随机100个数,统计被7整除的个数VB-搜答案-百度作业网

privatesubform_click()dimnasinteger,iasintegerfori=7to1000ifimod7=0thenn=n+1nextprintnendsu

3515再问：求过程，c语言的再答：你大一？再问：中职再答：是妹子？再问：不是再问：大哥帮帮忙再问：大哥帮帮忙再答：按说应该很简单的啊再答：你怎么学的？再问：刚学再答：看书啊再答：看书啊再问：而且现在

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

21X5=10521X21=44121-5+1=17个

Dimi,tjAsIntegertj=0Fori=100To10000IfiMod7=0AndiMod3=0Thentj=tj+1NextMsgBox"个数为"&tj

PrivateSubCommand1_Click()Dima(50),i,nRandomizeFori=1To50a(i)=Int(Rnd*191)+10Printa(i);IfiMod10=0The

$num=array('9','21','12','55','100');foreach($sumas$s){if($s%3==0||$s%7==0){echo$s;}}

还要写吗?再问：不会啊，怎么写呢？再答：你等下，我帮你写一个再答：参考代码：#include#includeintmain(intargc,char*argv[]){\x09inti,j=0;\x09

很简单.#includevoidmain(void){inti;intcount=0;/*计数*/for(i=100;i

//C#代码privatevoidForm1_Load(objectsender,EventArgse){//取出符合条件的数Listnums=newList();for(inti=100;i

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

voidmain(){inti;intmark=0;for(i=1;i

PrivateSubCommand1_Click()Fori=1To100IfiMod7=0Thenn=n+1EndIfNextiPrintnEndSu

#include"stdio.h"voidmain(){inta[10];inti=0;for(i=0;i{scanf("%d",&a[i]);}for(i=0;i{if(a[i

这个程序不难,不知你要统计什么呢?

intcount=0;for（i=1;i

#include#includemain(){inta[30];intnCount=0,i=0;for(i=1;i

#includeintmain(){intcount=0,i;for(i=0;i{if(i%7==0&&i%11!=0){count++;}}printf("%d\n",count);return0;

Java程序：importjava.util.Random;publicclassTest{\x05\x05\x05publicstaticvoidmain(String[]args){\x05\x0

能被4整除的有(2008-1000)/4+1=253个其中能被100整除的有(2000-1000)/100+1=11个所以能被4整除但不能被100整除的数有253-11=242个