POJ - 3070 Fibonacci 矩阵快速幂取模

最新推荐文章于 2020-03-23 16:35:12 发布

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最新推荐文章于 2020-03-23 16:35:12 发布

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原文链接：http://www.cnblogs.com/kunsoft/p/5312748.html

本文介绍了一种使用矩阵快速幂算法高效计算斐波那契数列的方法，并通过一个示例题目展示了如何实现该算法。具体地，文章给出了完整的C++代码实现，能够快速计算出指定项数的斐波那契数并返回最后四位数字。

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9776		Accepted: 6964

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:

刚刚学习了快速幂和矩阵快速幂，然后就找了个例题，根据题目的一个快速计算斐波那契数列的方法在利用矩阵快速幂模板，0ms水过

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;

const int MAXN = 2;
struct Mat
{
    int a[MAXN][MAXN];
} ;

Mat Multi(const Mat x, const Mat y, const int Mod)
{
    Mat res;
    memset(res.a, 0, sizeof(res.a));
    for(int i = 0; i < MAXN; i++)
        for(int j = 0; j < MAXN; j++)
            for(int k = 0; k < MAXN; k++)
                res.a[i][j] = (res.a[i][j] + x.a[i][k] * y.a[k][j]) % Mod;
    return res;
}

Mat Power(Mat base, int k, const int Mod)
{
    Mat res;
    memset(res.a, 0, sizeof(res.a));
    for(int i = 0; i < MAXN; i++)
        res.a[i][i] = 1;
    while(k > 0) {
        if(k & 1)
            res = Multi(res, base, Mod);
        base = Multi(base, base, Mod);
        k = k >> 1;
    }
    return res;
}

int main()
{
    int mod = 10000;
    int n;
    Mat base;
    base.a[0][0] = base.a[1][0] = base.a[0][1] = 1;
    base.a[1][1] = 0;
    while(cin >> n, n!=-1)
    {
        Mat res = Power(base, n, mod);
        printf("%d\n", res.a[0][1]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/kunsoft/p/5312748.html