灵鹊做题网编写一个递归函数来计算斐波那契数列
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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
为用了很没有效率的递归,所以出结果有点慢#includeiostream.h
帮你写好了.unsigned int fib(unsigned int n) {\x09if (n == 1
#includelongintfn(int);voidmain(){printf("%d",fn(10));}longintfn(intm){longinttemp;if((1==m)|(2==m))
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
#includelongintf(intn){if(n==0)return0;elseif(n==1)return1;elsereturnf(n-1)+f(n-2);}intmain
// C++int F(int n) {if (n == 0) return 1;else if
#include "stdafx.h"#include <iostream>using namespace std;int&nb
//递归intfun(intn){if(n==1||n==2)return1;elsereturnfun(n-1)+fun(n-2);}//非递归intfun(){intans[41];ans[0]=
#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(n-
#includevoidfib(intn,intf0,intf1){intf;//当前项inti=0;if(n=2)printf("%8d,%8d",f0,f1);//f0,f1for(i=2;i
#includelongfac(intn){inti;longx=1;for(i=2;i再问：谢谢咯！可是我说的是递归法哦！再答：#includelongfac(intn){if(n==0)retur
C描述functionttt(n){ returnn>1?n*ttt(n-1):1;}使用方法:ttt(21);
#includevoidfun(intn,int*s)///求斐波那契序列中第n位的值{intf1,f2;if(n==0||n==1)*s=1;else{fun(n-1,&f1);fun(n-2,&f
functiongqj=erfen(p,a,b,e)ifabs(b-a)
intSumNums(intnum){if(num
#includeintfibo(intn){if(nreturn1;elsereturnfibo(n-1)+fibo(n-2);}intmain(){intn;scanf("%d",&n);print
cleardimea[20]a[1]=1a[2]=1fori=3to20a[i]=a[i-1]+a[i-2]endforfori=1to30?a[i]if(i%5=0)?endifendfo
#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(n-