Evaluate 0 ~ The total number of 1 in all num

  # Include  <  Stdio. h  >  
# Include  <  Math. h  >  
  Int  Fun (  Int  Num)
{
  Int  K =  0  ;
  Int  Sum  =  0  ;

  Int  Pw_k0  =  (  Int  ) Pow (  10  , K );  //  The value of pw_k0 is 10 ^ K. 
    Int  Pw_k1  =  (  Int  ) Pow (  10  , K  +  1  );  //  Pw_k1 is 10 ^ (k + 1)  
  
  While (Num  -  Pw_k0)  > =  0  )  //  If num is less than 10 ^ K, it indicates that there is no value on the K bit, that is, output.  
  {
  Int  X  =  (Num  %  Pw_k1  - Num  %  Pw_k0)  /  Pw_k0;  //  Take the number on the K-bit,  
    If  (  0     =  X)
{
Sum  =  Sum +  Num  /  Pw_k1  *  Pw_k0;  //  When the number on K bit is 0, the number of all numbers on this bit is 1.  
  }
  Else     If  (  1     = X)
{
Sum  =  Sum  +  Num  /  Pw_k1  *  Pw_k0  +  Num  %  Pw_k0  +     1  ;  // When the number on the K bit is 1, the number of all numbers on this bit is 1.  
  }
  Else  
{
Sum  =  Sum  +  Num  /  Pw_k1  *  Pw_k0  +  Pw_k0;  // When the number in the K bit is greater than 1, the number of all numbers in this bit is 1.  
  }
K  ++  ;
Pw_k0  =  (  Int  ) Pow (  10  , K );
Pw_k1  =  (  Int  ) Pow (  10  , K +  1  );
}
  Return  SUM;
}

  Int  Main ()
{
Printf (  "  % D \ n  "  , Fun (  11  ));
  Return     0  ;
}