编写斐波拉契数列1,1,2,3,5-搜答案-搜搜考试网

#includeintmain(){inta[40],sum=2,i;a[0]=1;a[1]=1;for(i=2;i

#!/bin/bashi=1j=0for((k=0;k

floatfunction(intn){floatave,a[100],sum=0;inti;if(n==1)return1;elseif(n==2)return1.5;elseif

#includemain(){floatsum=0;inti;for(i=1;i

#include#includeintmain(void){intn;doublei,j,k;doubleres=0;printf("pleaseinputn:");scanf("%d",&n);i=

你需要一个分数式的结果?再问：什么结果都可以再答：#include"stdio.h"#include"math.h"voidPrintInfo();intmain(){doublesum,buf;in

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

#includeintmain(){inti=1,j=1,n;floatsum=0;scanf("%d",&n);for(intk=0;k

#includevoidmain(){inti,a[5];a[0]=a[1]=0;a[2]=1;i=3;while(i

#includeusingnamespacestd;intmain(){inta[]={0,0,1,};inti=3,n;coutn;cout

#includeintmain(){inta=2;intc=3;intsum=2;inti;for(i=1;i

Fibonacci其实就是不断求和,前两个数的和形成第三个数.为方便说明,假设两个变量A和B,A+B=C,C就是和.#!/bin/shfirst=1;echo“$first”#打印第一个数A=1sec

[root@localhostbaidu]#morefibonacci.shnum1=0num2=1echo$num1echo$num2couter=$(expr$1-2)#echo$numwhile

#include"stdio.h"intfun(intm){if(m==1){return1;}elseif(m==2){return1;}else{returnfun(m-1)+fun(m-2);}

求Fibonacci数列前20项不需要递归intPrint_Fib_Nums(intsize){if(size

vc写的#includevoidmain(){inti;doubles=0;for(i=1;i

程序如下F:>>clearn=input('pleaseinputn=');%输入前n项的n值..fori=1:(n-2)a(1)=1;a(2)=1;a(i+2)=a(i+1)+a(i);endM=s

doublenum1,num2,num3,sum;num1=num2=num3=sum=0;for(intii==0;ii再问：看的不是太明白哦，能写的再清楚一点吗？再答：doublenum1,num

#includeintsum(intn){intj,s=0;for(j=1;j

求Fibonacci数列前20项不需要递归intPrint_Fib_Nums(intsize){if(size