编程求fibonacci数列第28项的值

#includeunsignedintFibonacci(intn);intmain(void){inti;for(i=1;i

dimf()asdoublen=inputbox("in","NO.")redimf(n)asdoublef(1)=1f(2)=1fori=3tonf(i)=f(i-1)+f(i-2)nextprin

#includeintFibonacci(intn){if(n==0)return1;elseif(n==1)return1;elsereturnFibonacci(n-2)+Fibonacci(n-

#includevoidmain(){inta[40],i;a[0]=a[1]=1;for(i=2;i

PrivateSubForm_Click()DimnAsIntegern=Val(InputBox("请输入N:"))Dima,bAsLonga=1:b=1Fori=1TonPrinta&""&b&"

#include#include#includeusingnamespacestd;#defineM100inta[M];voidmain(){inti,j;a[0]=1;a[1]=2;i=1;

OptionExplicitDimf(40)AsLongPrivateSubCommand1_Click()DimiAsByteDimsAsLongf(1)=1f(2)=1s=2Print"No1:"

inta[101],i;a[0]=0,a[1]=1;floatt;for(i=2;i

#includeusingnamespacestd;intmain(void){\x09intn,i,*fib;\x09cin>>n;\x09if(n==1||n==2)\x09\x09cout

#include#defineCOL10//一行输出10个longscan(){//输入求fibonacci函数的第N项intn;printf("InputtheN=");scanf("%d",&n)

非递归：staticvoidf(intn){longp1=1,p2=1,p=1;for(inti=1;i

回答过了啊……Dimf1,f2,f3AsLongDimi,jAsIntegerf1=1f2=1j=3        &n

Private Sub Form_Load()Dim I As IntegerForm1.AutoRedraw = TrueFor

楼上的程序会慢死人的.给一个非递归实现.functionFibonacci(byvalnasLong)asLongdiml1aslong,l2aslong,l3aslongl1=1l2=1ifn

我给你代码：#include <stdio.h>#include <stdlib.h>#define N 47int fibo

#includeintFib(intm){if(m==1||m==2)return1;returnFib(m-1)+Fib(m-2);}voidmain(){intn,i;printf("请输入n的值

某个数等于前两个数之和,一个一个加就好了,第40个是726,当然也可以求出通项公式,不过很麻烦还可以编程：publicclassFibonacci{publicstaticvoidmain(Strin

publicclassFibonacci1{publicstaticlongfib(intn){longf1=1,f2=1;longm=0;if(n

递归法求作Fibonacci数列写生产fibonacci数列前20项;11235813.

下面的程序可以修改宏定义N的值来确定输出的数的个数#include#defineN30voidmain(){unsignedlonginta[N];inti,j;a[0]=1;a[1]=1;for(i