You can solve this using a method like greedy. The
rule of this game is very easy. You always try to put a ball in the left most possible peg. If you can not put a
ball in the given number of peg then the last ball number is the result.
i.e. First try to
put ball 1 on peg 1 then try to put the ball 2 in peg 1 if possible,
otherwise try on peg 2. For the ball 3 try in peg 1 and so on. A
interesting thing of this problem is, all the possible input have a
finite number of balls. So you do not need to print �1 for any of the
input !!!!!!!!!!!! believe me�. There was no line in my code like
printf(�-1\n�); J
And all the output fit in 32-bit integer.
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