Java求斐波那契数列前20项和

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

publicclassFibonacci{publicstaticvoidmain(Stringargs[]){inti=1,j=1;for(intn=1;n

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

publicclassFibonacci{\x09publicstaticvoidmain(Stringargs[]){intn,fn;//n为第n项,fn为第n项的值java.util.Scanne

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

publicclassFibonacci{\x09publicstaticvoidmain(Stringargs[]){intn,fn;//n为第n项,fn为第n项的值java.util.Scanne

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

//测试过的,能用publicclassFibonact{publicstaticint[]fib(){inta=1;//第一个数intb=1;//第二个数intc=0;//第三个数intn=46;/