26 Free float is the amount of time that an activity may be delayed with out affecting the _____.A. Early start of the succeeding activities.B. Late start of the succeeding activities.C. Project finish.D. Cost of the project.E. None of the above

题目

26 Free float is the amount of time that an activity may be delayed with out affecting the _____.

A. Early start of the succeeding activities.

B. Late start of the succeeding activities.

C. Project finish.

D. Cost of the project.

E. None of the above


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  • 第1题:

    请在下面程序中的每条横线处填写一个语句,使程序的功能完整。程序的输出结果为:

    TIME=1867

    b=9

    i=123

    注意:请勿改动main()主方法和其他已有的语句内容,仅在横线处填入适当的语句。

    源程序文件代码清单如下:

    import java.io.*;

    public class DataType

    {

    public ______ void main(String args[])

    {

    byte b;

    int i;//定义变量i为int类型

    ______//定义常量TIME为整型,且值为1867

    i=123;//给变量i赋值,该值为十进制的123

    ______//给变量赋值,改值为八进制的11,十进制的9

    System.out.println("TIME"+TIME);

    System.out.println("b="+b);

    System.out.println("i="+i);

    }

    }


    正确答案:static final int TIME=1867 b=011
    static final int TIME=1867 b=011 解析:本题主要考查main()主方法的使用以及如何定义常量。static修饰符是说明main()方法是静态方法,final int TIME=1867语句的功能是定义一个整型的常量TIME,并且其值为1867,b=011语句是给变量b赋值为八进制的11,十进制的9。

  • 第2题:

    当在下列()情况时,ULTBF的释放延迟。

    A.没有并发的ULTBF

    B.T_DELAYED_FINAL_PUAN=0

    C.T_DELAYED_FINAL_PUAN<>0

    D.没有并发的DLTBF


    参考答案:C, D

  • 第3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4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.*this

    B.this

    C.&amount

    D.amount


    正确答案:D
    解析: 此题考查的是“+”运算符重载和this指针。语句amount+=a.amount;实现3和7的求和,得到amount=10,要使程序的输出结果为10,需要把amount的值作为函数的返回值,所以横线处应填入amot。

  • 第5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6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