编写程序生成随机数并求得随机数组的最大值

公式是=RAND()*0.03-0.02选中生成的数据,设置小数位数为3位.

可以试试这个公式=ROUND(RAND()+LOOKUP(9E+307,$A$2:A2)-1,2)如果需要有一个数与基础数相同,则可=ROUND(IF(A2="",RAND()+LOOKUP(9E+3

char[] chs = { '0', 'a' };Random random =&

#include#includevoidmain(void){floaty[]={0.18,0.00,0.16,0.14,0.04,0.06,0.10,0.10,0.14,0.08};floats;i

#include　　#include　　#include/*用到了time函数,所以要有这个头文件*/　　intmain(void)　　{　　intnumber[15]={1,2,3,4,5,6,7,

while(a){case‘1‘intn=rand（）%10;break;case‘1‘intn=10+rand（）%100;break;case‘1‘intn=100+rand（）%1000;bre

importjava.util.Random;publicclassDemo{publicstaticvoidmain(Stringargs[]){//定义一个数组存放随机数：int[]randoms

#include#include#includeintmain(void){srand((unsignedint)time(NULL));printf("%d%%\n",3+rand()%7);

PrivateSubCommand1_Click()'IntegerRandomizeLabel1.Caption=Int(Rnd*(101)-50)EndSubPrivateSubCommand2_

importjava.awt.*;importjava.applet.*;importjava.util.*;publicclassdrawTestextendsApplet{publicvoidpa

#include#include#include#defineN10voidmain(){inta[20]={0},b[N],x,i,j,t;srand((unsignedint)time(NUL

dima1,a2a1=int（10+rnd（99））a2=a1=int（10+rnd（99））a3=(a1+a2)/2printa3;

intnum=(int)(Math.random()*100);这是生存的随机数,边界值用判断去掉你要有数组存放1-100数值对应的英文然后根据得到的随机数在数据中去英文然后输出.

publicclassTest{publicstaticvoidmain(String[]args)throwsException{intone=(int)Math.ceil(Math.random(

intlen=10;int[]arr=newint[len];for(inti=0;iintt=(int)(Math.random()*10000);System.out.println(t);arr

publicclassRandomTest{publicstaticvoidmain(String[]args){inta;java.util.Randomr=newjava.util.Random(

两种方法1补充你的算法,PrivateSubCommand1_Click()Dima(10)AsInteger,nAsIntegerDimiAsInteger,kAsIntegerRandomizea

#include"stdio.h"//#include//#include//voidmain(void){inta[10]={0,},i,max,min,n=100000000,ave=n/10;s

我HI你吧,等一下

这个没什么难度吧t=1:20;r=3*rand(1,20);plot(t,r)a=polyfit(t,r,5);y=polyval(a,t);holdonplot(t,y,'r')legend('随机