// C++ program to get number of Parallelograms we   // can make by given points of the plane   #include          using     namespace     std  ;   // Returns count of Parallelograms possible   // from given points   int     countOfParallelograms  (  int     x  []     int     y  []     int     N  )   {      // Map to store frequency of mid points      map   <  pair   <  int       int  >       int  >     cnt  ;      for     (  int     i  =  0  ;     i   <  N  ;     i  ++  )      {      for     (  int     j  =  i  +  1  ;     j   <  N  ;     j  ++  )      {      // division by 2 is ignored to get      // rid of doubles      int     midX     =     x  [  i  ]     +     x  [  j  ];      int     midY     =     y  [  i  ]     +     y  [  j  ];      // increase the frequency of mid point      cnt  [  make_pair  (  midX       midY  )]  ++  ;      }      }      // Iterating through all mid points      int     res     =     0  ;      for     (  auto     it     =     cnt  .  begin  ();     it     !=     cnt  .  end  ();     it  ++  )      {      int     freq     =     it  ->  second  ;      // Increase the count of Parallelograms by      // applying function on frequency of mid point      res     +=     freq  *  (  freq     -     1  )  /  2  ;      }      return     res  ;   }   // Driver code to test above methods   int     main  ()   {      int     x  []     =     {  0       0       2       4       1       3  };      int     y  []     =     {  0       2       2       2       4       4  };      int     N     =     sizeof  (  x  )     /     sizeof  (  int  );      cout      < <     countOfParallelograms  (  x       y       N  )      < <     endl  ;      return     0  ;   }   

   /*package whatever //do not write package name here */   import     java.io.*  ;   import     java.util.*  ;   public     class   GFG     {          // Returns count of Parallelograms possible      // from given points      public     static     int     countOfParallelograms  (  int  []     x       int  []     y       int     N  )      {      // Map to store frequency of mid points      HashMap   <  String       Integer  >     cnt     =     new     HashMap   <>  ();      for     (  int     i  =  0  ;     i   <  N  ;     i  ++  )      {      for     (  int     j  =  i  +  1  ;     j   <  N  ;     j  ++  )      {      // division by 2 is ignored to get      // rid of doubles      int     midX     =     x  [  i  ]     +     x  [  j  ]  ;      int     midY     =     y  [  i  ]     +     y  [  j  ]  ;      // increase the frequency of mid point      String     temp     =     String  .  join  (  ' '       String  .  valueOf  (  midX  )     String  .  valueOf  (  midY  ));      if  (  cnt  .  containsKey  (  temp  )){      cnt  .  put  (  temp       cnt  .  get  (  temp  )     +     1  );      }      else  {      cnt  .  put  (  temp       1  );      }      }      }      // Iterating through all mid points      int     res     =     0  ;      for     (  Map  .  Entry   <  String       Integer  >     it     :     cnt  .  entrySet  ())     {      int     freq     =     it  .  getValue  ();      // Increase the count of Parallelograms by      // applying function on frequency of mid point      res     =     res     +     freq  *  (  freq     -     1  )  /  2  ;      }      return     res  ;      }          public     static     void     main  (  String  []     args  )     {      int  []     x     =     {  0       0       2       4       1       3  };      int  []     y     =     {  0       2       2       2       4       4  };      int     N     =     x  .  length  ;      System  .  out  .  println  (  countOfParallelograms  (  x       y       N  ));      }   }   // The code is contributed by Nidhi goel.    

   # python program to get number of Parallelograms we   # can make by given points of the plane   # Returns count of Parallelograms possible   # from given points   def   countOfParallelograms  (  x     y     N  ):   # Map to store frequency of mid points   cnt   =   {}   for   i   in   range  (  N  ):   for   j   in   range  (  i  +  1     N  ):   # division by 2 is ignored to get   # rid of doubles   midX   =   x  [  i  ]   +   x  [  j  ];   midY   =   y  [  i  ]   +   y  [  j  ];   # increase the frequency of mid point   if   ((  midX     midY  )   in   cnt  ):   cnt  [(  midX     midY  )]   +=   1   else  :   cnt  [(  midX     midY  )]   =   1   # Iterating through all mid points   res   =   0   for   key   in   cnt  :   freq   =   cnt  [  key  ]   # Increase the count of Parallelograms by   # applying function on frequency of mid point   res   +=   freq  *  (  freq   -   1  )  /  2   return   res   # Driver code to test above methods   x   =   [  0     0     2     4     1     3  ]   y   =   [  0     2     2     2     4     4  ]   N   =   len  (  x  );   print  (  int  (  countOfParallelograms  (  x     y     N  )))   # The code is contributed by Gautam goel.    

   using     System  ;   using     System.Collections.Generic  ;   public     class     GFG   {      // Returns count of Parallelograms possible      // from given points      public     static     int     CountOfParallelograms  (  int  []     x       int  []     y       int     N  )      {      // Map to store frequency of mid points      Dictionary   <  string       int  >     cnt     =     new     Dictionary   <  string       int  >  ();      for     (  int     i     =     0  ;     i      <     N  ;     i  ++  )      {      for     (  int     j     =     i     +     1  ;     j      <     N  ;     j  ++  )      {      // division by 2 is ignored to get      // rid of doubles      int     midX     =     x  [  i  ]     +     x  [  j  ];      int     midY     =     y  [  i  ]     +     y  [  j  ];      // increase the frequency of mid point      string     temp     =     string  .  Join  (  ' '       midX  .  ToString  ()     midY  .  ToString  ());      if     (  cnt  .  ContainsKey  (  temp  ))      {      cnt  [  temp  ]  ++  ;      }      else      {      cnt  .  Add  (  temp       1  );      }      }      }      // Iterating through all mid points      int     res     =     0  ;      foreach     (  KeyValuePair   <  string       int  >     it     in     cnt  )      {      int     freq     =     it  .  Value  ;      // Increase the count of Parallelograms by      // applying function on frequency of mid point      res     +=     freq     *     (  freq     -     1  )     /     2  ;      }      return     res  ;      }      public     static     void     Main  (  string  []     args  )      {      int  []     x     =     {     0       0       2       4       1       3     };      int  []     y     =     {     0       2       2       2       4       4     };      int     N     =     x  .  Length  ;      Console  .  WriteLine  (  CountOfParallelograms  (  x       y       N  ));      }   }   

   // JavaScript program to get number of Parallelograms we   // can make by given points of the plane   // Returns count of Parallelograms possible   // from given points   function     countOfParallelograms  (  x       y       N  )   {      // Map to store frequency of mid points      // map  int> cnt;      let     cnt     =     new     Map  ();      for     (  let     i  =  0  ;     i   <  N  ;     i  ++  )      {      for     (  let     j  =  i  +  1  ;     j   <  N  ;     j  ++  )      {      // division by 2 is ignored to get      // rid of doubles      let     midX     =     x  [  i  ]     +     x  [  j  ];      let     midY     =     y  [  i  ]     +     y  [  j  ];      // increase the frequency of mid point      let     make_pair     =     [  midX       midY  ];      if  (  cnt  .  has  (  make_pair  .  join  (  ''  ))){      cnt  .  set  (  make_pair  .  join  (  ''  )     cnt  .  get  (  make_pair  .  join  (  ''  ))     +     1  );      }      else  {      cnt  .  set  (  make_pair  .  join  (  ''  )     1  );      }      }      }      // Iterating through all mid points      let     res     =     0  ;      for     (  const     [  key       value  ]     of     cnt  )      {      let     freq     =     value  ;      // Increase the count of Parallelograms by      // applying function on frequency of mid point      res     =     res     +     Math  .  floor  (  freq  *  (  freq     -     1  )  /  2  );      }      return     res  ;   }   // Driver code to test above methods   let     x     =     [  0       0       2       4       1       3  ];   let     y     =     [  0       2       2       2       4       4  ];   let     N     =     x  .  length  ;   console  .  log  (  countOfParallelograms  (  x       y       N  ));   // The code is contributed by Gautam goel (gautamgoel962)