作业帮linux求斐波那契数列的前10项以及总和.
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/23 00:24:53
#!/bin/bashi=1j=0for((k=0;k
/>#include<stdio.h>//the nest function used to calculate the&nbs
添加一个文本框输入前N项的N值,再添加一个命令按钮即可PrivateFunctionF(NAsLong)AsLongIfN>2ThenF=F(N-1)+F(N-2)ElseF=1EndIfEndFun
a=1b=1printa,bfori=1to6s=a+bprintsa=bb=snexti
为用了很没有效率的递归,所以出结果有点慢#includeiostream.h
因为用了很没有效率的递归,所以出结果有点慢#includef(int);main(){inti,s=0;for(i=1;i
PrivateFunctionF(nAsLong)AsLongIfn>2ThenF=F(n-1)+F(n-2)ElseF=1EndIfEndFunctionPrivateSubCommand1_Cli
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
#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
PrivateFunctionbq(ByValsAsLong)AsLongSelectCasesCase1bq=1Case2bq=1CaseIs>=3bq=bq(s-1)+bq(s-2)EndSele
Private Sub Command1_Click()Dim F(11), i As LongF(0) =
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
1123581321345589144就是新的项前两个连续项相加
PrivateFunctionbq(ByValsAsLong)AsLongSelectCasesCase1bq=1Case2bq=1CaseIs>=3bq=bq(s-1)+bq(s-2)EndSele
#includea,stdio>main(){intn,i=1,j,a;scanf("%d",&n);for(i=1,i