codeforce; avoid using ceil when looking for integer division

If you want to compute n / k such that:
	* if n % k == 0 (n = k * m), then n / k = m
	* if n % k != 0 and m is the biggest integer: k * m < n, then
	  n / k = m + 1

better use n / k = (n + k - 1) / k, instead of
cast_to_integer(ceil(n / cast_to_double(k)))