线性求逆元
\[\begin{align*} i \cdot k + r &= 0\pmod p \\ k + r \cdot i^{-1} &= 0 \pmod p \\ k \cdot r^{-1} + i^{-1} &= 0 \pmod p \\ i^{-1} &= -k \cdot r^{-1}\pmod p \end{align*} \] inv[1] = 1;
for(int i = 2; i < N; ++i)
inv[i] = (MOD - (MOD / i) * inv[MOD % i] % MOD) % MOD;
标签:cdot,inv,板子,pmod,速查,align,MOD
From: https://www.cnblogs.com/Aria-Math/p/17813686.html