编写一个递归函数,输入若干个整数并累加求和,输入0则算法结束

publiclongpower(intm,intn){if(nreturnm;elsereturnpower(m,n--)*m;}

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

#include#defineSIZE20voidmain(){printf("请输入10个整数：\n");inta[SIZE];intn=10,i,j,k,b;for(i=0;i

很高兴回答你的问题!#includeintfun(intn,intx){if(n==0)return1;intmul=x;for(inti=n;i>1;i--)mul*=x;returnfun(n-1

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

doublemax(doublea,doubleb,doublec){doublem;if(a>b)m=a;elsem=b;if(c>m)m=c;returnm;}

#includemain(){intn;printf("Pleaseenteranumber:");scanf("%d",&n);if(n%2==0)printf("Thenumber%dis偶数\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

publicclassA{publicstaticvoidmain(Stringargs[]){intn=9;//任意自然数System.out.println(f(n));}publicintf(i

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

先定义n1到n5五个数来计数,初始化为0然后使用If语句来判断输入的是哪个数,并在相应的ni上加1最后就可看到输入的数的个数了!祝你编程顺利!

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

#includeclasscircle{private:intr;public:voidinput(){cout

自带函数mean.还需要自己写函数么?如果是你就把mean函数调出来看看,然后简化一下就可以了.

#include#includeinti=0;voidisPro(intn){if(n==0)return;else{i*=10;i+=n%10;isPro(n/10);}}v

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