用数组随机输出一个为数不多于六位的正整数num,反向输出

voidmain(){intm[3][3];inti,j;for(i=0;i{for(j=0;j{printf("请输入第%d行第%d列的数：",i+1,j+1);scanf("%d",&m[i][j

#includeintmain(){floata[10],min,max;inti;printf("pleaseenter10num\n");for(i=0;i

#include#defineCOUNT100main(){inta[COUNT],i,x,no=0;for(i=0;i

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

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

int[]a=new[]{4,7,6,9,3,1,5,8,0};Listb=newList(a);b.Sort(delegate(intx,inty){return(y-x);});最大值：b[0]最

Dima(10)Fori=0To9a(i)=Int(100+100*Rnd)NextiFori=0To8Forj=iTo9Ifa(i)t=a(i)a(i)=a(j)a(j)=tEndIfNextjNe

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

'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

publicclassT{\x05/**\x05*@paramargs\x05*/\x05publicstaticvoidmain(String[]args){\x05\x05int[]arr={\x

functionsj(i%,a%,b%)aslongDimx%x=Int(Rnd*(b-a+1))+asj=xEndfunctionPrivateSubCommand1_Click()Dimn%,a%

#includeintmain(){inta[4][4];for(inti=0;i

publicstaticvoidmain(String[]args){String[]a={"1","2","3","4","5","6","7","8","9","10","j","q","k"};

dima(30)fori=1to30a(i)=int(rnd(97+3))ifisprime(a(i))=truethenprinta(i);nextifunctionisprime(xasinteg

/>很高兴为你解答答案是：#include <stdio.h> #include <stdlib.h>#include <time

其实二维数组在电脑内存中是连续的.例如:inta[][5]={{1,2,3,4,5},{6,7,8,9,10}};for(inti=0;i

晕.不用那么麻烦.先转化为list（为什么不一开始就用List呢?）例：String[]arr=newString[]{"1","2"};Listlist=Arrays.asList(arr);直接调

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(