[原创]POJ 3070 Fibonacci 【矩阵快速幂】

2016-07-17 09:37:27 Tabris_ 阅读数：261

博客爬取于2020-06-14 22:44:13
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/51930813

题目连接 ： http://poj.org/problem?id=3070

Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12738 Accepted: 9065
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

.这里写图片描述

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
这里写图片描述
.

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

.这里写图片描述

题目大意 ： 这个的大意应该不用说了吧

题解 ： 这个题目大意里已经说得很清楚了吧 。。

就是裸的矩阵快速幂 没什么难度 主要是用来熟悉下 矩阵快速么的代码实现 并顺手撸了一发模板

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const int M = 2;
const int MOD = 1e4;

struct Matrix
{
    LL m[M][M];
};

Matrix operator * (Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0; i<M; i++) //初始化矩阵
        for(int j=0; j<M; j++)
            c.m[i][j]= 0;

    for(int k=0; k<M; k++)
        for(int i=0; i<M; i++) //实现矩阵乘法
        {
            if(a.m[i][k] <= 0)  continue;
            for(int j=0; j<M; j++)
            {
                if(b.m[k][j] <= 0)    continue;
                c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j]+MOD)%MOD;
            }
        }
    return c;
}

Matrix operator ^ (Matrix a,LL b)
{
    Matrix c;
    for(int i=0; i<M; i++) //初始化单位矩阵
        for(int j=0; j<M; j++)
            c.m[i][j]= ( i == j );

    while(b)
    {
        if(b&1) c= c * a ;
        b >>= 1;
        a = a * a ;
    }

    return c;
}

直接附本题代码吧

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# include <iostream>
# include <stdio.h>
# include <string.h>
# include <string>
# include <algorithm>
# include <math.h>

using namespace std;

# define LL long long int
# define _LL __int64

/******************************/


/*****************************/

const int M = 2;
const int MOD = 1e4;

struct Matrix
{
    LL m[M][M];
};

Matrix operator * (Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0; i<M; i++) //初始化矩阵
        for(int j=0; j<M; j++)
            c.m[i][j]= 0;

    for(int k=0; k<M; k++)
        for(int i=0; i<M; i++) //实现矩阵乘法
        {
            if(a.m[i][k] <= 0)  continue;
            for(int j=0; j<M; j++)
            {
                if(b.m[k][j] <= 0)    continue;
                c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j]+MOD)%MOD;
            }
        }
    return c;
}

Matrix operator ^ (Matrix a,LL b)
{
    Matrix c;
    for(int i=0; i<M; i++) //初始化单位矩阵
        for(int j=0; j<M; j++)
            c.m[i][j]= ( i == j );

    while(b)
    {
        if(b&1) c= c * a ;
        b >>= 1;
        a = a * a ;
    }

    return c;
}

void init(Matrix &a)
{
    a.m[0][0]=1;
    a.m[0][1]=1;
    a.m[1][0]=1;
    a.m[1][1]=0;
}
int main()
{
    ios::sync_with_stdio(false);

    Matrix f;
    int n;
    while(~scanf("%d",&n)&&n!=-1)
    {
        init(f);
        f=f^n;


        printf("%lld/n",f.m[0][1]);
    }
    return 0;
}