灵鹊做题网编写涵数fun, 函数功能:求1到m之间的偶数之积
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/08 09:19:24
intfun(intn){inta=n,b=0;while(a>0){b=b*10;b=b+a%10;a=a/10;}printf("%d",b);getch();return0;}或者把后三行删掉,
floatfun(intn){\x09inta[30];\x09floatsum=2/1;\x09a[0]=2;\x09a[1]=3;\x09for(inti=2;i再问：floati;doublea
PrivateSubfun()ifimod2=0thenforx=itoastep2s=s*xnextxelseforx=(i+1)toastep2thens=s*xnextxendif试试看行不行吧
#includedoublefun(doubleh){returnfloor(h*100+0.5)/100;}
includeincludeddoublefun(inta,intb,intc){intp;p=(a+b+c)/2;returnsqrt(p*(p-a)*(p-b)*(p-c));}再问：ok再问：
#include#defineN10voidmain(){inta[N];inti,j=0;printf("输入原数组：\n");for(i=0;i
PrivateFunctionSumA(ByValaAsInteger)AsLongDimiAsInteger,NAsIntegerFori=500To600IfiModa=0ThenN=N+iNex
#include#includeintfun(intn){intsum=0,i;for(i=1;i
上面的错了应该是intfun(){inti,sum=1;for(i=1;i
#include#includevoidswap(intc[],intlen){inti=0;inttmp;for(;i{tmp=c[i];c[i]=c[len];c[len]=tmp;}}intmu
#include <stdio.h>#include <stdlib.h>#include <math.h>double f
#includeinta[50];intfun(intm){intn=0;for(inti=1;i
请楼主参考采纳intfun(intt){inti;intcurrent=0;for(i=1;totalt)break;current+=i;}returncurrent;}
Functions(ByValaAsInteger,ByValbAsInteger)AsIntegerDimiAsIntegerFori=1To1000IfiModa=0AndiModb=0Thens
1.intf1=0,f2=1,f3;2.returnf3;继续写：f3=f1+f2;f2=f3;f1=f2;
#include#includedoublefun(intn)//计算1!+2!+3!+.+n!,并赋值给fun{inti;doubles=0,t=1;for(i=1;i
#include#includefloatfun(inta,intb){floatc;c=sqrt(a)+sqrt(b);returnc;}intmain(){inta=12,b=20;floatc;
functionfun(d,h){if(d
intfun(w){intsum=0;for(inti=1;i