Problem A of the recently held CodeJam Round 1A has the following official analysis. I however coded an O(1) solution.
Did anyone else do this directly like this?
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cin >> n >> pd >> pg;
bool possible;
if ( pd != 0 && pg == 0 || pd != 100 && pg == 100) possbile = false;
else possible = (100/__gcd(pd, 100) <= n);
But it's described in analysis