Editorial for HHPC1 P4 - Yet Another A+B Problem - Olympiads Online Judge

Editorial for HHPC1 P4 - Yet Another A+B Problem

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Author: dxke01

The answer for some $c$ ~c~ is equal to $d(c^2)$ ~d(c^2)~ where $d$ ~d~ is the divisor function.

Proof

Rearrange the equation as follows:
$\frac{1}{x} + \frac{1}{y} = \frac{1}{c}$
$\frac{x+y}{xy} = \frac{1}{c}$
$xy = cx + cy$
$-cx - cy + xy = 0$
Using Simon's Favourite Factoring Trick:
$(x - c)(y - c) = c^2$
From here its obvious the answer is just the number of factors of $c^2$

Time Complexity: $\mathcal{O}(T * \sqrt{C})$ ~\mathcal{O}(T * \sqrt{C})~

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