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She returned home from the office , only ()the door open and something() .A. finding ; missedB. to find; missedC. to find ; missingD. finding ; losing

题目

She returned home from the office , only ()the door open and something() .

A. finding ; missed

B. to find; missed

C. to find ; missing

D. finding ; losing


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更多“She returned home from the office , only ()the door open and something() .A. finding ”相关问题
  • 第1题:

    He ______ that he had been in prison twice before.

    A、revealed

    B、revised

    C、rewrote

    D、returned


    参考答案:A

  • 第2题:

    Mary is looking for the book she lost yesterday.

    A:trying to find
    B:looking up
    C:looking at
    D:finding

    答案:A
    解析:

  • 第3题:

    设ADOrs为Recordset对象,从“学生基本情况表”中获取所有记录的语句是()。

    AADOrs.Open“select * From 学生基本情况表”

    BADOrs.Execute“select * From 学生基本情况表”

    CADOrs.New“select * From 学生基本情况表”

    DADOrs.Select “select * From 学生基本情况表”


    A

  • 第4题:

    阅读以下说明和C++代码。

    【说明】

    传输门是传输系统中的重要装置。传输门具有Open(打开)、Closed(关闭)、Opening (正在打开)、StayOpen(保持打开)和Closing(正在关闭)五种状态。触发传输门状态转换的事件有click、complete和timeout三种。事件与其相应的状态转换如下图所示。

    下面的C++代码1与C++代码2分别用两种不同的设计思路对传输门进行状态模拟,请填补代码中的空缺。

    【C++代码1】

    const int CLOSED=1; const int PENING=2;

    const int PEN=3; const int CLOSING=4;

    const int STAYOPEN=5; //定义状态变量,用不同整数表示不同状态

    class Door {

    Private:

    int state; //传输门当前状态

    void setState(int state){ this->state=state;} //设置当前状态

    public:

    Door():state(CLOSED){};

    void getState(){ //根据当前状态输出相应的字符串

    switch(state){

    case OPENING: cout<<"OPENING"<<endl; break;

    case CLOSED: cout<<"CLOSED"<<endl; break;

    case OPEN: cout<<"OPEN"<<endl; break;

    case CLOSING: cout<<"CLOSING"<<endl; break;

    case STAYOPEN:cout<<"STAYOPEN"<<endl; break;

    }

    };

    void click() { //发生click事件时进行状态转换

    if ((1)) setState(OPENING);

    else if ((2)) setState(CLOSING);

    else if ((3)) setState(STAYOPEN);

    }

    void timeout(){ //发生timeout事件时进行状态转换

    if (state == OPEN) setState(CLOSING);

    }

    void complete(){ //发生complete事件时进行状态转换

    if (state == OPENING) setState(OPEN);

    else if (state == CLOSING) setState(CLOSED);

    }

    };

    int main(){

    Door aDoor;

    aDoor.getState();aDoor.click(); aDoor.getState();

    aDoor.complete();aDoor.getState(); aDoor.click();

    aDoor.getState();aDoor.click(); aDoor.getState(); return 0;

    }

    【C++代码2】

    class Door {

    public:

    DoorState *CLOSED,*OPENING,*OPEN,*CLOSING,*STAYOPEN,*state;

    Door();

    virtual~Door(){……//释放申请的内存,此处代码省略);

    void setState(DoorState *state) { this->state = state;}

    void getState(){

    //此处代码省略,本方法输出状态字符串,

    //例如,当前状态为CLOSED时,输出字符串为“CLOSED”

    };

    void click();

    void timeout();

    void complete();

    };

    Door::Door(){

    CLOSED = new DoorClosed(this); OPENING = new DoorOpening(this);

    PEN = new DoorOpen(this); CLOSING = new DoorClosing(this);

    STAYOPEN = new DoorStayOpen(this);state = CLOSED;

    }

    void Door :: click() {(4);)

    void Door :: timeout() {(5);)

    void Door :: complete() {(6);}

    class DoorState//定义一个抽象的状态,它是所有状态类的基类

    {

    protected:Door *door;

    public:

    DoorState(Door *door) {this->door = door;}

    virtual~DoorState(void);

    virtual void click() {}

    virtual void complete(


    正确答案:(1)state == CLOSED || state == CLOSING (2)state == OPENING || state == STAYOPEN (3)state == OPEN (4)state->click() (5)state->timeout() (6)state->complete() (7)door->setState(door->OPENING)
    (1)state == CLOSED || state == CLOSING (2)state == OPENING || state == STAYOPEN (3)state == OPEN (4)state->click() (5)state->timeout() (6)state->complete() (7)door->setState(door->OPENING) 解析:本题考查的是状态转换图的程序设计与实现。
    空(1)、(2)和(3)需要根据状态转换图来填写,空(1)、(2)和(3)所在的方法为click,表示当发生click事件时应该发生什么状态转换。根据代码可知,发生click事件时,状态分别跳转到OPENING,CLOSING和STAYOPEN,则发生click前的状态由状态转换图可以得到,分别为CLOSED或CLOSING、STAYOPEN或OPENING以及OPEN。
    代码2中空(4)、(5)和(6)考查当发生click、timeout以及complete事件的时候,状态应该如何迁移。类Door的state成员变量用于记录类Door所处的状态,而state变量的类型为DoorState *,DoorState中分别具有click、timeout和complete方法用来响应对应的事件,因此,空(4)、(5)和(6)分别为:state->click()、state->timeout()和 state->complete()。
    空(7)主要考查门的当前状态为CLOSED时,发生click事件时状态的迁移。根据状态图可知,CLOSED状态的在click事件下将迁移到OPENING,因此,此处应该将传输门状态设置为OPENING,DoorState变量存储了当前其存储的传输门的实例,因此,可直接调用其方法setState来设置状态,由于传输门状态采用类的实例变量表示,所以此处应该填写door->setState(door->OPENING)。
    代码1和代码2的区别是:代码2将状态间的转换规则封装到具体的类中,当状态转换图的转换规则发生变化时,只需更改部分对应类中的状态迁移规则,而代码1中的迁移规则散落在程序中,维护起来较为困难。

  • 第5题:

    There was an inclination to treat geography as a less important subject.

    A:point
    B:tendency
    C:result
    D:finding

    答案:B
    解析:
    本句意思:曾经有-种倾向认为地理是一门次要的学科。inclination意思为“倾向,趋势”,与tendency(趋势,倾向)意思相近。point论点,观点,要点;result后果,结果;finding调查发现,调查结果。