fibonacci

我有我难倒一个家庭作业。 我想编写一个程序，输出Fibonacci序列了第n个号码。 这是我到目前为止有：高清谎（）：N = INT（输入（“请...(I have a homework assignment that I'm stumped on. I'm trying to write a program that outputs the fibonacci sequence up the nth number. Here's what I have so far: def fib(): n = int(input("Please ...)

我有一些麻烦，在项目欧拉这个问题。 这里的问题问什么：在Fibonacci序列中的每个新名词是通过将前两个方面产生。 通过从1开始...(I'm having some trouble with this problem in Project Euler. Here's what the question asks: Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 ...)

斐波那契堆数据结构在其名称中的“黄金分割”，但没有在数据结构似乎使用斐波那契数。 根据维基百科的文章：斐波那契的名字......(The Fibonacci heap data structure has the word "Fibonacci" in its name, but nothing in the data structure seems to use Fibonacci numbers. According to the Wikipedia article: The name of Fibonacci ...)

确定所述斐波那契序列是足够容易弄清楚：INT NUM = 0; INT NUM2 = 1; INT循环; INT斐波那契; 是System.out.print（NUM2）; 对于（循环= 1;环<= 10;循环++）{斐波纳契= ...(Determining the Fibonacci sequence is easy enough to figure out: int num = 0; int num2 = 1; int loop; int fibonacci; System.out.print(num2); for (loop = 1; loop <= 10; loop ++) { fibonacci = ...)

我只是通过斐波纳契数列算法的迭代版本去。 我发现这个下面的代码INT斐波那契（INT N）{INT F1 = 0; INT F2 = 1; INT FN; 对（INT I = 2; I <N; ...(I was just going through the iterative version of fibonacci series algorithm. I found this following code int Fibonacci(int n) { int f1 = 0; int f2 = 1; int fn; for ( int i = 2; i < n; ...)

我有一个很难理解为什么的#include <iostream的>使用命名空间std; INT FIB（INT X）{如果（X == 1）{返回1; }否则{返回FIB（X-1）+ FIB（X-2）; }} ...(I'm having a hard time understanding why #include <iostream> using namespace std; int fib(int x) { if (x == 1) { return 1; } else { return fib(x-1)+fib(x-2); } } ...)

我感兴趣的一个迭代算法斐波那契数，所以我发现维基公式...它看起来直截了当所以我尝试在Python ......它没有问题编译和公式...(I am interested in an iterative algorithm for Fibonacci numbers, so I found the formula on wiki...it looks straight forward so I tried it in Python...it doesn't have a problem compiling and formula ...)

我想计算N个即非常大的值的斐波那契。 10 ^ 6与O（logN）的复杂性。 这里是我的代码，但它给出了30秒10的6次方的结果，这是非常时期consuming.Help我...(I want to calculate the Fibonacci of very large value of N ie. 10^6 with a complexity of O(logN). Here is my code but it gives the result for 10^6 in 30 seconds which is very time consuming.Help me ...)

我做的受试者的任务是FIB（0）被定义为= 1。但是，这不可能是正确的？ FIB（0）为0？ 用FIB程序（0）= 1; 吐出FIB（4）= 5程序与FIB（0）= 0; 吐出FIB（3）= 3是什么？(I'm doing a task in a subject were fib(0) is defined to = 1. But that can't be right? fib(0) is 0? Program with fib(0) = 1; spits out fib(4) = 5 Program with fib(0) = 0; spits out fib(3) = 3 What ...)

我试图回忆斐波纳契递归算法。 以下内容：公众诠释斐波纳契（INT N）{如果（N == 0）返回0; 否则，如果（N == 1）返回1; 否则返回斐波那契（N - 1）+ ...(I'm trying to recall an algorithm on Fibonacci recursion. The following: public int fibonacci(int n) { if(n == 0) return 0; else if(n == 1) return 1; else return fibonacci(n - 1) + ...)