在具有10个元素的一维整形数组中,讲数组的元素逆序.要求:初始化一维数组
来源:学生作业帮助网 编辑:作业帮 时间:2024/04/20 04:45:14
#include#includevoidmain(){\x05inti,j;\x05floatsuma,sumb;\x05floatavga,avgb;\x05intmaxa,maxb;\x05int
#includeintmain(){\x05inta[10]={1001,1234,2234,2534,4444,6767,7896,6435,1098,8796};\x05inti;\x05intj
voidmain(){inta[30]={1,2};\x05for(inti=2;i
intfun(int*x){inti,k=0;for(i=0;iif(x[i]>x[k])k=i;returnx[k];}再问:运行不出来····再答:#includestaticin
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
OptionExplicitPrivatea(9)AsIntegerPrivateiMaxIndexAsIntegerPrivateSubCommand1_Click()Dimi,j,k,iMaxAs
voidmain(){intarray[]={23,14,36,1,-4,388,90,2000};inti=1;min=array[0];for(;i{if(min>array[i])min=arr
/>privatesubcommand1_click()dima(1to10)asdoubledimminasdoublefori=1to10a(i)=val(inputbox("请输入数组元素值:"
PrivateSubCommand1_Click()Dima(1To10)AsIntegerRandomizeFori=1To10a(i)=Int(Rnd*100)+1Printa(i);NextPr
fun(){inti,average,sun=0,max,min,aa[6]={10,100,20,5,15,30};min=aa[0];max=aa[0];for(i=0;imax)max=aa[i
#includemain(){inta[10];inti;for(i=0;i
是不是要这样啊.#includeintmain(){inta[3][4]={1,2,3,4,5,6,7,8,9,0,1,2};intb[3][4]={11,21,13,14,51,61,17,18
//程序运行时,请输入10个数.如:0123456789#include#defineN10intmax(int*a,intn){\x09inti,m=a[0];\x09for(i=1;im)m=a[
#includevoidmain(){inta[10]={1,2,3,4,5,6,7,8,9,10};intb[6]={2,3,4,5,6,7};intc[10],i,j,k=0,n;for(i=0;
int[]arr={2,4,6,8,10,12,14,16};intsum=0;for(inti=0;i
#includevoidmain(){inta[100],max,min,n,i;scanf("%d",&n);for(i=0;i
#includevoidmain(){inta[10]={2,3,5,1,8,4,7,9,0,6};inti,s;intmax,min;max=min=a[0];for(i=s=0;imax)max=
intfun(intx[N]) {inti,k=0; for(i=0;i
#includevoidinput(inta[],intn)//输入{inti;printf("请输入%d个数,用空格隔开\n",n);for(i=0;i
PrivateSubCommand1_Click()Dima1()AsLongDima2()AsLongDimiAsLongDimjAsLongDimoAsLong,pAsLongDimdistanc