1.编写一个递归函数,计算下面的级数:f(n)=1 2 2 3 n n 1
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/12 01:18:28
你应该定义成doublegetPower(doublex,inty)就可以了目前你的定义的情况下getPower(b,m)找不到最匹配的就是(double,int)只能找次匹配的,找到了(double
帮你写好了.unsigned int fib(unsigned int n) {\x09if (n == 1
// C++int F(int n) {if (n == 0) return 1;else if
doublef(doublex){doubles=.0;if(x
longadd(intn){intt=n-1;if(t>1){longresult=n*t;longsum=result+add(t);returnsum;}else{returnn;}}楼上的方法,
#includedoubleH(intn,doublex){if(x>1){if(n==0)return1.0;//H0(x)=1;if(n==1)return2.0*x;//H1(x)=2x;//直
cludestdio.hvoidmain(){intmax_4(inta,intb,intc,intd);inta,b,c,d,max;printf("Pleaseenterintergernumbe
#includeusingnamespacestd;doublefun(intn,doubleh){if(n==1)returnh;elseif(n再问:usingnamespacestd;这句
#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(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
#include <stdio.h>char* dg(char* instr, char* outstr, char* 
C描述functionttt(n){ returnn>1?n*ttt(n-1):1;}使用方法:ttt(21);
#includeintfact(int);main(){inti,sum=0;for(i=1;i
functiongqj=erfen(p,a,b,e)ifabs(b-a)
intSumNums(intnum){if(num
1.#include"stdio.h"//#defineRECURSION1#ifdefRECURSIONlongfact(intn){if(n
#includemain(){intn,i,j,k;while(scanf("%d",&n)==1){if(n==1||n==2){printf("%d\n",1);cont
if(n==1)\x05\x05return1;你可以改成n==0或者改成returnx取其中一种就可以了
#include"stdio.h"doublesum(intn){if(n>1){return(double)n/(double)(2*n+1)+sum(n-1);}else{return(doubl
#include <iostream>using namespace std;long fact(int n){ &