随机产生10个50以内的正整数从大到小排序

Dima%(1To20)Fori=1To20a(i)=Int(Rnd*90+10)Printa(i);NextPrintFori=1To20Ifa(i)Mod2=1ThenPrinta(i);EndI

OptionBase1OptionExplicitDima(11)AsIntegerPrivateSubCommand1_Click()DimiAsIntegerPicture1.ClsPicture

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

command1完成第一题,command2完成第二题PrivateSubCommand1_Click()Dimx%(1To10),n%,i%n=1DoWhilen

#include <stdio.h>#include <stdlib.h>#include <time.h>int mks(

这样才对初始化应加在For循环里s要变为双精度（这样平均数才有小数点后几位）楼主试我这个：PrivateSubCommand1_Click()Dimi%,j%,max%,min%,s%Fori=1To

你拉三个文本框,一个按钮text1和text1拉高一点,MultiLine设置为true,排列如图所示Private Sub Command1_Click()  

我的是最简洁的程序算法,哈哈,整个算法只有一个20次的循环,精简明朗快速DimRanValue(100)AsInteger'100个临时数据来计算重复DimiValueAsIntegerDimiCou

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

PrivateSubCommand1_Click()Dima(10)RandomizeDimsumAsInteger,minAsInteger,maxAsIntegersum=0:min=100:ma

一下代码!你参考一下!PrivateSubCommand1_Click()ClsDima()AsInteger'定义一个动态数组Dimb(10)AsInteger'用于统计Randomizen=4:m

#include<stdio.h>#include<math.h>#include<stdlib.h>#include<time.h>int 

你是郑大的吧,这道题我也刚做完哦PrivateSubCommand1_Click()Dima(20)AsIntegerDimb()Print"20个随机数：";Fori=1To20a(i)=Int(R

subform_click()dima%(20)fori=1to20a(i)=int(rnd()*10+1)printa(i),nextendsu

importjava.util.Random;publicclassTest{publicstaticvoidmain(String[]args){Randomr=newRandom();int[]a

publicstaticvoidmain(String[]args){int[]numbers=newint[10];intsum=50;Randomrandom=newRandom();for(in

#include <stdio.h>#include <stdlib.h>#include <time.h>bool isp

C++?C#?再问：C++再答：#include#include#include#include#includeusingnamespacestd;voidgenerator(intnums[],un

kk(i,1)=rnd*90+10'生成20个100以内的两位正整数10

vb6测试通过PrivateSubCommand1_Click()Dima(10)AsInteger,tempAsInteger,iAsInteger,jAsIntegerRandomizetemp=