试用递归的方法编写一个返回长整形的函数,以计算菲波那切数列的前20项

#includeintdigit(intn,intk){returnk>1digit(n/10,k-1):n%10;}intmain(){printf("%d",digit(12345,3));}

#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

#includelongfun(longx){intn=10,m=0;while(x){intt=x%10;if(t%2==0){m+=t*n/10;n*=10;}x/=10;}returnm;}vo

帮你写好了.unsigned int fib(unsigned int n) {\x09if (n == 1 

#includelongintf(intn){if(n==0)return0;elseif(n==1)return1;elsereturnf(n-1)+f(n-2);}intmain

c:intfib(intn){return(n

// C++int F(int n) {if (n == 0) return 1;else if

解法一：如图，AB=AC=5，BC=4，过A点作AD⊥BC，垂足为D，cosB=BDAB=25，∴B≈65°，A=180°-2B=50°，∴cosA≈0.68；解法二：如图，AB=AC=5，BC=4，

cludestdio.hvoidmain(){intmax_4(inta,intb,intc,intd);inta,b,c,d,max;printf("Pleaseenterintergernumbe

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

代码如下:OptionExplicitPrivateSubCommand1_Click()MsgBoxP(2,2)EndSubFunctionP(ByValnAsInteger,ByValxAsDou

#include <stdio.h>char* dg(char* instr, char* outstr, char* 

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

#include#includefloatmyfunction(intn,intx){if(0==n){return1;}elseif(1==n){returnx;}else{return((2*n-

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

1.#include"stdio.h"//#defineRECURSION1#ifdefRECURSIONlongfact(intn){if(n

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

#includeintprime(inta){inti;if(a

用递归法计算n!用递归法计算n!可用下述公式表示：n!=1(n=0,1)n×(n-1)!(n>1)按公式可编程如下：longff(intn){longf;if(n