编写程序输出1-100之间能被7或者3整除的整数-搜答案-智慧作业帮

#include<stdio.h>intmain(){ printf("%d",count(100)); &nb

PrivateSubCommand1_Click()DimiAsIntegerDimnAsIntegern=0Fori=1To100IfiMod3=0Thenn=n+1Printn;iEndIfNex

privatesubcommand1_click()dimsumasintegerdimiasintegersum=0fori=1to50ifimod3=0thenmsgboxi&"能被3整除",Vb

clears=0t=0fori=1to100step1ifi%5=0thens=s+it=t+1endifnexti?s,t

两句代码就够了：>> A=100:200;>> A(mod(A,3)~=0)ans = Columns 1&nbs

SubMain()MsgBoxFunc(100,2,5)'按照要求写的例子EndSubRem参数(最大值,被整除的数的数组)FunctionFunc(n&,ParamArrayx())Dimi&,b&

我用C给你编一个for（i=100;i

#include<iostream.h>void main(){ int n; &nbs

declare@iint,@Jsumint,@Osumintset@i=1set@Jsum=0set@Osum=0while(@i

#includevoidmain(){ints=0;for(inti=0;i

#includevoidmain(){inti,sum=0;for(i=3;i

解法1:#includeintF1(){return50*(1+99)/2;}intF2(){return50*(2+100)/2;}voidmain(){printf("Oddnumbersum=%

用C++语言么?那应该很简单for(inti=1;icout

*用vfp编写程序,输出100到200之间能被3整除不能被7整除的所有自然数.SETTALKOFFCLEARFORi=100TO200\x05IFMOD(i,3)=0.and.MOD(i,7)0\x0

inta3=0,a7=0,a37=0;for(inti=1;i

#include"stdio.h"intmain(){intcountou,countji=0;inti=0;for(i=1;i

clears=0fori=0to5000step2ifi%5=0s=s+iendififs>500exitendifendfo

给个代码你publicvoidtest(){intnum=0;intsum=0;for(inti=0;i

用什么写的程序?c语言:#includevoidmain(){inti,j=0;for(i=99;i

这是vb6Dimn%,x%Forn=3To100Step1IfnMod3=0AndnMod10=6Thena=nPrintaEndIfNextn