用递归fibonacci数列前20项.1,1,2,3,5法

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

clc,cleara(1)=1;a(2)=1;n=input('n=');k=2;whilea(k)

dimf()asdoublen=inputbox("in","NO.")redimf(n)asdoublef(1)=1f(2)=1fori=3tonf(i)=f(i-1)+f(i-2)nextprin

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

#includeintGetFibonacci(intn){if(n==1||n==2)return1;elsereturnGetFibonacci(n-1)+GetFibonacci(n-2);}v

#include#defineN20main(){intf(intn);inti;for(i=1;i

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

#include#includevoidfun(intn){\x09inti;\x09int*a=(int*)malloc(n*sizeof(int));\x09a[0]=a[1]=1;\x09for

#include#defineCOL10//一行输出10个longscan(){//输入求fibonacci函数的第N项intn;printf("InputtheN=");scanf("%d",&n)

这是数学题?!确定不是程序设计...1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,搞定,希望没算错...其实斐波

public static int fibonacci(int n) {  if (n <= 

递归算法intfib(intn){//求fibonacci数列第n个数if(n==1||n==2)return1;elsereturnfib(n-1)+fib(n-2);}非递归intfib(intn

写一函数intFibonacci(intn){if(n==1||n==2){return1;}returnFibonacci(n-1)+Fibonacci(n-2);}

用递归多好啊,我用的递归,你参考一下!#includeusingnamespacestd;intfib(intn){if(n==0||n==1)return1;returnfib(n-1)+fib(n

#includeintf(inti){if(i==1||i==2)return1;elsereturnf(i-1)+f(i-2);}voidmain(){intn,j;scanf("%d",&n);f

编程首先计算Fibonacci数列1,1,2,3,5,8,13,21,.的前n项（n不超过40）存入一维整型数组f中,再按%12d的格式输出每项的值,每6项换一行.说明：（1）输入数列项n,在scan

longfib(intn)　　{　　if(n==0)return0;　　if(n==1)return1;　　if(n>1)returnfib(n-1)+fib(n-2);　　}

publicclassFibonacci1{publicstaticlongfib(intn){longf1=1,f2=1;longm=0;if(n

#includevoidmain(){inta[20];int*p=a;*p=0;*(p+1)=1;for(inti=2;i

//fibonacci数列:11235813213455...#includedoublefib_val[100]={0};doublefibonacci_1(intn)//递归,计算时间长,n最好不