百度作业网编写调用递归函数的程序,输出斐波那契数列前20项
来源：学生作业帮助网 编辑：作业帮 时间：2024/08/08 07:12:31
上面是对的,f1是求体积的,f2是求表面积的.
#includeusingnamespacestd;boolIsPrime(intn){if(n
#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#includeintmax(inta,intb){if(a>b)returna;if(a
这个不难,注意看好了.#includeusingnamespacestd;voidsum(intx,inty);//声明一个函数sumvoidmain(){inta,b;couta>>b;sum(a,
#includeintsushu(intn){intj;for(j=2;j
floatfun(intn){floatsum=0;inti;for(i=1;i
/>#includeusingnamespacestd;longunsignedfun(intn){if(n>1)returnn*fun(n-1);return1;}voidmain(){intn;c
#include <stdio.h>char* dg(char* instr, char* outstr, char*
intFibona(intn){intm;if(n==1)return(1);elseif(n==2)return(1);else{m=Fibona(n-1)+Fibona(n-2);return(m
#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(){\
functiongqj=erfen(p,a,b,e)ifabs(b-a)
首先你那个a[i]=a[i]*(n-j)/j公式有点问题吧,这样会输出全是1的啊...如果还是按照你这样可以改成如下代码：#include"stdafx.h"#include#include#incl
1.#include"stdio.h"//#defineRECURSION1#ifdefRECURSIONlongfact(intn){if(n
#includeusingnamespacestd;longunsignedfun(intn){if(n>1)returnn*fun(n-1);return1;}voidmain(){intn;cou
#includedoublef(doublex,intn){if(n==1)returnx;else{doubled=1.0;inti;for(i=1;i
#includeusingnamespacestd;boolpanduan(intn){intm=n%2;if(m==0)returntrue;elsereturnfalse;}intmain(){i
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
//今天被百度吞了4份代码,上帝保佑这个别被吞了,哥的分数啊.#includeintmain(){intn,i,j;scanf("%d",&n);for(i=1;i