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I've become good friends with several of the students in my school I met in theEnglish speech contest last year.A. whoB. whereC. whenD. which

题目

I've become good friends with several of the students in my school I met in theEnglish speech contest last year.

A. who

B. where

C. when

D. which

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3.乳牙萌出的顺序为A．I→Ⅱ→Ⅳ→Ⅲ→VB．I→Ⅱ→Ⅲ→]V→VC．I→Ⅲ→Ⅱ→Ⅳ→VD．Ⅱ→I→Ⅲ→Ⅳ→VE．Ⅱ→I→Ⅳ→Ⅲ→V

4.I, my, will, true, come, hope, dream__________________________________________________________________________.

更多“I've become good friends with several of the students in my school I met in theEnglish ”相关问题

  • 第1题：

    I've _____ these lines again and again but I still keep forgetting them.

    A. gone off

    B. gone about

    C. gone over

    D. gone through


    正确答案：C

  • 第2题：

    下面程序输出的结果是()。includeusing namespace std;void main(){ char ch[][8]={"g

    下面程序输出的结果是( )。 #include＜iostream＞ using namespace std; void main() { char ch[][8]={"good","better","best"}; for(int i=1;i＜3;++i) { cout＜＜ch[i]＜＜endl； } }

    A．good better

    B．better best

    C．good best

    D．good


    正确答案：B
    解析：二维数组ch共3行8列，for循环语句输出第2、3行的数组元素

  • 第3题：

    ――I've got your invitation．Oh，good．――＿＿＿＿

    A.Can you come?
    B.Thanks a lo
    C.I'll take i
    D.May I help you?

    答案：A
    解析：
    暂无解析

  • 第4题：

    I won’t discuss the ______ report since time is limited. I’ll just hit the high spots.

    A. short

    B. new

    C. entire

    D. good


    参考答案：C

  • 第5题：

    B 宽度优先（种子染色法）

    5．关键路径

    几个定义： 顶点1为源点，n为汇点。

    a. 顶点事件最早发生时间Ve[j], Ve [j] = max{ Ve [j] + w[I,j] },其中Ve (1) = 0;

    b. 顶点事件最晚发生时间 Vl[j], Vl [j] = min{ Vl[j] – w[I,j] },其中 Vl(n) = Ve(n);

    c. 边活动最早开始时间 Ee[I], 若边I由<j,k>表示，则Ee[I] = Ve[j];

    d. 边活动最晚开始时间 El[I], 若边I由<j,k>表示，则El[I] = Vl[k] – w[j,k];

    若 Ee[j] = El[j] ，则活动j为关键活动，由关键活动组成的路径为关键路径。

    求解方法：

    a. 从源点起topsort,判断是否有回路并计算Ve;


    正确答案：

     

     

  • 第6题：

    3. Good. I'll _________ （有空）then.


    B

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