算法——矩阵快速幂求第N个斐波那契数

「爱情、让人受尽委屈。」 2022-06-07 04:35 140阅读 0赞

Fibonacci

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;11123</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;7913</td> 
  </tr> 
 </tbody> 
</table>

Description

In the Fibonacci integer sequence, *F*0 = 0, *F*1 = 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

![3070_1.png][].

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

![3070_2.png][].

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

![3070_3.gif][].

题意：给你一个求解第N个斐波那契数的公式。 让你求出Fn % 10000。

![3070_1.png][]

构造单位矩阵，然后就是矩阵快速幂了。

**快速幂的思路就是：**

**设A为矩阵，求A的N次方，N很大，1000000左右吧。。。**

**先看小一点的，A的9次方**

**A^9**

**= A\*A\*A\*A\*A\*A\*A\*A\*A 【一个一个乘，要乘9次】**

**= A\*(A\*A)\*(A\*A)\*(A\*A)\*(A\*A)【保持格式的上下统一，所以加上这句】**

**= A\*(A^2)^4 【A平方后，再四次方，还要乘上剩下的一个A，要乘6次】**

**= A\*((A^2)^2)^2【A平方后，再平方，再平方，还要乘上剩下的一个A，要乘4次】**

**也算是一种二分思想的应用吧，1000000次幂，暴力要乘1000000次，快速幂就只要（log2底1000000的对数） 次，大约20次**

//
    //  main.cpp
    //  斐波那契进阶
    //
    //  Created by 徐家兴 on 2017/10/9.
    //  Copyright © 2017年 徐家兴. All rights reserved.
    //
    
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #define MAXN 3
    #define LL long long
    #define MOD 10007
    
    using namespace std;
    
    struct Matrix
    {
        LL a[MAXN][MAXN];
        int r, c;//行数 列数
    };
    Matrix ori, res;//初始矩阵 和 结果矩阵
    void init()//初始化矩阵
    {
        memset(res.a, 0, sizeof(res.a));
        res.r = 2; res.c = 2;
        for(int i = 1; i <= res.r; i++)//构造单位矩阵
            res.a[i][i] = 1;
        ori.r = 2; ori.c = 2;
        ori.a[1][1] = ori.a[1][2] = ori.a[2][1] = 1;
        ori.a[2][2] = 0;
    }
    Matrix multi(Matrix x, Matrix y)
    {
        Matrix z;
        memset(z.a, 0, sizeof(z.a));
        z.r = x.r, z.c = y.c;//新矩阵行数等于x矩阵的行数 列数等于y矩阵的列数
        for(int i = 1; i <= x.r; i++)//x矩阵的行数
        {
            for(int k = 1; k <= x.c; k++)//矩阵x的列数等于矩阵y的行数 即x.c = y.r
            {
                if(x.a[i][k] == 0) continue;//优化
                for(int j = 1; j<= y.c; j++)//y矩阵的列数
                    z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;
            }
        }
        return z;
    }
    void Matrix_mod(long long n)
    {
        while(n)//N次幂
        {
            if(n & 1)
                res = multi(ori, res);
            ori = multi(ori, ori);
            n >>= 1;
        }
        printf("%lld\n", res.a[1][2] % MOD);
    }
    int main()
    {
        long long N;
        while(~scanf("%lld", &N))
        {
            init();//初始化单位矩阵
            Matrix_mod(N);//矩阵快速幂
        }
        return 0;
    }

[3070_1.png]: /images/20220607/0dc18c71429443989450ace9b2eb87fe.png
[3070_2.png]: /images/20220607/3dc2c6cf009041aebbd28253b856d610.png
[3070_3.gif]: /images/20220607/489b387ea7174789aaacbd2ef0c8dbd1.png