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https://www.zhihu.com/question/28062458 http://blog.csdn.net/hikean/article/details/9749391 对于Fibonacci数列,1,1,2,3,5,8,13,21...    F(0) = 1, F(1) = 1, F(i) = F(i-1) + F(i-2) 求解第n项. 1.递归 long fib(int n) { || n == ) { ; } ) + fib(n-); }     这是最好写,也是效率最低…
app-node.js ; var http = require('http'); var htutil = require('./htutil'); var server = http.createServer(function(req, res){ htutil.loadParams(req, res, undefined); if (req.requrl.pathname === '/'){ require('./home-node').get(req, res); }else if (r…
本文按以下顺序讲解JUnit4的使用 下载jar包 单元测试初体验 自动生成测试类 执行顺序 @Test的属性 下载jar包## 下载地址 在github上,把以下两个jar包都下载下来.     下载junit-4.12.jar,junit-4.12-javadoc.jar(文档),junit-4.12-sources.jar(源码).     下载hamcrest-core-1.3.jar,hamcrest-core-1.3-javadoc.jar(文档),hamcrest-core-1.3…
Fibonacci Time Limit: 2000MS Memory limit: 131072K 题目描述 Fibonacci numbers are well-known as follow: Now given an integer N, please find out whether N can be represented as the sum of several Fibonacci numbers in such a way that the sum does not inclu…
http://scottsievert.github.io/blog/2015/01/31/the-mysterious-eigenvalue/ The Fibonacci problem is a well known mathematical problem that models population growth and was conceived in the 1200s. Leonardo of Pisa aka Fibonacci decided to use a recursiv…
/*===================================== 1978 Fibonacci数列 3 题目描述 Description 斐波纳契数列是这样的数列: f1 = 1 f2 = 1 f3 = 2 f4 = 3 .... fn = fn-1 + fn-2 输入一个整数n 求fn 输入描述 Input Description 一个整数n, n<= 40 输出描述 Output Description 一个整数fn 样例输入 Sample Input 3 样例输出 Sampl…
Pandigital Fibonacci ends The Fibonacci sequence is defined by the recurrence relation: F[n] = F[n-1] + F[n-2], where F[1] = 1 and F[2] = 1. It turns out that F541, which contains 113 digits, is the first Fibonacci number for which the last nine digi…
废话不多说,直接上代码 #include "stdio.h" #include "queue" #include "math.h" using namespace std; /////////////////////////////////////////////////////////////////////////// //一:递归实现 // 使用公式f[n]=f[n-1]+f[n-2],依次递归计算,递归结束条件是f[1]=1,f[2]=1…
  入门训练 Fibonacci数列   时间限制:1.0s   内存限制:256.0MB        问题描述 Fibonacci数列的递推公式为:Fn=Fn-1+Fn-2,其中F1=F2=1. 当n比较大时,Fn也非常大,现在我们想知道,Fn除以10007的余数是多少. 输入格式 输入包含一个整数n. 输出格式 输出一行,包含一个整数,表示Fn除以10007的余数. 说明:在本题中,答案是要求Fn除以10007的余数,因此我们只要能算出这个余数即可,而不需要先计算出Fn的准确值,再将计算的…
斐波那契(Fibonacci)数列 问题描述 递归算法: package chapter2shuzizhimei.fibonacci; /** * Fibonacci数列递归求解 * @author DELL * */ public class Fibonacci1 { public static int fibonacci(int n){ if(n<=0) return 0; else if(n==1) return 1; else return fibonacci(n-1)+fibonacc…