

34

626

6875

#include <stdio.h>

struct Matrix
{
    Matrix( int a=1, int b=1, int c=1, int d=0 ):a11(a),a12(b),a21(c),a22(d){}
    void set( int a=1, int b=1, int c=1, int d=0 ){a11=a;a12=b;a21=c;a22=d;}
    int a11; int a12; int a21; int a22;
};

Matrix MatrixMultiply( Matrix& a, Matrix& b )
{
    Matrix result ;
    result.a11 = ( a.a11*b.a11 + a.a12*b.a21 ) % 10000 ;
    result.a12 = ( a.a11*b.a12 + a.a12*b.a22 ) % 10000 ;
    result.a21 = ( a.a21*b.a11 + a.a22*b.a21 ) % 10000 ;
    result.a22 = ( a.a21*b.a12 + a.a22*b.a22 ) % 10000 ;

    return result ;
}

Matrix Fibonacci( Matrix& a, int n )
{
    Matrix result ;
    if( n==0 )
    {
        result.set(1,0,0,1) ;
        return result ;
    }
    if( n==1 )
        return result ;

    Matrix tmp ;
    tmp = Fibonacci( a, n/2 ) ;
    result = MatrixMultiply( tmp, tmp ) ;
    if( n%2==0 )
        return result;
    else
        return MatrixMultiply(a,result) ;
}

void run3070()
{
    Matrix a,result ;
    int n ;

    while( scanf( "%d", &n ) && n!=-1 )
    {
        result = Fibonacci( a, n ) ;
        printf( "%d\n", result.a12 ) ;
    }
}