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PDM includes four types of dependencies or precedence relationships.( ),the initiation of the successor activity,depends upon the initiation ofthe predecessor activity. A.Start-to-Start B.Finish-to-Finish C.Start-to-Finish D.Finish-to—Start

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PDM includes four types of dependencies or precedence relationships.( ),the initiation of the successor activity,depends upon the initiation ofthe predecessor activity. A.Start-to-Start B.Finish-to-Finish C.Start-to-Finish D.Finish-to—Start


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  • 第1题:

    在一个NSD的Definition文件夹中,会有哪两个文件,用来定义网络业务的节点类型和属性?()

    A.TOSCA-Normative-Types-Definitions.yaml

    B.vercfg.xml

    C.NSD-Types-Definitions.yaml

    D.service.desc


    答案:AC

  • 第2题:

    有以下程序: #include<stdio.h> main( ) { char a[5][10]={"one","tw0","three","four","five"}; int i,j; char t: 、 for(i=0;i<4;i++) for(j=i+1;j<5;j++) if(a[i][O]>a[j][0]) {t=a[i][O];a[i][O]=a[j][O];a[j][O]=t;) puts(a[1]); } 程序运行后的输出结果是( )。

    A.fwo

    B.fix

    C.two

    D.owo


    正确答案:A
    For循环完成的功能是把二维数组a的第一列的字母按从小到大排序,其他列的字母不变。

  • 第3题:

    The question "How fast did he drive when he ran the red light?" __________ "He ran the red light".

    A.presupposes
    B.entails
    C.contradicts
    D.includes

    答案:A
    解析:
    考查语句之间的涵义关系。A项表示预设关系,预设通常被认为是一种背景信息,也就是说话者假设听者已经知道的信息。题干中第二句he ran the red light(他闯红灯了)是第一句How fast did he drive when he ran the red light (他闯红灯时车速多少 )的预设(即前提条件),预设信息是他闯红灯了。故答案选A。B项表示蕴含关系,C项表示自相矛盾,D项表示内包关系。

  • 第4题:

    If you take the()to learn a new skill, you will grasp it quickly.

    A. Initiating

    B. initiative

    C. initiation


    参考答案:B

  • 第5题:

    Which Behavior Aggregate (BA) is used to classify traffic arriving on an LSP?()

    A. EXP

    B. DSCP

    C. 802.1P

    D. inet-precedence


    参考答案:A

  • 第6题:

    下列给字符串二维数组进行赋值的语句中,错误的是()。

    A.String s[ ] [ ] = new String [ ] [ ] { { “One “ , “ Two “ }, { “ Three “ , “ Four “ } } ;

    B.String s[ ] [ ] = { { “ One “ , “Two “},{ “ Three “ , “ Four “ } } ;

    C.String s[ ] [ ] = new String [ ] [ ] { { “Zero”} , { “ One ” , “Two” , “ Three” , “ Four” } } ;

    D.String s[ 2] [2 ] = { { “ One ” , “Two”},{“ Three” , “ Four” } } ;


    String s[ 2] [2 ] = { { “ One ” , “Two”},{“ Three” , “ Four” } } ;