I completely forgot () the front door last night and feel fortunate that nothing is stolen.A.lockingB.being lockedC.to lockD.to have locked

题目
I completely forgot () the front door last night and feel fortunate that nothing is stolen.

A.locking

B.being locked

C.to lock

D.to have locked

相似考题

1.– Oh, God! I’ve made such a mistake! –()(A) Oh, nothing!(B) Never mind. We all make mistakes.(C) You should blame yourself.(D) Don’t worry.

2.______________[A] shared [B] forgot [C] attained [D] rejected

3.4．I __________my teacher carefully，but l __________nothing．A． listened，listenedB． heard，listened toC．listened to，heardD． listened to，listened

4.to, I, how, last, winter, skate, learned__________________________________________________________________________.

更多“I completely forgot () the front door last night and feel fortunate that nothing is stolen. ”相关问题

  • 第1题：

    I _____ but _____ nothing.

    A was listened;was hearing

    B listened;heard

    C have listened;heard

    D listened; heard of


    参考答案：B

  • 第2题：

    What a nice surprise? What brings you here? ( )

    A、It’s really a surprise.

    B、I’m here on vacation.

    C、Nothing.

    D、My brother.


    参考答案：B

  • 第3题：

    2、在顺序结构表示的线性表中，删除第i个元素（数组下标为i-1)，需要把后面的所有元素都往前挪一位，相应的语句是： for （___________ ) PtrL-＞Data[j-1]=PtrL-＞Data[j]; 其中空缺部分的内容应该是

    A．j = i; j＜ = PtrL-＞Last; j++

    B．j =PtrL-＞Last; j＞= i; j--

    C．j = i-1; j＜ = PtrL-＞Last; j++

    D．j =PtrL-＞Last; j＞= i-1; j--


    O(n)

  • 第4题：

    整箱货的交接方式有()。

    A.CY-CY

    B.CY-DOOR

    C.DOOR-CY

    D.DOOR-DOOR


    正确答案：ABCD

  • 第5题：

    template int SeqList::Insert(Type &x, int i) {

    if (i<0 || i>last+1 || last== MaxSize-1) return 0;

    else {

    Last++;

    for(int j=last;j

    data[i]=x;

    return 1;

    }

    }

    对于结点类型为SeqList的顺序表，以上算法的功能为：()


    正确答案：：顺序表插入算法

  • 第6题：

    在顺序结构表示的线性表中，删除第i个元素（数组下标为i-1)，需要把后面的所有元素都往前挪一位，相应的语句是： for （___________ ) PtrL-＞Data[j-1]=PtrL-＞Data[j]; 其中空缺部分的内容应该是

    A．j = i; j＜ = PtrL-＞Last; j++

    B．j =PtrL-＞Last; j＞= i; j--

    C．j = i-1; j＜ = PtrL-＞Last; j++

    D．j =PtrL-＞Last; j＞= i-1; j--


    O(n)

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