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It can be done also without floats. Let x be the difference between the two watches after one day (in seconds) x = |a-b| So the difference between two watches after one second (y) will be y = x / 86400 Both watches will show the same time after the difference between their times will be 12h, that is...

I haven't check if 'long' is enough, but I've used instead 'long long' (it is 64bit integer, 2x bigger than normal long which is 32bit), and I got AC. Some parts of my program: [c] long long dog_x, dog_y; long long gopher_x, gopher_y; int can_escape( long long x, long long y) { long long dog_l, goph...

I haven't analyzed code too hard, but it is possible that
you haven't consider two equal circles, one over second.

Hauser

I've used long longs, not doubles, and I've got AC.
(You can simply operate on numbers multiplied by 1000).
Also, you don't have to use sqrt function, or something. You can simply
compare squares of numbers.


Hauser