Counting game (ultraviolet)

counting game

There are n people standing in a line, playing a famous game called ''counting ". when the game begins, the leftmost person says ''1 "loudly, then the second person (people are numbered 1 to n from left to right) says ''2 "loudly. this is followed by the 3rd person saying ''3 "and the 4th person says ''4", and so on. when the n -th person (I. e. the rightmost person) said'' n "loudly, the next turn goes to his immediate left person (I. e. the ( n -1) -th person ), who shoshould say '' n + 1 " loudly, then the ( n -2) -th person shoshould say '' n + 2 " loudly. after the leftmost person spoke again, the counting goes right again.

There is a catch, though (otherwise, the game wocould be very boring !) : If a person shoshould say a number who is a multiple of 7, or its decimal representation contains the digit 7, he shoshould clap instead! The following tables shows us the counting processN= 4('X' represents a clap). When the 3rd Person Claps for the 4th time, he's actually counting 35.

person	1	2	3	4	3	2	1	2	3
action	1	2	3	4	5	6	x	8	9
person	4	3	2	1	2	3	4	3	2
action	10	11	12	13	x	15	16	x	18
person	1	2	3	4	3	2	1	2	3
action	19	20	x	22	23	24	25	26	x
person	4	3	2	1	2	3	4	3	2
Action	X	29	30	31	32	33	34	X	36

GivenN,MAndK, Your task is to find out, whenM-Th Person Claps forK-Th time, what is the actual number being counted.

Input

There will be at most 10 test cases in the input. Each test case contains three IntegersN,MAndK(

2N100,

1MN,

1K100) In a single line. The last test case is followed by a lineN=M=K= 0, Which shoshould not be processed.

Output

For each line, print the actual number being counted, whenM-Th Person Claps forK-Th time. If this can never happen, print'-1'.

Sample Input

 
4 3 14 3 24 3 34 3 40 0 0
 

Sample output

 
17212735
 

The seventh Hunan Collegiate Programming Contest
Problemsetter:Rujia Liu,Special thanks:Yiming Li & Jane Alam Jan

 

 

 # Include  <  Stdio. h  >  
# Include  < String  . H  >  
  Bool  AA [  1000000  ];
  Bool  Solve (  Int  N)
{
  If  (N  %  7  =  0 )  Return     True  ;
  Int  T  =  N;
  While  (T)
{
  If  (T  %  10  =  7  ) Return     True  ;
T  /=  10  ;
}
  Return     False  ;
}
  Void  Calc ()
{
  For  (  Int  I =  1  ; I  <  1000000  ; I  ++  )
{
  If  (Solve (I) AA [I]  =  True  ;
  Else  AA [I]  =  False ;
}

}
  Int  Main ()
{
  Int  N, m, K;
  Int  T;
  Int  CNT;
Calc ();
  While  (Scanf (  "  % D  "  ,  &  N, &  M,  &  K)  ! =  EOF)
{
  If  (N  =  0  &&  M  =  0  &&  K  =  0 )  Break  ;
T  =  M;
CNT  =  0  ;
  While  (  1  )
{
  If  (AA [T]) CNT  ++  ;
  If (CNT  =  K)
{
Printf (  "  % D \ n  "  , T );
  Break  ;
}
  If  (N  =  M)
{
T  + =  2  * (M  -  1  );
  Continue  ;
}
  If  (M  =  1  )
{
T  + =  2  *  (N  -  M );
 Continue  ;
}
T  + =  2  *  (N  -  M );
  If  (AA [T]) CNT  ++  ;
  If  (CNT  =  K)
{
Printf ( "  % D \ n  "  , T );
  Break  ;
}

T  + =  2  *  (M  -  1  );
}
}
  Return     0  ;
}