无穷数列1,1,2,3,5,8,13,21,34,55...称为Fibonacci数列,它可以递归地定义为F(n)=1 ...........(n=1或n=2)F(n)=F(n-1)+F(n-2).....(n>2)现要你来求第n个斐波纳奇数.(第1个.第二个都为1): 输入第一行是一个整数m(m<5)表示共有m组测试数据每次测试数据只有一行,且只有一个整形数n(输出对每组输入n,输出第n个Fibonacci数 #include<stdio.h> long int fib(long

今天这篇博客就聊聊几种常见的查找算法,当然本篇博客只是涉及了部分查找算法,接下来的几篇博客中都将会介绍关于查找的相关内容.本篇博客主要介绍查找表的顺序查找.折半查找.插值查找以及Fibonacci查找.本篇博客会给出相应查找算法的示意图以及相关代码,并且给出相应的测试用例.当然本篇博客依然会使用面向对象语言Swift来实现相应的Demo,并且会在github上进行相关Demo的分享. 查找在生活中是比较常见的,本篇博客所涉及的这几种查找都是基于线性结构的查找.也就是说我们的查找表是一个线性表,我

Difficulty: Easy Topic: Fibonacci seqs Write a function which returns the first X fibonacci numbers. ;; 首先实现一个求fibonacci数的函数 ;;最简单的实现,就是通过定义来实现递归函数,(如下的fibonacci-number),但是这样缺点很明显,首先这不算尾递归,代码里面有大量的重复计算.其次,jvm不支持尾调用优化,因此,即使是尾递归,当嵌套层侧过深时,也会出现stackoverf

经典的Fibonacci数的问题 主要想展示一下迭代与递归,以及尾递归的三种写法,以及他们各自的时间性能. public class Fibonacci { /*迭代*/ public static int process_loop(int n) { if (n == 0 || n == 1) { return 1; } int a = 1, b = 1; int i = 1; while (i < n) { i++; int t = b; b = a + t; a = t; } return

第一种:利用for循环 利用for循环时,不涉及到函数,但是这种方法对我种小小白来说比较好理解,一涉及到函数就比较抽象了... >>> fibs = [0,1] >>> for i in range(8): fibs.append(fibs[-2] + fibs[-1]) >>> fibs [0, 1, 1, 2, 3, 5, 8, 13, 21, 34] 或者说输入一个动态的长度: fibs = [0,1] num = input('How many

自己没动脑子,大部分内容转自:http://www.jb51.net/article/37286.htm 斐波拉契数列,看起来好像谁都会写,不过它写的方式却有好多种,不管用不用的上,先留下来再说. 1.递归公式:f[n]=f[n-1]+f[n-2],f[1]=f[2]=1;(比较耗时,效率不高) 代码: int fib(int n) //递归实现 { ) { ; } || n==) ; )+fib1(n-); } 2.数组实现:空间复杂度和时间复杂度都是0(n),效率一般,比递归来得快.代码:

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 sequenc

UVA - 1025 A Spy in the Metro Secret agent Maria was sent to Algorithms City to carry out an especially dangerous mission. After several thrilling events we find her in the first station of Algorithms City Metro, examining the time table. The Algorit

/************************************************* * Fibonacci 数列算法分析 *************************************************/ #include<iostream> #include<stdio.h> #include<vector> #include<cmath> #include<time.h> using namespace s

Copyright © 1900-2016, NORYES, All Rights Reserved. http://www.cnblogs.com/noryes/ 欢迎转载,请保留此版权声明. ----------------------------------------------------------------------------------------- 对于Fibonacci数列,1,1,2,3,5,8,13,21...    F(0) ＝ 1, F(1) ＝ 1, F(i)

巨大的斐波那契数 The i'th Fibonacci number f (i) is recursively defined in the following way: f (0) = 0 and f (1) = 1 f (i+2) = f (i+1) + f (i)  for every i ≥ 0 Your task is to compute some values of this sequence. Input begins with an integer t ≤ 10,000, th

Buge's Fibonacci Number Problem Description snowingsea is having Buge’s discrete mathematics lesson, Buge is now talking about the Fibonacci Number. As a bright student, snowingsea, of course, takes it as a piece of cake. He feels boring and soon com

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

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

1)低级版本 var fibonacci = function(n) { if (n == 0 || n == 1) { return n; } else { return fibonacci(n - 1) + fibonacci(n - 2); } } 基本上算到fibonacci(50), 就不行了, 老半天. 2) 进阶版一 var fibonacci = function(n) { var cache = {}; return function(n) { if (!cache[n]) c

#include<iostream>using namespace std;int fibonacci(int n){if(n<=0) return 0; else if(n==1) return 1; else return fibonacci(n-1)+fibonacci(n-2);} int main(){ int n,s; cout<<"please input a n to be the max of fibonacci\n"; cin>&

求fibonacci数列前N个数的K次方和. 通项公式:F[n]=((1+sqrt(5))/sqrt(5)-(1-sqrt(5))/sqrt(5))/sqrt(5). 有点乱,不过由于可以保证最后的结果是一个整数,所有所有的根号都可以化为整数进行取模和逆元运算. 首先解二次同余方程,X^2=n (mod M),显然,当n=5的时候,X就可以相当于5了. 后面的都可以替换掉. 然后就变成了 F[n]=(A^n+B^n)*C. 这里通过二项式展开,分别对每一项进行计算,同时又可以发现,每一项求和其实

Problem Introduction The Fibonacci numbers are defined as follows: \(F_0=0\), \(F_1=1\),and \(F_i=F_{i-1}+F_{i-2}\) for $ i \geq 2$. Problem Description Task.Given two integers \(n\) and \(m\), output \(F_n \ mod \ m\)(that is, the remainder of \(F_n

Problem Introduction The Fibonacci numbers are defined as follows: \(F_0=0\), \(F_1=1\),and \(F_i=F_{i-1}+F_{i-2}\) for $ i \geq 2$. Problem Description Task.Given an integer \(n\),find the last digit of the \(n\)th Fibonacci number \(F_n\)(that is ,

Problem Introduction The Fibonacci numbers are defined as follows: \(F_0=0\), \(F_1=1\),and \(F_i=F_{i-1}+F_{i-2}\) for $ i \geq 2$. Problem Description Task.Given an integer \(n\),find the \(n\)th Fibonacci number \(F_n\). Input Format.The input con

原题目: Write a recursive program that extends the range of the Fibonacci sequence.  The Fibonacci sequence is 1, 1, 2, 3, 5, 8, etc., where each element is the sum of the previous two elements. For this problem, instead of only adding the last two elem