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There is a nice way to do it in O(n^(2/3)) time, which is less than 1s for n=2*10^10.
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Project Euler #251: Cardano Triplets
You are viewing a single comment's thread. Return to all comments →
There is a nice way to do it in O(n^(2/3)) time, which is less than 1s for n=2*10^10.