POJ 3070 Fibonacci(矩阵快速幂)

最新推荐文章于 2021-07-08 09:30:58 发布

原创最新推荐文章于 2021-07-08 09:30:58 发布 · 345 阅读

0 ·

CC 4.0 BY-SA版权

数学专栏收录该内容

14 篇文章

订阅专栏

本文介绍了一种利用矩阵快速幂方法高效计算斐波那契数列第n项，并对其结果进行10000取模的算法实现。通过这种方式可以有效地解决大整数运算的问题，适用于竞赛编程等场景。

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13147		Accepted: 9348

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:

题意：求解斐波那契数列第n项对1e4取模的结果。

思路：运用矩阵快速幂即可

<pre name="code" class="cpp">#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<cctype>
using namespace std;
typedef long long ll;
const int M = 1e4;

struct Mar
{
    int a[2][2];
    Mar(int ha = 0) {
       memset(a, 0, sizeof a);
    }
    void init() {
       a[0][0] = a[0][1] = a[1][0] = 1;
       a[1][1] = 0;
    }
    void dan() { //矩阵单位化
       memset(a, 0, sizeof a);
       a[0][0] = a[1][1] = 1;
    }
    Mar& operator = (const Mar &Rhs);
    friend Mar operator * (const Mar &A, const Mar &B);
};

Mar& Mar::operator = (const Mar &Rhs)
{
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++)
        a[i][j] = Rhs.a[i][j];
    return *this;
}

Mar operator * (const Mar &A, const Mar &B)
{
    Mar C;
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++)
          C.a[i][j] = (C.a[i][j] + A.a[i][k] * B.a[k][j] % M) % M;
    return C;
}

Mar my_pow(Mar& A, ll n)
{
    Mar Ans;
    if (n == 0) {
        Ans.dan();
        return Ans;
    }
    Ans = my_pow(A, n/2);
    Ans = Ans * Ans;
    if (n % 2) Ans = Ans * A;
    return Ans;
}

void solve(ll n)
{
    Mar A;
    A.init();
    Mar Ans = my_pow(A,n);
    cout << Ans.a[1][0] << endl;
}

int main()
{
    ll n;
    while(cin >> n && n+1) {
        solve(n);
    }
}