求斐波那契数列前20项

为用了很没有效率的递归,所以出结果有点慢#includeiostream.h

PrivateFunctionF(nAsLong)AsLongIfn>2ThenF=F(n-1)+F(n-2)ElseF=1EndIfEndFunctionPrivateSubCommand1_Cli

现成的

//分别使用两个递归求分子分母即可：代码如下：usingSystem;namespace数列求和{classProgram{staticvoidMain(string[]args){intresult

斐波那契数列前13项为1,1,2,3,5,8,13,21,34,55,89,144,2331+1+2+3+5+8+13+21+34+55+89+144+233=609

dima()aslong,nasintegern=inputbox("请输入n的值：")redima(1ton)callFibonaccia()subFibonacci(a()aslong)dimia

intnum=1;intprev=0;for(inti=0;i

267914295,用EXCEL很简单的

#include#defineCOL5//一行输出5个longfibonacci(intn){//fibonacci函数的递归函数if(0==n||1==n){//fibonacci函数递归的出口re

1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121

#include#includevoidsolve(){inti;inta[100],n=20;//保存数列,可以更改大小a[0]=0;a[1]=1;for(i=2;i再问：这个运行结果对着没再答：对

PrivateFunctionbq(ByValsAsLong)AsLongSelectCasesCase1bq=1Case2bq=1CaseIs>=3bq=bq(s-1)+bq(s-2)EndSele

#includevoidmain(){inta[21];a[0]=0;a[1]=1;for(inti=2;i

 Private Sub Command1_Click()Dim F(11), i As LongF(0) = 

#includevoidfib(intn,intf0,intf1){intf;//当前项inti=0;if(n=2)printf("%8d,%8d",f0,f1);//f0,f1for(i=2;i

1123581321345589143232375607……

方法1：斐波那数列前30项是1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,4

#!/bin/bash#fibo.sh:Fibonaccisequence(recursive)#Author:M.Cooper#License:GPL3######----------algorit

n＝1,2,3,4,.第n项的数值an：an＝﹙1／√5﹚×﹛[﹙1＋√5﹚／2]^n－[﹙1－√5﹚／2]^n﹜.1,1,2,3,5,8,.再问：捣乱自重，不要通项公式，是前n项和公式再答：唉，那还

PrivateFunctionbq(ByValsAsLong)AsLongSelectCasesCase1bq=1Case2bq=1CaseIs>=3bq=bq(s-1)+bq(s-2)EndSele