如何产生10个元素的随机数组VB

#include#includevoidmain(){\x05inti,j;\x05floatsuma,sumb;\x05floatavga,avgb;\x05intmaxa,maxb;\x05int

窗体中加命令按钮,单击事件代码如下：PrivateSubCommand1_Click()  Randomize  Dima(1To20)AsInteger&nb

OptionBase1OptionExplicitDima(11)AsIntegerPrivateSubCommand1_Click()DimiAsIntegerPicture1.ClsPicture

PrivateSubForm_Load()Dima(10),i,x,jAsIntegerDimyAsStringFori=0To9Randomizea(i)=Int(Rnd(1)*101)x=x+a(

a1:a10=randbetween(30,100)最大值=max(a1:a10)最小值=min(a1:a10)平均值=average(a1:a10)

Print"平均值是："&平均改为Print"平均值是："&平均(a)再问：为什么这么改啊？原因是什么？再答：PrivateFunction平均(a()AsInteger)AsInteger你这个函数

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

将问题补充完全Dimab(10)AsIntegerDimiAsIntegerFori=1To1ab(i)=Int(Rnd()*100)+1Nextprint"原数组值"fori=1to10printa

#include#include#includeintmain(){srand((unsigned)time(0));intArray[10];inti=0;doublesum=0;intMax=-1

一个两重循环搞定,随机数的函数是rand(),一个例子：/*产生介于1到10间的随机数值,此范例未设随机数种子,完整的随机数产生请参考srand（）*/#includemain(){inti,j;fo

PrivateSubCommand1_Click()Dima(1To10)AsIntegerRandomizeFori=1To10a(i)=Int(Rnd*100)+1Printa(i);NextPr

Dimm(99)AsInteger,n(9)AsInteger,iAsInteger,kAsIntegerRandomizeFori=0To99m(i)=Int(Rnd*100)Printm(i);k

OptionExplicitPrivateSubForm_Click()'定义i循环变量,imax存放最大数,imin存放最小数,数组ishuz下标为1-10DimiAsLong,imaxAsLong

'vb6测试成功OptionExplicitPrivateFunctionmax(b()AsInteger)AsIntegerDimiAsIntegerDimjAsIntegerFori=1To6Fo

#include#include#includeintmain(){inti,a[10],m,t=0,s=0;floatn;srand((unsigned)time(NULL));for(i=0;i

#include#include#includeintmain(){inti,a[10],m,t=0,s=0;floatn;srand((unsigned)time(NULL));for(i=0;i

#include#includevoidmain(){inta[10];inti,j,temp;cout

网上找段冒泡排序

a=round(100*rand(4,4));%生成0-100的随机4*4矩阵a(1,:)%输出第一行...a(:,1)%输出第一列...sum([a(1,:)a(2,1)a(2,4)a(3,1)a(

A=0:10;%产生0到10index=randperm(11);%随机生成0-10在数组中的位置A=A(index);%0到10之间的随机数组Ab1=A(A3)%关系运算找到大于3的数A2=A(A3