c 编程2. 编写一递归函数,计算1 2 3 4 - n.n为正整数

#includefun(intx,intn){if(n==0)return(1);elsereturn(x*fun(x,n-1));}main(){intx,n;scanf("%d,%d",&x,&n

为用了很没有效率的递归,所以出结果有点慢#includeiostream.h

#includemain(){doublex,y;scanf("%lf",&x);if(x

#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;

#includedoubleH(intn,doublex){if(x>1){if(n==0)return1.0;//H0(x)=1;if(n==1)return2.0*x;//H1(x)=2x;//直

#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)

#includeusingnamespacestd;doublefun(intn,doubleh){if(n==1)returnh;elseif(n再问：usingnamespacestd;这句

publicf(byvalnasinteger)aslongifn=1thenf=1elsef=f(n-1)+nendifendfunction

intfib(n){if(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

利用c++中的函数实现,程序很简单该注释的都注释上了,不用多说了吧!源程序如下所示：#include#include#includeusingnamespacestd;intmain(){charhs

#includemain(){inti,j,n,k,t;doublesum=2;printf("inputanumber:\n");scanf("%d",&n);j=2;k=1;for(i=1;i再问

程序已经调试过了,你看下吧,可以的话给个好评#include"iostream"usingnamespacestd;floatpower(floata,intn){if(n==1)returna;el

#includeintfact(int);main(){inti,sum=0;for(i=1;i

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

#includedoublepow(doublex,intn){if(1==n){returnx;}else{doubletemp=pow(x,n-1);return(x*temp);}}voidma

intSumNums(intnum){if(num

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

#include <stdio.h>#include <stdlib.h>#include <math.h>double f

cleardimea[20]a[1]=1a[2]=1fori=3to20a[i]=a[i-1]+a[i-2]endforfori=1to30?a[i]if(i%5=0)?endifendfo