用java编程,利用数组定义两个3*3的矩阵,并计算这两个矩阵的乘积.

publicclasstest3{publicstaticvoidmain(String[]args){Scannerscan=newScanner(System.in);System.out.pri

importjava.util.Date;importjavax.swing.JFrame;importjavax.swing.JTextField;publicclassTest{\x05publi

publicclassTest{publicstaticvoidmain(String[]args){intarr[]=newint[11];for(inti=10;i

时间很紧,给个思路：主要是两个公式,一个是两点间距离求法,根据坐标得出三角形边长.然后利用海伦公式求面积（只需要边长即可）其他正余弦定理也可.

publicclassaaa{publicstaticvoidmain(String[]args){String[]s={"George","Albert","Tina","Xavier","Roge

PrivateSubCommand1_Click()RandomizeDima(10)AsInteger,imasasinteger,iminasinteger,Averageasintegerima

publicclassDemo{publicstaticvoidmain(Stringargs[]){int[]nums={5,2,45,11,13,32,7,24,63,48};intodd=0,e

intfunc(int*a;intb)//这里a表示数组名,b表示数组的元素个数{intc=0;inti;for(i=0;i

用C11的变长数组,但是最终还是需要确定数组a里面的n值,变长数组最主要是用来做数值计算的.我觉得你提出的这个问题其实没有多大的实际意义,可以考虑用链表.

定义和初始化要分清int[][]array;array=newint[][]{{1,1}{2,2}};int[][]array=newint[2][2];int[][]array=newint[][]

char*s[12]={"Jan.","Feb.","Mar.","Apr.","May.","Jun.","Jul.","Aug.","Sep.","Oct.","Nov.","Dec."};int

importjava.util.Scanner;publicclassFirst{publicstaticvoidmain(String[]args){System.out.print("Please

package com.ectrip.swing.main;import java.util.Scanner;public class Test {\

#include <stdio.h>#include <stdlib.h>int main(){  char *a[

import java.util.Scanner;public class ex4_9{   public static 

看看谁的效率高过我importjava.util.Arrays;importjava.util.InputMismatchException;importjava.util.Scanner;publi

classClock{privateinthourse;privateintminute;privateintsecond;publicClock(){this.hourse=0;this.minut

Mapmap=newHashMap();for(inti:num){Integerc=map.get(i);if(c==null){map.put(i,1);}else{map.put(i,c+1);

看的出来你是在考试呢.呵呵intsum=0;for(inti=1;i再问：嘻嘻……再答：娶我吧~啦啦啦珍竹意坊你的最佳答案在这呢啦啦啦再问：好的……呵呵我都不知道选哪个好

publicclassTest{publicstaticintfib(intn){if(n==1||n==2)return1;elsereturnfib(n-1)+fib(n-2);}publicst