p3306-solution

P3306 Solution

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\(x_{i+1}\equiv a\times x_i+b \pmod p\)

\(x_{i+1}\equiv a(ax_{i-1}+b)+b \pmod p\)

\(x_{i+1}\equiv a(a(ax_{i-2}+b)+b)+b \pmod p\)

\(...\)

\(\displaystyle x_{i+1}\equiv a^ix_1+b\sum\limits_{j=0}^{i-1}a^j \pmod p\)

即求关于 \(i\) 的方程 \(\displaystyle t\equiv a^{i-1}x_1+b\sum\limits_{j=0}^{i-2}a^j \pmod p\) 的解。

\(a^{i-1}\equiv \dfrac{t-b\sum\limits_{j=0}^{i-2}a^j}{x_1} \pmod p\)

\(\displaystyle\sum\limits_{j=0}^{i-2}a^j\) 是一个等比数列求和，它等于

\(\dfrac{a^{i-1}-1}{a-1}\)，代回去得到

\(a^{i-1}\equiv \dfrac{t-b\times\frac{a^{i-1}-1}{a-1}}{x_1} \pmod p\)

\(a^{i-1}\equiv \dfrac{t(a-1)-b(a^{i-1}-1)}{x_1(a-1)} \pmod p\)

右边 \(a^{i-1}\) 的系数是 \(\dfrac{-b}{x_1(a-1)}\)。于是

\((1+\dfrac{b}{x_1(a-1)})a^{i-1}\equiv \dfrac{at-t+b}{x_1(a-1)} \pmod p\)

\(\dfrac{x_1(a-1)+b}{x_1(a-1)}a^{i-1}\equiv \dfrac{at-t+b}{x_1(a-1)} \pmod p\)

\((x_1(a-1)+b)a^{i-1}\equiv at-t+b \pmod p\)

\(a^{i-1}\equiv \dfrac{at-t+b}{x_1(a-1)+b} \pmod p\)

现在右边的东西我们都知道了，BSGS求解即可。