随机产生100*100的由0和1组成的矩阵

在前面includesrand(time(0));for(i=1;i

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)

import java.util.Random;public class EchoTest {    public&nb

A=randint(1,1000)randint（M,N）是产生由0和1两个数组成的M*N维矩阵这样的随机数产生函数还有randn(M,N)产生标准正太分布的随机数rand(M,N)产生均匀分布随机数

?STR(A(I),3)改为?STR(A(i),3)?"最大数是："+Mmax+"最小数是：",Mmin改为：?"最大数是：",Mmax,"最小数是：",Mmin

帮你写好了,这个是工程文件,下载下来就可以了.以下是源码:PrivateSubCommand1_Click()DimA(10)AsDouble,LAsInteger,BAsDouble,CAsDoub

PrivateSubCommand1_Click()DimIAsInteger,NAsInteger,MaxAsInteger,MinAsIntegerMin=100RandomizeForI=1To

#includevoidmain(){inta[100],i;for(i=0;i

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

privatesubcommand1_click()dimjs,os,i,afori=1to20a=int(Rnd*99)+1ifamod2=0thenjs=js+ielseos=os+inextpr

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

PrivateSubCommand1_Click()Dima(100)AsInteger,xAsIntegerFori=1To100Ifsushu(i)ThenPrinti;IfiMod10=0The

RND默认产生0至1之间的随机数,既0

PrivateSubCommand1_Click()Dima(29)AsInteger,tmpAsIntegerDimiAsInteger,jAsIntegerRandomizeTimeFori=0T

#include<stdio.h>#include<stdlib.h>#include<time.h>#defineM20voidmain(){inta[M],i;

Int(101*Rnd)

选择法：PrivateSubForm_Click()Dimi%,j%,a%(1To10)RandomizeFori=1To10a(i)=Int(101*Rnd)NextiFori=1To9k=iFor

回答：随机生成1次出现那个组合的概率是1/(2^5000)≈1x10^(-1505.15).随机生成100次,出现那个组合的次数的均值是100/(2^5000)≈1x10^(-1503.15).这个数

Java程序：importjava.util.Random;publicclassTest{\x05\x05\x05publicstaticvoidmain(String[]args){\x05\x0