作业帮下面是前1000项斐波那契数列的和的VB代码
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/28 10:41:48
/>#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
#includevoidmain(){inti,count=0,num[30]={1,1};for(i=2;i
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
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
staticvoidMain(string[]args){doublei=1;doublej=1;doublen=1;while(true){Console.WriteLine("a{0}:a{1}=
1112233455861372183495510891114412233133771461015987161597172584184181196765201094621177112228657234
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……
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
设第一项是a那么前12项依次是：a,2,a+2,a+4,2a+6,3a+10,5a+16,8a+26,13a+42,21a+58,34a+100,55a+158所以55a+158=122所以S10=2
1123581321345589144就是新的项前两个连续项相加