java求1-999之间能被3整除的数的个数,以及平均值-搜答案-神马作文网

publicclassTest{publicstaticvoidmain(String[]args){intsum=0;for(inti=1;i

publicclassA3{publicstaticvoidmain(String[]args){for(inti=1;i

我简单写了一个应该符合你的要求：publicclasstests{publicstaticvoidmain(String[]args){intn=0;for(inti=0;i

如果是85个数字满足的话,那么应该是100-999,不是1-999#include <stdio.h>#include <stdlib.h>//求被3整除且

importjava.math.BigDecimal;publicclassTest{publicstaticvoidmain(String[]args){BigDecimalmul=newBigDe

publicclassTest{publicstaticvoidmain(Stringargs[]){for(inti=2,j;i<=100;i++){intk=(int)Math.sqrt(i

clearn=0;forx=1:100ifmod(x,3)==0n=n+1;disp(x)endenddisp(n)

publicstaticvoidmain(String[]args){ListarrayList=newArrayList();for(inti=1;i

--求1到50之间能被3整除的数的和DECLARE@iINT,@sumINTSELECT@i=1,@sum=0WHILE@i

我的代码：publicclassTest{publicstaticvoidmain(String[]args){inti=1;intsum=0;while(i

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

public class Test{ public static void main(String[] a

publicclassSumOfEven{\x09publicstaticvoidmain(String[]args){\x09\x09intsum=0;\x09\x09for(inti=1;i

classMain{publicMain(){}publicstaticvoidmain(Stringagrs[]){inti,sum=0;for(i=1;i

800/3=266余2能被3整除的数有266个800/8=100能被8整除的数有100个800/（3*8）=33余8能同时被3和8整除的数有33个所以能被3,8整除的数有：266+100-33=333

importjava.util.Random导入Random包Randomrandom=newRandom();inta=random.nextInt(6)+1;nextInt是产生在0（包括）和指定

publicclassqiuhe{publicstaticvoidmain(String[]args){intsum1=0;intsum2=0;for(inti=1;i

import java.util.ArrayList;public class Test {public static void

intsum=0;for(inti=1;i

publicclassSum{intsum(){inti=1;intsum=0;while(i