POJ3070 Fibonacci[矩阵乘法]

时间：2022-07-02 00:17:45
Fibonacci
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 13677   Accepted: 9697

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

POJ3070 Fibonacci[矩阵乘法].

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

POJ3070 Fibonacci[矩阵乘法].

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

POJ3070 Fibonacci[矩阵乘法].

Source

矩阵乘法的应用
一个有趣的理解：
结果矩阵第m行与第n列交叉位置的那个值，等于第一个矩阵第m行与第二个矩阵第n列，对应位置的每个值的乘积之和
白书上的一句：
把一个向量v变成另一个向量v1，并且v1的每一个分量都是v各个分量的线性组合，考虑使用矩阵乘法
对于斐波那契数列，构造矩阵
1 1
1 0
然后
让矩阵
A     1 1
B     1 0
相乘，就是得到
A+B
A
就是斐波那契数列啊
快速幂来优化到logn
遇到一个n很大的DP/递推关系，都可以考虑用这种方法
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

const int MOD=1e4;

typedef long long ll;

inline int read(){

    char c=getchar();int x=,f=;

    while(c<''||c>''){if(c=='-')f=-;c=getchar();}

    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}

    return x*f;

}

int n;

struct mat{

    int r,c;

    int m[][];

    mat(){r=;c=;memset(m,,sizeof(m));}

}im,f;

mat mult(mat x,mat y){

    mat t;

    for(int i=;i<=x.r;i++)

        for(int k=;k<=x.c;k++) if(x.m[i][k])

            for(int j=;j<=y.c;j++)

                t.m[i][j]=(t.m[i][j]+x.m[i][k]*y.m[k][j]%MOD)%MOD;

    return t;

}

void init(){

    for(int i=;i<=im.c;i++)

        for(int j=;j<=im.r;j++)

            if(i==j) im.m[i][j]=;

    f.m[][]=;f.m[][]=;

    f.m[][]=;f.m[][]=;

}

int main(){

    init();

    while((n=read())!=-){

        mat ans=im,t=f;

        for(;n;n>>=,t=mult(t,t))

            if(n&) ans=mult(ans,t);

        printf("%d\n",ans.m[][]);

    }

}

相关文章