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[ ] Petty cash is a large amount of money kept in an office.
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更多 “ [ ] Petty cash is a large amount of money kept in an office. ” 相关考题
考题
Do you think it necessary ______ an English diary every day?A. keepB. to keepC. keepingD. kept
考题
有下列程序:includeUsing namespace std;Class Amount{ int amount;public; Amount(i
有下列程序: #include<iostream> Using namespace std; Class Amount{ int amount; public; Amount(int n=O):amount(n){} Int getAmount()const{return amount;} Amount operator+=(AmountA) {A.*thisB.thisC.amountD.amount
考题
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
考题
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
}
考题
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 DESCB.SUM(Amount) DESCC.COUNT(Amount) DESCD.ADD(Amount) DESC
考题
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 DESCB.SUM(Amount) DESCC.COUNT(Amount) DESCD.ADD(Amount) DESC
考题
【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 DESCB.SUM(Amount) DESCC.COUNT(Amount) DESCD.ADD(Amount) DESC
考题
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
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