用java输出输出1000以内的数中能被3整除,且至少有一位数字是5的所有整数.

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

publicclassTest{publicvoiddisplay(){intsum=0;for(inti=0;i

publicclassPrintUpperChar{publicstaticvoidmain(String[]args){for(charM='Z';M>='A';M--){System.out.pr

PrivateSubCommand1_Click()DimiAsLongDimsAsStringFori=1To1000DoEventsIf(iMod37)=0Thens=s&CStr(i)&""Ne

?1!这有什么值?没明白.longsum=0;for(inti=1;i

publicclassCat{publicstaticvoidmain(String[]args){intnum=2;drawDiamond(num);}privatestaticvoiddrawDi

publicclassTest{publicstaticvoidmain(String[]args){intsum=0;inti=2;while(i

按照你的要求写的如下代码,精短而高效,可以直接运行publicclassLuck{\x09publicstaticvoidmain(String[]args){\x09\x09intcount=123

publicclassFibonacci{publicstaticvoidmain(Stringargs[]){inti=1,j=1;for(intn=1;n

publicclassTest{publicstaticvoidmain(String[]args){inta=1203;intsum=0;while(a/10!=0){sum

staticvoidBubbleSort(inta[]){inttemp=0;for(inti=0;ifor(intj=0;jif(a[j]>a[j+1]){//把这里改成大于,就是升序了temp=a

最简单的：publicclassHelloJava{publicstaticvoidmain(String[]args){System.out.println("*******************

PrivateSubCommand1_Click()Fori=1To100Forj=1ToiFork=1TojIfi^2=j^2+k^2ThenPrinti;j;kNextkNextjNextiEnd

import java.awt.*;import java.awt.event.ActionEvent;import java.awt.event.ActionListe

importjava.util.Scanner;publicclassTest{publicstaticvoidmain(String[]args){Scannerinput=newScanner(S

inth=5;//行数intx=65;//开始字符for(inti=1;i

#include "stdio.h"int isPrime(int n){   int i; &nb

packagelogicjava;importjava.util.Scanner;publicclassNumberJudge{publicstaticvoidmain(String[]args){

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

for(inti=10;i再问：if里面应该有三个公式吧再答：呵呵，那不是挪上面去了吗，相当于是for(inti=（2*5）;i