As soon as I () the tools, I shall begin with the work. A.haveB.am havingC.have been havingD.have been had
As soon as I () the tools, I shall begin with the work.
A.have
B.am having
C.have been having
D.have been had
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【C程序】#includestdio.h／*此处为栈类型及其基本操作的定义，省略*／int main()｛STACK station;int state[1000];int n; ／*车厢数*／int begin, i, j, maxNo; ／*maxNo为A端正待入栈的车厢编号*／printf("请输入车厢数：");scanf("%d",n);printf(“请输入需要判断的车厢编号序列（以空格分隔）：”);if（n＜1）return-1;for (i＝0; in; i++) ／*读入需要驶出的车厢编号序列，存入数组state[]*／scanf("%d",state[i]);（1） ; ／*初始化栈*／maxNo＝1;for(i=0; i＜n; ){ ／*检查输出序列中的每个车厢号state[i]是否能从栈中获取*／if( （2） ){ ／*当栈不为空时*／if (state[i]=Top(station)) { ／*栈顶车厢号等于被检查车厢号*／printf("%d",Top(station));Pop(station);i++;｝elseif ( （3） ) {printf(“error\n”）;return 1;｝else｛begin= （4） ;for(j＝begin+l;j =state [i];j++)｛Push(station, j);}｝｝else｛ ／*当栈为空时*／begin＝maxNo；for(j＝begin; j=state[i];j++) {Push(station, j);｝maxNo＝ （5） ;｝｝printf("OK");return 0;｝
33． —Excuse me，where are we going to have our．class meeting?—I'm not sure． Ask our monitor，please． He _________ know．A． canB． mayC． needD． shall
–– Jack: Mommy, when will we have dinner? I'm starving.–– Mother:————————? We have to wait for daddy.A: Quickly, honeyB: OK, honeyC: Soon, honeyD: All right, honey
Ben: Mommy, when will we have dinner? I’m starving.Mother: _______.We have to wait for daddy.A: Quickly, honeyB: All right, honeyC: Soon, honeyD: O.K., honey
WhichofthefollowingisthecommandthatinstallsRecoveryConsoleontheBootmenufromtheWindows2000CD?() A.DRIVELETTER：\TOOLS\RCINSTALL.EXEB.DRIVELETTER：\i386\WINNT32.EXE/RECCONSC.DRIVELETTER：\i386\CMDCONS.EXED.DRIVELETTER：\i386\WINNT32.EXE/CMDCONS
YouarethenetworkadministratorandneedtoinstallWindows2000Professionalon35computers.Youneedtoinstalltherecoveryconsoletheinstallations.YoucreateadistributionfolderonaserverandcopyWindows2000ProfessionalSupportfoldertothedistributionfolder.YoucreateanetworkbootfloppytoinstallW2kProfromthedistributionfolder.Youneedtomakeabatchfile,whichthenetworkbootfloppywillusetostarttheinstallation.Whichcommandwouldyouincludeinthebatchfile:（）A.Winnt32/cmd：z：\i386\winnt\aB.Winnt32/cmd：z：\support\tools\setup.exeC.Winnt/e：z：\i386\winnt32/cmdconsD.Winnt/e：z：\support\tools\setup.exe
阅读下列说明和C代码，回答下列问题。[说明]?? ?采用归并排序对n个元素进行递增排序时，首先将n个元素的数组分成各含n/2个元素的两个子数组，然后用归并排序对两个子数组进行递归排序，最后合并两个已经排序的子数组得到排序结果。?? ?下面的C代码是对上述归并算法的实现，其中的常量和变量说明如下：?? ?arr:待排序数组?? ?P,q,r:一个子数组的位置从P到q，另一个子数组的位置从q+1到r?? ?begin,end:待排序数组的起止位量?? ?left,right:临时存放待合并的两个子数组?? ?n1,n2:两个子数组的长度?? ?i,j,k:循环变量?? ?mid:临耐变量?? ?[C代码]?? ?#inciude＜stdio, h＞?? ?#include＜stdlib, h＞?? ?Define MAX 65536?? ?void merge(int arr [ ],int p,int q,int r) {?? ?int * left,* right;?? ?int n1,n2,I,j,k;?? ?n1=q-p+1;?? ?n2=r-q;?? ?If(left=(int *)malloc((n1+1) * sizeof(int)))=NULL) {?? ?Perror( "malloc error" );?? ?exit11?? ?}?? ?If((right = (int *)malloc((n2+1) * sizeof(int)))=NULL)?? ?Perror("malloc error");?? ?exit 11;?? ?}?? ?for(i=0;i＜n1;i++){?? ?left[i]=arr [p+i];?? ?}?? ?left[i]=MAX;?? ?for(i=0;i＜n2;i++){?? ?right[i]=arr[q+i+1]?? ?}?? ?right[i]=MAX;?? ?i=0;j=0;?? ?For(k=p;______;k++){?? ?If(left[i]＞right[j] {?? ?______?? ?j++;?? ?}else{?? ?arr[k1]=left[i];?? ?i++;?? ?}?? ?}?? ?}?? ?Void merge Sort(int arr[ ], int begin, int end) {?? ?int mid;?? ?if(______){?? ?mid=(begin + end)/2;?? ?merge Sort(arr,begin,mid);?? ?______;?? ?Merge(arr,begin,mid,end);?? ?}?? ?}
下列代码的输出结果是1 2 3 4 5 6 7 8 （） ＜% int[] a=new int[] {1,2,3,4,5,6,7,8}; pageContext.setAttribute（"a",a); %＞ ＜c:forEach items="${a }" var="i" begin="3" end="5" step="2" ＞ ${i } ＜/c:forEach＞