Fibonacci （POJ - 3070 ）（矩阵快速幂）

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;
}
}