编程求Fibnacci数列的前20项
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#includeintmain(){intA[20];inti;A[0]=A[1]=0,A[2]=2,A[3]=5;for(i=4;i
PrivateSubForm_Click()Dima(1To40)AsLongFori=1To40Ifi=1Ori=2Thena(i)=1Elsea(i)=a(i-2)+a(i-1)EndIfPrin
functionsum(nasinteger)dima(30)asintegerdimsasintegers=0a(1)=0a(2)=0a(3)=1fori=4to30a(i)=a(i-1)+a(i-
#includeintFibonacci(intn){if(n==0)return1;elseif(n==1)return1;elsereturnFibonacci(n-2)+Fibonacci(n-
Dima1,a2,a3,a4,iAsIntegera1=3:a2=4:a3=5fori=4to26a4=a1+a2+a3a1=a2a2=a3a3=a4next输出a4即可
#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&"
OptionExplicitDimf(40)AsLongPrivateSubCommand1_Click()DimiAsByteDimsAsLongf(1)=1f(2)=1s=2Print"No1:"
#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)
用VB写的~PrivateSubForm_Load()Dimfn&,f1&,f2&,n%f1=1:f2=1n=Val(InputBox("请输入n"))fn=f1+f2Printf1Printf2Pr
菲波那契数列指的是这样一个数列:1,1,2,3,5,8,13,21……这个数列从第三项开始,每一项都等于前两项之和.它的通项公式为:(1/√5)*{[(1+√5)/2]^n-[(1-√5)/2]^n}
#includeintmain(){inti=0;floatsum=0;intn;intx[n],y[n];printf("请输出计算的项数:");scanf("%d",&n);x[0]=2;x[1]
#include"stdio.h"voidmain(){intn;inta,b,c;scanf("%d",&n);a=0;b=1;if(n==1)printf("0\n");elseif(
intfib(n){if(n
//fibonacci数列:1123581321...#include#includeintmain(void){longa=1;longb=1;intn;intk;printf("inputnumb
你写的有问题,斐波那契(Fibonacci)数列指的是这样一个数列:f(1)=1,f(2)=1,f(3)=f(1)+f(2)=2,f(4)=f(2)+f(3)=3,…这个数列从第三项开始,每一项都等于
Private Sub Form_Load()Dim I As IntegerForm1.AutoRedraw = TrueFor
你的意思是:第1组1,2;第2组3,5,8,第3组,13,21,34,55;计算2014所在的那一组?再问:是的。编程......再答:对的话多个点分clccleara=[];a(1)=1;a(2)=
#includeintFib(intm){if(m==1||m==2)return1;returnFib(m-1)+Fib(m-2);}voidmain(){intn,i;printf("请输入n的值
cleara=1b=1s=0fori=1to18c=a+bs=s+ca=bb=cendfor"斐波拉契数列前30项的和:",s再问:不是求前30项的和吗》为什么fori=1to18呢?再答:哦,笔误,