given an array consists of $$$n$$$ integers and an integer $$$p$$$, find number of pair($$$i,j$$$):
so that $$$i<j$$$ and $$$(a[i]*a[j])$$$ $$$mod$$$ $$$p$$$ $$$=$$$ $$$(a[i]+a[j])$$$ $$$mod$$$ $$$p$$$
$$$n<=10^5$$$
$$$1<p<=10^9$$$
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given an array consists of $$$n$$$ integers and an integer $$$p$$$, find number of pair($$$i,j$$$):
so that $$$i<j$$$ and $$$(a[i]*a[j])$$$ $$$mod$$$ $$$p$$$ $$$=$$$ $$$(a[i]+a[j])$$$ $$$mod$$$ $$$p$$$
$$$n<=10^5$$$
$$$1<p<=10^9$$$
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Im not sure if this really works, but this is what I thought: rearranging the expression, we have $$$a_i \cdot a_j - (a_i+a_j) \equiv 0 \mod{p}$$$, which is equivalent to $$$(a_i-1)(a_j-1) \equiv 1 \mod{p}.$$$ This implies that $$$a_i \equiv a_j \equiv 0$$$ or $$$2 \mod{p}$$$. Then, let $$$x$$$ be the number of $$$a_k$$$ such that $$$a_k \equiv 0 \mod{p}$$$ and $$$y$$$ the number of $$$a_l$$$ such that $$$a_l \equiv 2 \mod{p}$$$. Answer is $$${x \choose 2} + {y \choose 2}$$$.
actually, this is not correct. You can pair every $$$a_i$$$ with any $$$a_j \equiv (a_i-1)^{-1}+1 \mod{p}$$$.
please give the problem link
http://online.vku.udn.vn/problem/tongtich
you can translate it into english