斐波那契数列poj（矩阵快速幂）

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18040		Accepted: 12541

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:

.

Source

Stanford Local 2006

 

       以前写斐波那契数列用递推公式写，对以前的n不大，但碰见n比较大的时候就会超时，比较尴尬，这个时候矩阵快速幂来了。

1. 首先构造f1,f0矩阵（此题比较特殊，不用构造）
2. 然后类似于1构造一个单位矩阵（单位矩阵就是那么一个矩阵 这个矩阵乘以任何一个矩阵都等于他乘的那个矩阵）sec
3. 如下图，构造矩阵fir
4. 然后判断n如果n<2直接输出,否则fir的（n-1)次方，在这个过程中注意取余操作；然后再乘以初始矩阵（此题比较特殊，你会发现直接输出乘方后的pos[0][0]就行了。
5. 第一次写这个东西，多多指教，代码很烂。

 

 

 

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
struct mat
{
    int pos[2][2];
    mat()
    {
        memset(pos,0,sizeof(pos));
    }
};
mat fir,sec;
mat Cheng(mat a,mat b)
{
    mat c;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
            c.pos[i][j]=(c.pos[i][j]+a.pos[i][k]*b.pos[k][j])%10000;
    return c;
}
void mul(mat b,int n)
{
    while(n>0)
    {
        if(n%2==1)
            sec=Cheng(sec,b);
        n=n/2;
        b=Cheng(b,b);
    }
}
int main()
{
    int n;
    while(scanf("%d",&n)&&n>=0)
    {
    if(n==0)
        cout<<0<<endl;
    else if(n==1)
        cout<<1<<endl;
    else
    {
        sec.pos[1][1]=1,sec.pos[0][0]=1,sec.pos[0][1]=0,sec.pos[1][0]=0;
        fir.pos[0][0]=1,fir.pos[0][1]=1,fir.pos[1][0]=1,fir.pos[1][1]=0;
        mul(fir,n-1);
        cout<<sec.pos[0][0]<<endl;
    }
    }
    return 0;
}