---- I’m sorry I can’t help you----- ___________.A. Not at all. B. It doesn’t matterC. Thank you all the same D. I’m sorry to hear that

题目

---- I’m sorry I can’t help you

----- ___________.

A. Not at all. B. It doesn’t matter

C. Thank you all the same D. I’m sorry to hear that

相似考题

1.－ May I help you, madam? － ().A、Sorry, I have no ideaB、Yes, I'd like 2 kilos of orangesC、You'd better give me a hand

2.“Must I finish my homework now?” “No, you() . But you() finish it this evening. ”A. mustn’ t…mustB. needn’ t…mustC. can’ t…canD. can’ t…must

3.6． —Let’ s play volleyball．—Sorry， I _________a volleyball．A． haveB． don’t haveC． not haveD． doesn’t have

4.－－Must we finish our homework now? －－No, you ______. You _______ finish it after school.A.needn't , mayB.mustn't , mayC.can't , canD.needn't , must

更多“－－－－ I’m sorry I can’t help you －－－－－ ___________.A. Not at all. B. It doesn’t matterC. ”相关问题

  • 第1题：

    -Mayweleavetheclassroomnow?

    -No,you______.You______toleaveuntilthebellrings.

    A.mustn’t,areallowed

    B.don’thaveto,aresupposed

    C.needn’t,aren’tallowed

    D.can’t,aren’tsupposed


    正确答案：D

  • 第2题：

    The price you quoted is so high that we () help ()this transaction.

    A、can…to cancel

    B、can't...canceling

    C、can…canceling

    D、can't…to cancel


    参考答案：B

  • 第3题：

    I put my coat here? Sorry, you（）

    AMay; mustn't

    BDo; don't

    Ccan; needn't

    DMay; can't


    D
    略

  • 第4题：

    Can you go skating with us this afternoon?

    Sorry,I can't. I()take care of my little sister at home.

    A. could

    B. may

    C. have to


    参考答案：C

  • 第5题：

    下列程序段的时间复杂度为()。for(i=0；i＜m；i++)for(j=0；j＜t；j++)e[i][j]=0；for(i=0；i＜m；i++)for(j=0；j＜t；j++)for(k=0；k＜n；k++)c[i][j]_c[i][j]+a[i][k]×b[k][j]；

    A.O(m×n×t)
    B.O(m+n+t)
    C.O(m×t+n)
    D.O(m+n×t)

    答案：A
    解析：
    在程序段中，有两段循环程序，第一段是一个双层嵌套循环，另一个是三层嵌套循环，所以基本操作是c[i][j]=c[i][j]+a[i][k]×b[k][j]，此基本操作共执行m×t×n次。

  • 第6题：

    下列程序段的时间复杂度是（）。 for（i=0; i＜m; i++) for（j=0; j＜t; j++) c[i][j]=0; for（i=0; i＜m; i++) for（j=0; j＜t; j++) for（k=0; k＜n; k++) c[i][j]=c[i][j]+a[i][k]*b[k][j];

    A．O(m+n+t)

    B．O(m+n*t)

    C．O(m*t+n)

    D．O(m*n*t)


    O(n^2)

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