编写函数,求数列的第n项的值指针函数

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

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

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

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))

main(){inti,n,s=1,f[]={0,1,1};printf("Pleaseinputthenumberofterms:");scanf("%d",&n);if(n==0){s=0;f[2

很高兴回答你的问题!#includeintfun(intn,intx){if(n==0)return1;intmul=x;for(inti=n;i>1;i--)mul*=x;returnfun(n-1

PrivateSubForm_Click()DimnAsIntegern=Val(InputBox("请输入N:"))Dima,bAsLonga=1:b=1Fori=1TonPrinta&""&b&"

intfib(n){if(n

//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-

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);

楼上的程序会慢死人的.给一个非递归实现.functionFibonacci(byvalnasLong)asLongdiml1aslong,l2aslong,l3aslongl1=1l2=1ifn

1）a1=1,a2=1,a(n+2)=a(n+1)+an,a(n+2)+[(√5-1)/2]a(n+1)=[(√5+1)/2][a(n+1)+(√5-1)/2*an]==.=[(√5+1)/2]^n[

%编成M函数文件运行后,在命令窗口输入要知道的自然数n,即可求得对应项的Fibonacci数列%有哪步有疑问请问user_entry=input('Pleaseenterthenumberyouwan

#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

#includefib(intn){if(n==0)return(0);elseif(n==1)return(1);elsereturn(fib(n-1)+fib(n-2));}main(){intn

functionf=d(n)f(1)=1;f(2)=1;fori=3:nf(i)=f(i-1)+f(i-2);end

a(n+1)=Aa(n)^2+Ba(n)+C,求a(n)为方便计,引满足a1=Aa0^2+Ba0+C的项a0,称之为数列的零项.对于一般情况的ABC值,属于非线性递推式,据我所知,目前除了不动点方法外