循环输出1到100之间只能被1和这个数本身整除的数-搜答案-搜狗做题网

count==3时会打印两遍可以改成：if ( count == 3 ) { System.o

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

DimiAsIntegerFori=0To100i=i+1NextResponse.Write(i)

#includeusingnamesapcestd;voidmain(){inthe=0;for(inti=1;i

Console.WriteLine("-----------------100以内的质素-----------------------");ArrayListaryList=newArrayList(

inti=1;do{if(i%3==0){continue;}elseprintf("%d",i);i++;}while(i

dimi!,m!Fori=1To100IfkMod3=0ThenPrintk;m=m+1IfmMod5=0ThenPrintEndIfEndIfNextk其实用多一个累加器``就是每循环一次M就加一`

#includeusingnamespacestd;intmain(){intsum=0;for(inti=1;i

#includeintmain(){inti;for(i=1;i{if(i%2==0)printf("%d\n",i);}return0;}

下面是本人的实验的一个题目,经测试,运行成功,GoodLuck.DECLARE@sumint,@iintSET@sum=1SET@i=1WHILE@i

intjishuhe=0;intoushuhe=0;for(inti=1;i

先是for循环的：#include#includevoidmain(){inti,j,sign=0;doublek;for(i=1000;i

PrivateSubCommand1_Click()Fori=3To100Forj=2ToSqr(i)IfiModj=0ThenExitForNextIfj>Int(Sqr(i))Thens=s&i&

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

inti,sum;i=1,sum=0;while(i

用C++语言么?那应该很简单for(inti=1;icout

publicstaticvoidmain(String[]args){for(inti=0;i

main(){ints=0,a=2;while(1){s+=a*a;if(a>=100)break;a=a+2;}printf("%d",s);}//如果有问题就是S的值可能要LONG吧,如果是LON

PrivateSubForm_Click()DimaAsIntegerDimbAsIntegerDimcAsIntegerDimSAsIntegerS=2（循环无法找出2因为2是唯一一个能被自己整除的

includemain(){inti,j=0;for(i=1;i