Use onkeydown (or onkeyup), not onkeypress Use keyCode 90, not 122 function KeyPress(e) { var evtobj = window.event? event : e if (evtobj.keyCode == 90 && evtobj.ctrlKey) alert(“Ctrl+z”); } document.onkeydown = KeyPress; Online demo: http://jsfiddle.net/29sVC/ To clarify, keycodes are not the same as character codes. Character codes are for text (they differ depending on the … Read more
See scipy.special.comb (scipy.misc.comb in older versions of scipy). When exact is False, it uses the gammaln function to obtain good precision without taking much time. In the exact case it returns an arbitrary-precision integer, which might take a long time to compute.
$result = $data1 . $data2; This is called string concatenation. Your example lacks a space though, so for that specifically, you would need: $result = $data1 . ‘ ‘ . $data2;
This problem can be solved with a recursive combinations of all possible sums filtering out those that reach the target. Here is the algorithm in Python: def subset_sum(numbers, target, partial=[]): s = sum(partial) # check if the partial sum is equals to target if s == target: print “sum(%s)=%s” % (partial, target) if s >= … Read more
This answer missed one aspect: the OP asked for ALL combinations… not just combinations of length “r”. So you’d either have to loop through all lengths “L”: import itertools stuff = [1, 2, 3] for L in range(len(stuff) + 1): for subset in itertools.combinations(stuff, L): print(subset) Or — if you want to get snazzy (or … Read more
Art of Computer Programming Volume 4: Fascicle 3 has a ton of these that might fit your particular situation better than how I describe. Gray Codes An issue that you will come across is of course memory and pretty quickly, you’ll have problems by 20 elements in your set — 20C3 = 1140. And if … Read more