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城市變化的作文

時間:2025-11-28 11:12:06 變化 我要投稿

關(guān)于城市變化的作文(通用20篇)

  筆試題目篇一

關(guān)于城市變化的作文(通用20篇)

  1.判斷一年是否為閏年的方法

  2.非遞歸實現(xiàn)exists方法

  template

  class BinaryTree{

  protected:

  T* _data;

  BinaryTree * _left;

  BinaryTree * _right;

  };

  template

  class BST:public BinaryTree{

  public:

  virtual bool exists(const T& item> const;

  };

  3.找到下面的bugs

  #include

  #include

  void foo(int a ,char*b)

  {

  b=(char *)malloc(64);

  snprintf(b,"you are %d years old!/n",a);

  }

  int main (void)

  {

  char *f;

  foo(23,f);

  printf("%s/n",b);

  }

  4.Design a subway traffic control system.What objects/classes/modules would you best use for this

  purpose?Please draw diagrams and write up class definitions to better illustrate your

  design.Write anything additional to clarify your design.

  筆試題目篇二

  Question:

  We have an array representing customer’s shopping records.

  For example, it’s an array like this:

  custA, item1,

  custB, item1,

  custA, item2,

  custB, item3,

  custC, item1,

  custC, item3,

  custD, item2,

  This array indicates that customer A bought item 1, customer B bought item 1, customer A bought item 2, customer B bought

  item 3, etc..

  For a given item X and shopping records array, write code to find out what else (item Y) was bought mostly by the customers

  who bought item X.

  For example, in above example, if X is item 1 then Y should be item 3.

  Rules:

  1. One customer can only buy one item once.

  2. The mostly brought item should not be item X.

  3. If no customer brought item X, then return “None”

  4. If all the customers who brought item X only brought item X, then return “None”

  5. The first line of input is the item X. The second line of input is the shopping record array, this shopping record array is

  split by space.

  6. If there are many other mostly brought items which have equally brought times, then return any one of those items.

  Examples:

  Input1:

  item1

  custA item1 custB item1 custA item2 custB item3 custC item1 custC item3 custD item2

  Output1:

  item3

  Input2:

  item2

  custA item1 custB item1 custC item1 custA item2 custB item3 custA item3

  Output2:

  item1

  (The output2 can be item3 too)

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