网友您好, 请在下方输入框内输入要搜索的题目:

题目内容 (请给出正确答案)

I'm used to()up early now.

A. get

B. getting

C. to get


参考答案

更多 “ I'm used to()up early now.A. getB. gettingC. to get ” 相关考题
考题 在SQL SERVER中,下列哪些语句属于给局部变量赋值的语句() A.DECLAREB.SETC.USED.GET

考题 阅读以下说明和C++程序,将应填入(n)处的字句写在对应栏内。【说明】字符串在程序设计中扮演着重要角色。现需要设计字符串基类string,包含设置字 符串、返回字符串长度及内容等功能。另有一个具有编辑功能的串类edlt_string,派生于string,在其中设置一个光标,使其能支持在光标处的插入、删除操作。【程序】include <iostream.h>include <stdio.h>include <string.h>class string{int length;char *data;public:int get_length() {return length;}char *get_data() {return data;}~string() {delete data;}int set data(int in_length, char *in_data);int set_data(char *data);void print() {cout<<data<<endl;}};class edit_string: public string{int cursor;public:int get_cursor() {return cursor;}void move_cursor(int dis) {cursor=dis;}int add_data(string *new_data);void delete_data(int num);};int string::set_data(int in_length,char *in_data){length=in_length;if(!data)delete data;(1)strcpy(data,in_data);return length;}int string::set data(char *in_data){(2)if(!data)delete data;(1)strcpy(data,in_data);return length;}int edit_string::add_data(string *new_data){int n,k,m;char *cp,*pt;n=new_data->get_length();pt=new_data->get_data();cp=this->get_data();m=this->get_length();char *news=new char[n+m+1];for(int i=0; i<cursor; i++)news[i]=cp[i];k=i;for(int j=0; j<n; i++,j++)news[i]=pt[j];cursor=i;for(j=k; j<m; j++,i++)(3)news[i]='\0';(4)delete news;return cursor;}void edit string::delete_data( int num){int m;char *cp;cp=this->get_data();m=this->get_length();for(int i=cursor; i<m; i++)(5)cp[i]='\0';}

考题 —What do you think of the _______car?—I think it is much better than his one.A. latest, lastB. lately, lastlyC. early, latelyD. early, lastly

考题 I() packing in wooden cases. A、thinkB、preferC、get

考题 I_______ to get up late, but now I get up early every morningl.A、used toB、is used toC、useD、use to

考题 PHP中超级全局变量有()。 A.$_GETB.$_POSTC.$_PUTD.$GLOBALS

考题 m个人的成绩存放在score数组中,请编写函数fun(),它的功能是将高于平均分的人数作为函数值返回,将高于平均分的分数放在叩所指的数组中。例如,当score数组中的数据为24,35,88,76,90,54, 59,66,96时,函数返回的人数应该是5,up中的数据应为88, 76, 90, 66, 96。注意:部分源程序给出如下。请勿改动主函数main和其他函数中的任何内容,仅在函数fun的花括号中填入所编写的若干语句。试题程序:include <conio.h>include <stdio.h>include <string.h>int fun(int score[],int m,int up[]){}main(){int i,n,up[9];int score[9]={24,35,88,76,90,54,59,66,96};clrscr();n=fun(score,9,up);printf("\nup to the average score are:");for(i=0;i<n;i++)printf("%d",up[i]);}

考题 By the time l_________ (get) up,my mother _________ (get) the breakfast ready.

考题 3.—Mr Zhang,I'm not feeling well. I-d like to have this afternoon _________ to go to a doctor.—OK.A.onB.offC.downD.up

考题 下面的代码用于输出字符数组ch中每个字符出现的次数,应该填入的代码是()public static void main(String[] args) { char[] ch = { 'a', 'c', 'a', 'b', 'c', 'b' }; HashMap map = new HashMap(); for (int i = 0; i < ch.length; i++) { < 填入代码 > } System.out.println(map); }A.if (map.contains(ch[i])) { map.put(ch[i], map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }B.if (map.contains(ch[i])) { map.put(ch[i], (Integer) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }C.if (map.containsKey(ch[i])) { map.put(ch[i], (int) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }D.if (map.containsKey(ch[i])) { map.put(ch[i], (Integer) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }