矩阵快速幂--Fibonacci POJ3070

矩阵快速幂–Fibonacci POJ3070

题目连接
Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 17587 Accepted: 12257
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

解题思路
显然用递归定义肯定不行，即使将递归转化成递推优化10^18也会超时，所以用到矩阵来转化
1.显然常规递推时间复杂度和空间复杂度都会超
2.当遇到这种问题时显然是找公式和规律
3.比如f[1][1]=1,f[n][m]=a*f[n-1][m]+b*f[n-1][m-1]当n和m超级大时，显然要利用公式，找出了杨辉三角
4.进而找出了快速幂求模，组合快速幂求模等
5.这道题没有公式这是用矩阵快速幂来解决递推无限大问题

#include<iostream>
#include<memory.h>
#include<stdio.h>
#define N 2
using namespace std;
typedef long long LL;
LL p=1000000009;
LL res[N][N];
void mult(LL res[][N],LL a[][N]){
    LL tem[N][N];
    memset(tem,0,sizeof(tem));
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            for(int k=0;k<N;k++){
                tem[i][j]+=(res[i][k]*a[k][j])%p;
            }
        }
    }
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            res[i][j]=tem[i][j];
        }
    }
    return;
}
void quick_mod(LL a[][N],LL n){
    memset(res,0,sizeof(res));
    for(int i=0;i<N;i++){
        res[i][i]=1;
    }
    while(n){
        if(n&1){
            mult(res,a);
        }
        mult(a,a);
        n>>=1;
    }
    return;
}
int main(){
    LL n;
    LL a[2][2]={{1,1},{1,0}};
    cin>>n;
    quick_mod(a,n-1);
    cout<<res[0][0]%1000000009<<endl;
    return 0;
}