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It was great for a while

Name: Anonymous 2009-02-18 6:38

Please try to ignore troll posts.

http://userscripts.org/scripts/show/40415

Name: Anonymous 2009-02-19 20:54

>>34
well, basically what i'm trying to do is this:
f (g . h $ div a b) (h $ mod a b)
the best i've been able to come up with is:
(\(x,y) -> f (g $ h x) (h y)) $ divMod a b

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