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矩阵快速幂求斐波那契数列

news 来源：原创 2024/4/29 3:31:05

恕我愚昧，今天才知道斐波拉契还可以这样解
源于上次快手的笔试题，本来以为很简单的斐波拉契结果总是时间超时
矩阵快速幂精髓：比如A^19 => （A^16）*（A^2）*（A^1），显然采取这样的方式计算时因子数将是log(n)级别的(原来的因子数是n)，不仅这样，因子间也是存在某种联系的，比如A^4能通过(A^2)*(A^2)得到，A^8又能通过(A^4)*(A^4)得到，这点也充分利用了现有的结果作为有利条件。下面举个例子进行说明：现在要求A^156,而156(10)=10011100(2) 也就有A^156=>(A^4)*(A^8)*(A^16)*(A^128) 考虑到因子间的联系，我们从二进制10011100中的最右端开始计算到最左端。

while(N)
 {
          if(N&1)
                 res=res*A;
          N>>=1;
          A=A*A;
 }

例题：

//计算(x^y) % N; 注：(x^y)表示x的y次方
#if 0

int main()
{
    int x, y, N;

    cin >> x >> y >> N;

    long res = 1;
    x = x % N; //开始

    while (y > 0) {
        if (y % 2 == 1) //等价于 if(y&1)
            res = (res * x) % N;
        y /= 2; //y>>1; 分解y为二进制编码
        x = (x * x) % N;
    }

    cout << res << endl;

    return 0;
}

#endif

对于矩阵乘法与递推式之间的关系：

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_n_{− 1} + F_n_{− 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, …

如：在斐波那契数列之中

f[i] = 1*f[i-1]+1*f[i-2] f[i-1] = 1*f[i-1] + 0*f[i-2];即

所以：

POJ-3070Fibonacci（矩阵快速幂求Fibonacci数列）

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7241		Accepted: 5131

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_n_{− 1} + F_n_{− 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 F_n.

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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

Sample Output

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

#if 1

#include <cstdio>
#include <iostream>

using namespace std;

const int MOD = 10000;

struct matrix
{
    int m[2][2];
}ans, base;

matrix multi(matrix a, matrix b)
{
    matrix tmp;
    for(int i = 0; i < 2; ++i)
    {
        for(int j = 0; j < 2; ++j)
        {
            tmp.m[i][j] = 0;
            for(int k = 0; k < 2; ++k)
                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
        }
    }
    return tmp;
}
int fast_mod(int n)  // 求矩阵 base 的  n 次幂 
{
    base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;
    base.m[1][1] = 0;
    ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵 
    ans.m[0][1] = ans.m[1][0] = 0;
    while (n)
    {
        if (n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t 
        {
            ans = multi(ans, base);
        }
        base = multi(base, base);
        n >>= 1;
    }
    return ans.m[0][1];
}

int main()
{
    int n;
    while (scanf("%d", &n) && n != -1)
    {
        printf("%d\n", fast_mod(n));
    }
    return 0;
}

#endif