更多“[ ] Petty cash is a large amount of money kept in an office. ”相关问题
  • 第1题:

    阅读下列说明和C++代码,将应填入 (n) 处的字句写在答题纸的对应栏内。
    【说明】
    某大型购物中心欲开发一套收银软件,要求其能够支持购物中心在不同时期推出的各
    种促销活动,如打折、返利(例如,满300返100)等等。现采用策略(Strategy)模式实现该要求,得到如图5-1所示的类图。



    图5-1 策略模式类图

    【C++代码】
    #include
    using namespace std;
    enum TYPE{NORMAL, CASH_DISCOUNT, CASH_RETURN};
    class CashSuper{
    public:
    (1);
    };
    class CashNormal : public CashSuper { //正常收费子类
    public:
    double acceptCash(double money) {
    retum money; }
    };
    class CashDiscount : public CashSuper {
    private:
    double moneyDiscount; // 折扣率
    public:
    CashDiscount(double discount) { moneyDiscount=
    discount; }
    double acceptCash(double money) { retum money *
    moneyDiscount; }
    };
    class CashRetum : public CashSuper { // 满额返利
    private:
    double moneyCondition; // 满额数额
    double moneyReturn; // 返利数额
    public:
    CashRetnm(double
    motieyCondition, double moneyReturn) {

    this->moneyCondition=moneyCondition;

    this->moneyReturn=moneyRetum;
    }



    double acceptCash(double
    money) {
    double result =
    money;

    if(money>=moneyCondition)

    result=money-(int)(money/moneyCondition ) * moneyRetum;
    return
    result ;
    }
    };
    class CashContext {
    private:
    CashSuper *cs;
    public:
    CashContext(int type) {
    switch(type) {
    case
    NORMAL: //正常收费
    (2)
    ;
    break;
    case CASH_RETURN: //满300返100
    (3)
    ;
    break;
    case CASH_DISCOUNT: //打八折
    (4)
    ;
    break;
    }
    }
    double GetResult(double money) {
    (5) ;
    }
    };
    //此处略去main( )函数


    答案:
    解析:
    (1)virtual double acceptCash(double money) = 0
    (2)cs = new CashNormal()
    (3)cs = new CashReturn(300,100)
    (4)cs = new CashDiscount(0.8)
    (5)return cs->acceptCash(money)
    试题分析:
    策模式的结构图如下:

  • 第2题:

    7、BookStore数据库中有销售明细表OrderDetail(OrderCode, BookCode, Amount) 查询销售总量前20%的图书,并按总销量降序排列。SQL语句: SELECT TOP 20 PERCENT Bookcode As 书号, SUM(Amount)As总销量 ROM OrderDetail GROUP BY Bookcode ORDER BY ________

    A.Amount DESC

    B.SUM(Amount) DESC

    C.COUNT(Amount) DESC

    D.ADD(Amount) DESC


    SUM(Amount) DESC

  • 第3题:

    【单选题】BookStore数据库中有销售明细表OrderDetail(OrderCode, BookCode, Amount)查询销售总量前20%的图书,并按总销量降序排列。SQL语句: SELECT TOP 20 PERCENT Bookcode As 书号, SUM(Amount)As总销量 ROM OrderDetail GROUP BY Bookcode ORDER BY ________

    A.Amount DESC

    B.SUM(Amount) DESC

    C.COUNT(Amount) DESC

    D.ADD(Amount) DESC


    B

  • 第4题:

    阅读下列说明和Java代码,将应填入 (n) 处的字句写在答题纸的对应栏内。
    【说明】
    某大型购物中心欲开发一套收银软件,要求其能够支持购物中心在不同时期推出的各种促销活动,如打折、返利(例如,满300返100)等等。现采用策略(Strategy)模式实现该要求,得到如图6-1所示的类图。



    import javA.util.*;
    enum TYPE {
    NORMAL, CASH_DISCOUNT, CASH_RETURN};
    interface
    CashSuper {

    public (1) ;
    }
    class CashNormal
    implements CashSuper{ // 正常收费子类

    public double accptCash(double money){

    return money;
    }
    }
    class
    CashDiscount implements CashSuper {

    private double moneyDiscount;
    // 折扣率

    public CashDiscount(double moneyDiscount) {

    this moneyDiscount = moneyDiscount;
    }

    public double acceptCash(double money) {

    return money* moneyDiscount;
    }
    }
    class CashReturn
    implements CashSuper { // 满额返利

    private double moneyCondition;

    private double moneyReturn;

    public CashReturn(double moneyCondition, double moneyReturn) {


    this.moneyCondition =moneyCondition; // 满额数额

    this.moneyReturn =moneyReturn; // 返利数额
    }

    public double acceptCash(double money) {

    double result = money;

    if(money >= moneyCondition )

    result=money-Math.floor(money/moneyCondition ) *
    moneyReturn;

    return result;
    }
    }
    class
    CashContext_{

    private CashSuper cs;

    private TYPE t;

    public CashContext(TYPE t) {

    switch(t){

    case NORMAL: // 正常收费

    (2) ;

    break;

    case CASH_DISCOUNT: // 满300返100

    (3) ;

    break;

    case CASH_RETURN: // 打8折

    (4) ;

    break;

    }
    }

    public double GetResult(double money) {

    (5) ;
    }
    ∥此处略去main( )函数
    }


    答案:
    解析:
    (1)double acceptCash(double money) (2)cs = new CashNormal()(3)cs = new CashDiscount(0.8)(4)cs = new CashReturn(300,100)(5)return cs.acceptCash(money)

  • 第5题:

    BookStore数据库中有销售明细表OrderDetail(OrderCode, BookCode, Amount) 查询销售总量前20%的图书,并按总销量降序排列。SQL语句: SELECT TOP 20 PERCENT Bookcode As 书号, SUM(Amount)As总销量 ROM OrderDetail GROUP BY Bookcode ORDER BY ________

    A.Amount DESC

    B.SUM(Amount) DESC

    C.COUNT(Amount) DESC

    D.ADD(Amount) DESC


    INSERT INTO OrderDetail(OrderCode, Amount, BookCode) VALUES('08110801',3, '0701');INSERT INTO OrderDetail(OrderCode, BookCode, Amount) VALUES('08110801', '0701', 3);INSERT INTO OrderDetail VALUES('08110801', '0701', 3)

  • 第6题:

    在一个模型中,一些产品必须用手中的现金购买,该模型叫作

    A.货币效用(money-in-utility)模型

    B.现金先行(cash-in-advance)模型

    C.实际经济周期(real business cycle)模型

    D.货币意外模型(money surprise)模型


    现金先行 (cash-in-advance) 模型