hdu1249

纯规律题：找到n=3时，分割数为20，从而可发现2，8, 20之间差的关系正好是6的倍数。

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     #include
    
 
    
 
    
     
     
     
     "
    
 
    
 
    
     
     
     
     iostream
    
 
    
 
    
     
     
     
     "
    
 
    
 
    
     
     
     
     

    
 
    
 
    
     
     
     
     using
    
 
    
 
    
     
     
     
      
    
 
    
 
    
     
     
     
     namespace
    
 
    
 
    
     
     
     
      std;

    
 
    
 
    
     
     
     
     int
    
 
    
 
    
     
     
     
      a[
    
 
    
 
    
     
     
     
     10000
    
 
    
 
    
     
     
     
     ];

    
 
    
 
    
     
     
     
     int
    
 
    
 
    
     
     
     
      main()
{
 
    
 
    
 
    
     
     
     
     int
    
 
    
 
    
     
     
     
      t;
 
    
 
    
 
    
     
     
     
     int
    
 
    
 
    
     
     
     
      n;
 cin
    
 
    
 
    
     
     
     
     >>
    
 
    
 
    
     
     
     
     t;
 
    
 
    
 
    
     
     
     
     while
    
 
    
 
    
     
     
     
     (t
    
 
    
 
    
     
     
     
     --
    
 
    
 
    
     
     
     
     )
 {
 cin
    
 
    
 
    
     
     
     
     >>
    
 
    
 
    
     
     
     
     n;
 a[
    
 
    
 
    
     
     
     
     1
    
 
    
 
    
     
     
     
     ]
    
 
    
 
    
     
     
     
     =
    
 
    
 
    
     
     
     
     2
    
 
    
 
    
     
     
     
     ;
 
    
 
    
 
    
     
     
     
     for
    
 
    
 
    
     
     
     
     (
    
 
    
 
    
     
     
     
     int
    
 
    
 
    
     
     
     
      i
    
 
    
 
    
     
     
     
     =
    
 
    
 
    
     
     
     
     2
    
 
    
 
    
     
     
     
     ;i
    
 
    
 
    
     
     
     
     <=
    
 
    
 
    
     
     
     
     n;i
    
 
    
 
    
     
     
     
     ++
    
 
    
 
    
     
     
     
     )
 a[i]
    
 
    
 
    
     
     
     
     =
    
 
    
 
    
     
     
     
     a[i
    
 
    
 
    
     
     
     
     -
    
 
    
 
    
     
     
     
     1
    
 
    
 
    
     
     
     
     ]
    
 
    
 
    
     
     
     
     +
    
 
    
 
    
     
     
     
     6
    
 
    
 
    
     
     
     
     *
    
 
    
 
    
     
     
     
     (i
    
 
    
 
    
     
     
     
     -
    
 
    
 
    
     
     
     
     1
    
 
    
 
    
     
     
     
     );
 cout
    
 
    
 
    
     
     
     
     <<
    
 
    
 
    
     
     
     
     a[n]
    
 
    
 
    
     
     
     
     <<
    
 
    
 
    
     
     
     
     endl;
 }
 
    
 
    
 
    
     
     
     
     return
    
 
    
 
    
     
     
     
      
    
 
    
 
    
     
     
     
     0
    
 
    
 
    
     
     
     
     ;
}