编写程序输出1-100之间能被7或者3整除的整数
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![编写程序输出1-100之间能被7或者3整除的整数](/uploads/image/f/6763599-63-9.jpg?t=%E7%BC%96%E5%86%99%E7%A8%8B%E5%BA%8F%E8%BE%93%E5%87%BA1-100%E4%B9%8B%E9%97%B4%E8%83%BD%E8%A2%AB7%E6%88%96%E8%80%853%E6%95%B4%E9%99%A4%E7%9A%84%E6%95%B4%E6%95%B0)
#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