神马作文网Python 求斐波那契数列前20项和
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/11 23:59:25
/>#include<stdio.h>//the nest function used to calculate the&nbs
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
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
在Python里面是分为可变对象与不可变对象两类的.对于你这个问题就是strategy_do_nothing是list,而list是可变对象,所以在strategy.append(strategy_d
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
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
混份儿题主机智
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