poj 3070 Fibonacci(矩阵快速幂)

[align=center]Fibonacci[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7303		Accepted: 5186

Description

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:


.

题意：用矩阵乘法来求Fibonacci数列。
分析：矩阵快速幂

AC代码;

#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <iostream>

using namespace std;
typedef struct node
{
int m[2][2];
void init()
{
m[0][0]=m[0][1]=m[1][0]=1;
m[1][1]=0;
}
}matrix;
int n;
matrix matrix_multi(matrix a,matrix b)
{
matrix temp;
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
{
temp.m[i][j]=0;
for(int k=0;k<2;k++)
temp.m[i][j]+=a.m[i][k]*b.m[k][j];
if(temp.m[i][j]>10000)
temp.m[i][j]%=10000;
}
return temp;
}
int main()
{
while(cin>>n)
{
if(n==-1) break;
matrix r,a;
r.m[1][0]=r.m[0][1]=0;  //将r设为单位矩阵（主对角线为1，其余为0，任何矩阵与之相乘得到其本身）
r.m[0][0]=r.m[1][1]=1;
a.init();
while(n)
{
if(n&1)
r=matrix_multi(r,a);
a=matrix_multi(a,a);
n>>=1;
}
cout<<r.m[1][0]<<endl;
}
return 0;
}