Poj 3070 Fibonacci

Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 26063 Accepted: 17394

文章目录

Description题意：题解：代码： Poj 3070 Fibonacci

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 .

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 .

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

题意：

其实就是求第x位斐波那契数列（记得mod10000），不过题目给你介绍了矩阵的相关知识，以及矩阵与斐波那契数列的联系，就是让你用矩阵快速幂来求斐波那契数列。

题解：

斐波那契数列的博客 我在这个博客里详细介绍了三种斐波那契数列的求法，其中就有矩阵快速幂的方法 所以我在这就不详细介绍，简单说说： mul（a，b）是用来进行矩阵a与b的乘法运算 pow(mat a,ll n)//快速幂求矩阵a的n次幂

代码：

#include <iostream> #include <cstddef> #include <cstring> #include <vector> using namespace std; typedef long long ll; const int mod=10000; typedef vector<ll> vec; typedef vector<vec> mat; mat mul(mat &a,mat &b) { mat c(a.size(),vec(b[0].size())); for(int i=0; i<2; i++) { for(int j=0; j<2; j++) { for(int k=0; k<2; k++) { c[i][j]+=a[i][k]*b[k][j];//矩阵的运算法则 c[i][j]%=mod; } } } return c; } mat pow(mat a,ll n) { mat res(a.size(),vec(a.size())); for(int i=0; i<a.size(); i++) res[i][i]=1; while(n) { if(n&1) res=mul(res,a); a=mul(a,a); n/=2; } return res; } ll solve(ll n) { mat a(2,vec(2)); a[0][0]=1; a[0][1]=1; a[1][0]=1; a[1][1]=0; a=pow(a,n); return a[0][1]; } int main() { ll n; while(cin>>n&&n!=-1) { cout<<solve(n)<<endl; } return 0; }