Luogu P5277 题解

\(\texttt{Description}\)

给 \(\deg = n-1\) 的多项式 \(f\)，求多项式 \(g\) 满足
$$
g \equiv \sqrt f \pmod {x^n}
$$

\(\texttt{Solution}\)

考虑牛顿迭代：有 \(h(g(x)) = g(x)^2 - f(x) \equiv 0 \pmod {x^n}\)
$$
g(x) \equiv \dfrac {g_0(x) - g_0^2(x) - f(x)} {2g_0(x)} \pmod {x^n}
$$
$$
g(x) \equiv \dfrac {g_0^2(x)+f(x)} {2g_0(x)} \pmod {x^n}
$$
其中 \(g_0(x)\) 是在模 \(x^{ \lceil \dfrac {n} 2 \rceil}\) 时的答案。

在 \(n=1\) 时用 cipolla 算一下就可以了。

\(\texttt{Code}\)

inline auto Sqrt(vector<int> a) {
    if (a.size() == 1) return vector<int> (1,quad::cipolla(a[0])) ;
    const int len = a.size() ;
    auto ta = a;
    ta.resize((len + 1) >> 1) ;
    auto tb = Sqrt(ta);
    tb.resize(len) ;
    auto tc = tb * tb;
    tc.resize(len) ;
    for (auto &it : tb) it = 2ll * it % mod ;
    for (int i = 0; i < len; ++i) tc[i] = (tc[i] + a[i]) % mod ;
    return tc = tc * Inv(tb),tc.resize(len),tc ;
}