编写一递归函数求斐波纳契数列的前 0 40

#includeunsignedintFibonacci(intn);intmain(void){inti;for(i=1;i

intfunction(intn){if(n==0)return0;elseif(n==1)return1;elsereturnfunction(n-1)+function(n-2);}

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

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

#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/*非递归求：f(1)+f(2)+...+f(m)其中f(n)=n*(n+1)*/unsignedintsum_fn(unsignedintm){intn,sum=0;for(n=1;

把以下程序另存为fib.m%>>fib(8)Fibonacci=123456781123581321

用VB写的~PrivateSubForm_Load()Dimfn&,f1&,f2&,n%f1=1:f2=1n=Val(InputBox("请输入n"))fn=f1+f2Printf1Printf2Pr

//fibonacci数列:1123581321...#include#includeintmain(void){longa=1;longb=1;intn;intk;printf("inputnumb

#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(n-

#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

#includelongfib(intn){inta;if(n==1)a=1;elseif(n==2)a=1;elsea=fib(n-1)+fib(n-2);returna;}voidmain(){\

functiongqj=erfen(p,a,b,e)ifabs(b-a)

intSumNums(intnum){if(num

#includemain(){intn,i,j,k;while(scanf("%d",&n)==1){if(n==1||n==2){printf("%d\n",1);cont

#includeintfibo(intn){if(nreturn1;elsereturnfibo(n-1)+fibo(n-2);}intmain(){intn;scanf("%d",&n);print