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更多“Janet used to take part in labor in that village, ()? ”相关问题
  • 第1题:

    They are making a cake_____Janet-it' sher birthday.


    参考答案:for

  • 第2题:

    客(货)比例分摊过程中需要遵守的规则来自于客(货)分摊手册(  )中Section B的多边比例分摊协议(MPA)。
    A.Part 1
    B.Part 2
    C.Part 3
    D.Part 4


    答案:C
    解析:
    分摊手册的公布分为客货两个方面,即《客运分摊手册》(Prorate Manual Passenger,PMP)和《货运分摊手册》(Prorate Manual Cargo,PMC)。客(货)比例分摊过程中需要遵守的规则来自于客(货)分摊手册Part3中Section B的多边比例分摊协议(MPA)。

  • 第3题:

    矩阵A=Array[a,{4,5}],取第1,4两行,2,3,5列元素的子矩阵,使用语句是

    A.Part[A,{1,4},{2,3,5}]

    B.Take[A,{1,4},{2,3,5}]

    C.Part[A,{2,3,5},{1,4}]

    D.Take[A,{2,3,5},{1,4}]


    Part[A,{1,4},{2,3,5}]

  • 第4题:

    Given the following DDL for the PARTS table:CREATE TABLE parts (part_no INT(9) NOT NULL, part_name VARCHAR(24), part_remain INT(9));All part numbers entered will be different and all rows should be displayed in order of increasing part numbers whenever the table is queried. Which of the following create index statements will meet this criteria and require the least amount of storage for the index object?()

    A.CREATE UNIQUE INDEX idx_partno ON parts(part_no)

    B.CREATE UNIQUE INDEX idx_partno ON parts(part_name ASC)

    C.CREATE UNIQUE INDEX idx_partno ON parts(part_name, part_no ASC)

    D.CREATE UNIQUE INDEX idx_partno ON parts(part_no, part_name ASC)


    参考答案:A

  • 第5题:

    在Creo 中,需选用实体零件公制模板是()。

    A.mmns_part_solid

    B.mmns_harn_part

    C.solid_part_mmks

    D.inlbs_part_solid


    mmns_part_solid

  • 第6题:

    1、利用艾拉托斯特尼筛法无穷素数列表primes (参见讲义第8章) 获得介于100000和200000之间素数的方法是

    A.take 200000 (take 100000 primes)

    B.take 100000 (take 200000 primes)

    C.take 100000 (drop 100000 primes)

    D.takeWhile (\x -> x <= 200000) (dropWhile (\x -> x <= 100000) primes)

    E.takeWhile (\x -> x <= 200000) (take 200000 primes)

    F.[x | x <- primes, x >100000, x < 200000]

    G.[x | x <- take 200000 primes, x >100000, x < 200000]


    Sieve(n,P). 输入:正整数n. 输出:小于等于n的所有素数P. ①if n=1 then P← ,计算结束; ②P←{2}; ③a←2; ④if n=a then计算结束; ⑤b←min{a 2 ,n}; ⑥Q←{x|a<x≤b}; ⑦for P中的每一个x ⑧for Q中的每一个y ⑨ then从Q中删去y; ⑩P←P∪Q; (11)a←b; (12)转④.