用java输出1*2*3*4567

//提供一个方法片段,供参考!Eclipse通过的哦.//和=500500；i0=32.publicvoidshow0(){inti=1,i0=-1,sum=0;while(i500)&&(i0==-

publicclassTest{publicstaticvoidmain(String[]args){System.out.println(show());}publicstaticdoublesho

上面的程序错了publicclassT{publicstaticvoidmain(String[]args){intn=4;//可以修改intn2=n*n;int[][]arr=newint[n][n

importjava.io.*;publicclassTest{publicstaticvoidmain(String[]args)throwsException{BufferedReaderbr=n

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

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

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

//交集String[]a={"1","2","3","4","7","8"};String[]b={"1","2","5","6","9","8"};for(inti=0;i

test{publicstaticvoidmain(Stringargs[]){longsum=0,a=1;longi=1;while(i=1)sum=sum+a;i++;a=a*i;}System.

publicclassFactorial{privateintresult=1;publicintfactorial(intnumber){if(number==1)returnresult;else

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

这句话的意思就是参数J按照+2的数量递增,这个J比循环中的上一个参数要多2,比下一个要少2,如果i=1,那J就要大于等于1,并且小于等于3,以此来限制J,符合要求再问：谢谢,刚刚由于情急,忘记客气点了

inta=0;booleanfind=false;for(inti=0;i500&&!find){find=ture;System.out.println("当i="+i+"时,和大于500");}}

for(intx=1;x

publicstaticvoidmain(String[]args){String[]a={"1","2","3","4","5","6","7","8","9","10","j","q","k"};

给你写了个程序可以实现,比如在主函数输入5,则输出表示5!(120)的数组[0,2,1],其中0表示个位数,2表示十位数,1表示百位数,程序如下：public static voi

intsum=0;for(inti=1;i

int[]arr={1,3,6,10,15,2,5,9,14,4,8,13,7,12,11};inti=0;//使用while遍历while(iif(i!=arr.length-1){System.o

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

publicclassTestArrays{publicstaticvoidmain(String[]args){int[][]a=newint[3][3];int[][]b=newint[3][3]