【poj3070】 Fibonacci 【矩阵乘法】

讲解：点击打开链接

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15917 Accepted: 11188

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

099999999991000000000-1

Sample Output

0346266875

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:

.

题目描述 
斐波那契数列是由如下递推式定义的数列 
F0=0 
F1=1 
Fn+1=Fn+1+Fn 
求这个数列第n项的值对104取余后的结果。 
限制条件 
⋅ 0⩽n⩽1016

    我也是刚刚才搞懂了矩阵乘法(如果你不知道什么是矩阵乘法的话,右转百度百科)，于是来应用一下新知识，如有表述不到位的地方请见谅。 
下面进入正文

首先，我们先介绍一下对于斐波那契数列如何求解。把斐波那契数列的递推式表示成矩阵就得到下面的式子 
(Fn+2Fn+1)= (1110)(Fn+1Fn)

我们发现式子里有个固定的矩阵 (1110)

记这个矩阵为A，则有 
(Fn+1Fn)=An(F1F0)=An(10)

因此只要求出An就可以求出Fn了。关于An的计算我们可以采用类似快速幂的算法,在O(logn)时间里求出第n项的值。（转自：点击打开链接）

#include<cstdio> int n,a[2][2],b[2][2]; void mul(int a[2][2],int b[2][2],int ans[2][2]){    int t[2][2];    for (int i=0;i<2;i++)        for (int j=0;j<2;j++) {            t[i][j]=0;            for (int k=0;k<2;k++) t[i][j]=(t[i][j]+a[i][k]*b[k][j])%10000;        }    for (int i=0;i<2;i++)        for (int j=0;j<2;j++) ans[i][j]=t[i][j];}void pow(int k) {    while (k) {        if (k%2==1) mul(a,b,b);        k/=2;        mul(a,a,a);    }}int main() {    while (scanf("%d",&n)!=EOF) {        if (n==-1) break;        a[0][0]=a[0][1]=a[1][0]=1;a[1][1]=0;                b[0][0]=b[1][1]=1;        b[1][0]=b[0][1]=0;                pow(n);        printf("%d\n",b[1][0]);    }    return 0;}

这道题还碰见一种神奇的操作，我还没理解到底是什么规律。。

来自：（点击打开链接）

#include<stdio.h>int a[100050];void f() {a[0] = 0;a[1] = 1;for (int i  = 2; i <= 100050; i ++) {a[i] = (a[i - 1] + a[i - 2]) % 10000;}}int main () {int n;while (scanf("%d", &n) != EOF) {if(n == -1) break;f();printf("%d\n", a[n % 15000]);}return 0;}