Codeforces 1189E. Count Pairs Codeforces Round #572 (Div. 2)

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Let's transform condition a little bit. ai−aj≢0$a_i-a_j\not\equiv \mathrm{<span\; style="color:\; \#222222;"><span\; style="white-space:\; nowrap;">0</span> mod </span>}$p, so the condition is equivalent: That's why we just need to count the number of pairs of equal numbers in the array 

⇔(ai−aj)(ai+aj)(a2i+a2j)≡(ai−aj)k   //Both side multiple (ai−aj)
⇔a4i−a4j≡kai−kaj

⇔a4i−kai≡a4j−kaj

bi=(a4i−kai)$b_i=(a_i^4-k{a}_i)$ $mod$ p$p$. It's easy to do it, for example, using map. Complexity O(n)$O(n)$ or O(nlog(n))$O(nlog(n))$.

A solution in c++