Fibonacci （POJ

In 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.

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 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 Input
0
9
999999999
1000000000
-1
Sample Output
0
34
626
6875
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:

详情见http://blog.csdn.net/coldfresh/article/details/70948818
代码几乎一模一样，没有本质的改动，只是系数矩阵m不同
.
代码：

import java.util.Scanner;public class Main {    public static void main(String[]args)    {        Scanner sc=new Scanner(System.in);        M m=new M();        m.a[0][0]=1;        m.a[0][1]=1;        m.a[1][0]=1;        m.a[1][1]=0;        M o=new M();        o.a[0][0]=1;        o.a[1][1]=1;        for(;;)        {            int n=sc.nextInt();            if(n==-1)                break;            M k=o.copy();            M l=m.copy();            while(n>0)            {                if((n&1)==1)                {                    k=k.muip(l);                }                l=l.muip(l);                n>>=1;            }            System.out.println(k.a[1][0]);        }    }}class M{    long a[][]=new long[2][2];    M muip(M x)    {        M m=new M();        for(int i=0;i<2;i++)            for(int j=0;j<2;j++)            {                m.a[i][j]=(a[i][0]*(x.a[0][j]%10000)+a[i][1]*(x.a[1][j]%10000))%10000;            }        return m;    }    M copy()    {        M m=new M();        for(int i=0;i<2;i++)            for(int j=0;j<2;j++)            {                m.a[i][j]=a[i][j];            }        return m;    }}