They put me in mind of _______ good things I might have bought with the rest of the money.A. thatB. whichC. whatD. whichever

题目
They put me in mind of _______ good things I might have bought with the rest of the money.

A. that

B. which

C. what

D. whichever

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更多“They put me in mind of _______ good things I might have bought with the rest of the money. ”相关问题

  • 第1题：

    I ()have been here, but I()not find the time.

    A. could; could

    B. might; could

    C. should; could

    D. should; would


    参考答案：C

  • 第2题：

    I__________have been there,but I__________not find the time.

    A.should；would
    B.should；could
    C.might；could
    D.could；could

    答案：B
    解析：
    考查情态动词的用法。句意为：“我本应该去那里，但我没时间。”should have done表示“过去本应做的事而实际上并没有做”，常含有责备、遗憾的语气。

  • 第3题：

    28、下列代码的输出结果是（）。 function sum（x: number, ...rest: number[]): number { let result = x; for （let i = 0; i ＜ rest.length; i++) { result += rest[i]; } return result; } console.log（sum（1, 2, 3, 4, 5));

    A．运行错误

    B．运行正确，结果为5

    C．运行正确，结果为10

    D．运行正确，结果为15


    C

  • 第4题：

    —I'd met Smith several times before.—So (have) ( ) I.


    正确答案：had

  • 第5题：

    下面的代码用于输出字符数组ch中每个字符出现的次数，应该填入的代码是（）public static void main（String[] args) { char[] ch = { 'a', 'c', 'a', 'b', 'c', 'b' }; HashMap map = new HashMap（); for （int i = 0; i ＜ ch.length; i++) { ＜ 填入代码 ＞ } System.out.println（map); }

    A．if (map.contains(ch[i])) { map.put(ch[i], map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }

    B．if (map.contains(ch[i])) { map.put(ch[i], (Integer) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }

    C．if (map.containsKey(ch[i])) { map.put(ch[i], (int) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }

    D．if (map.containsKey(ch[i])) { map.put(ch[i], (Integer) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }


    if (map.containsKey(ch[i])) { map.put(ch[i], (Integer)map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }

  • 第6题：

    8、下列代码的输出结果是（）。 function sum（x: number, ...rest: number[]): number { let result = x; for （let i = 0; i ＜ rest.length; i++) { result += rest[i]; } return result; } console.log（sum（1, 2, 3, 4, 5));

    A．运行错误

    B．运行正确，结果为5

    C．运行正确，结果为10

    D．运行正确，结果为15


    0 1 4 9 16

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