编写程序求斐波那契数列前20项之和

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

因为用了很没有效率的递归,所以出结果有点慢#includef(int);main(){inti,s=0;for(i=1;i

PrivateFunctionF(nAsLong)AsLongIfn>2ThenF=F(n-1)+F(n-2)ElseF=1EndIfEndFunctionPrivateSubCommand1_Cli

main(){inti,n,s=1,f[]={0,1,1};printf("Pleaseinputthenumberofterms:");scanf("%d",&n);if(n==0){s=0;f[2

//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

dima()aslong,nasintegern=inputbox("请输入n的值：")redima(1ton)callFibonaccia()subFibonacci(a()aslong)dimia

#includeintmain(){inti,k,j,l,sum=0;i=1;k=0;for(j=0;j

staticvoidMain(string[]args){doublei=1;doublej=1;doublen=1;while(true){Console.WriteLine("a{0}:a{1}=

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

#include#includevoidsolve(){inti;inta[100],n=20;//保存数列,可以更改大小a[0]=0;a[1]=1;for(i=2;i再问：这个运行结果对着没再答：对

#includeusingnamespacestd;intmain(){intf[11]={0,1,1};inti,j,n;for(i=3;i{f[i]=f[i-1]+f[i-2];i

#includeusingnamespacestd;#include/*编写程序,计算并输出下面数列前20项的和,结果取3位小数.(C++)2/1,3/2,4/3,5/4,6/5,7/6,8/7,……

sum=sum+1/(5*i+1);这一句,1/(5*i+1)的值是整数的,所以它一直是0这样好像可以sum=sum+(double)1/(5*i+1);

#include"stdio.h"voidmain(){doublea,n,s=0;for(n=1;n

#include"stdafx.h"voidmain(){ints=0;for(intn=1;n

“i=1”---->"i==1","i=2"------>"i==2"

用什么语言呢?C还是PASCAL、VB?再问：vc++再答：#include<stdio.h>main(){ longa[30],i; a[0]=1;a[1]=1;&n

#include"stdio.h"intmain(){\x05inti,j,fenzi;\x05floatresult=0,fenmu;\x05for(i=1;i01:-1;\x05\x05for(f