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}{x y} = \frac{1}{c}

x y = c x + c y

- c x - c y + x y = 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: $\mathcal{O}(T * \sqrt{C})$ $O (T * \sqrt{C})$

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