• 2022-06-19 问题

    The keyword "unsigned" can modify the keyword [ ] A: signed B: long double C: long D: float

    The keyword "unsigned" can modify the keyword [ ] A: signed B: long double C: long D: float

  • 2022-06-19 问题

    [音频] Ithink the college life will become wonderful as long as you _____________.

    [音频] Ithink the college life will become wonderful as long as you _____________.

  • 2022-06-06 问题

    The tunnel is only ______. It's not very long.[音频] A: 8 meters B: 80 meters C: 18 meters D: 800 meters

    The tunnel is only ______. It's not very long.[音频] A: 8 meters B: 80 meters C: 18 meters D: 800 meters

  • 2022-06-06 问题

    Q 18 How long has the company been in business?Over ____.[音频]

    Q 18 How long has the company been in business?Over ____.[音频]

  • 2022-07-25 问题

    已知long a[3][8]; 则a数组中共有_____个元素.

    已知long a[3][8]; 则a数组中共有_____个元素.

  • 2022-06-11 问题

    ​表示水平线的大小的HTML代码是( )。‎ 未知类型:{'options': ['[hr size=?]', '[hr long=?]', '[hr height=?]', '[hr space=?]'], 'type': 102}

    ​表示水平线的大小的HTML代码是( )。‎ 未知类型:{'options': ['[hr size=?]', '[hr long=?]', '[hr height=?]', '[hr space=?]'], 'type': 102}

  • 2022-06-11 问题

    7.表示水平线的大小的HTML代码是( )。 未知类型:{'options': ['[hr size=?]', '[hr long=?]', '[hr height=?]', '[hr space=?]'], 'type': 102}

    7.表示水平线的大小的HTML代码是( )。 未知类型:{'options': ['[hr size=?]', '[hr long=?]', '[hr height=?]', '[hr space=?]'], 'type': 102}

  • 2022-06-25 问题

    This dog is a great pet.If you ____ ____ ____ of him,he can have avery long lifespan. [音频]

    This dog is a great pet.If you ____ ____ ____ of him,he can have avery long lifespan. [音频]

  • 2022-06-06 问题

    Q 18 How long has the company been in business? Over ____. [音频]/module/audioplay.html?objectid=8deb1d4a7593d98cc2de8c7aaf4840f5

    Q 18 How long has the company been in business? Over ____. [音频]/module/audioplay.html?objectid=8deb1d4a7593d98cc2de8c7aaf4840f5

  • 2022-07-24 问题

    利用递归方法求n的阶乘。 long Fac(int n) { if(n==1) __[填空1]__;//递归头 else __[填空2]_;//递归调用自身 }

    利用递归方法求n的阶乘。 long Fac(int n) { if(n==1) __[填空1]__;//递归头 else __[填空2]_;//递归调用自身 }

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