编写一个递归函数,计算下面的级数m(i)=1 1 2-搜答案-智慧作业帮

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

longpower(intm,intn){doublep=1;if(n>0){p=m*power(m,(n-1));returnp;}}voidmain(){intm,n;longk;scanf("%

doublef(doublex){doubles=.0;if(x

#includevoidmain(){intx,y;//你用的是浮点型,我改成整形的了...你如果真的要用浮点的话...要加上小数点,如2.0等...printf("x=");scanf("%d",&

#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

#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

intgetpower(intx,inty){if(y==1)returnx;elsereturnx*getpower(x,y-1);}doublegetpower(doublex,inty){if(