然后编写程序求斐波那契数列前20项之和.
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/28 07:16:21
#!/bin/bashi=1j=0for((k=0;k
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
viewplaincopytoclipboardprint?publicclassFibonacci{/***@paramargs*/publicstaticvoidmain(String[]args
#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
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
你要的应该是这个东西了!
#include"stdio.h"intfun(intm){if(m==1){return1;}elseif(m==2){return1;}else{returnfun(m-1)+fun(m-2);}
#includeusingnamespacestd;intmain(){intf[11]={0,1,1};inti,j,n;for(i=3;i{f[i]=f[i-1]+f[i-2];i
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);
“i=1”---->"i==1","i=2"------>"i==2"
#includeintsum(intn){intj,s=0;for(j=1;j
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