题解 P5265 多项式反三角函数

$\texttt{link}$

这个题很简单，几个板子套一套就做完了，需要了解一些简单微积分。

首先两个结论$$\arcsin'x=\dfrac{1}{\sqrt{1-x^2}}$$$$\arctan'x=\dfrac{1}{1+x^2}$$

证明也很好证，经典隐函数证明。

先看 $\arcsin'x$ 的，设 $\sin y=x$，则

$$\sin'y=x'$$$$y'\cos y=1$$$$y'=\dfrac{1}{\cos y}$$$$y'=\dfrac{1}{\sqrt{1-\sin^2y}}$$$$\arcsin'x=\dfrac{1}{\sqrt{1-x^2}}$$

$\arctan'x$ 同理，设 $\tan y=x$，则$$\tan'y=x'$$$$y'\sec^2y=1$$$$y'=\dfrac{1}{\sec^2y}$$$$y'=\dfrac{1}{1+\tan^2y}$$$$\arctan'x=\dfrac{1}{1+x^2}$$

回到这题，设 $G=\arcsin F$，则$$\begin{aligned}G'&=\arcsin'F\\&=\dfrac{F'}{\sqrt{1-F^2}}\end{aligned}$$

积回来

$$G=\int \dfrac{F'}{\sqrt{1-F^2}}dx$$

同理，设 $G=\arctan F$，则$$\begin{aligned}G'&=\arctan'F\\&=\dfrac{F'}{1+F^2}\end{aligned}$$

积回来

$$G=\int\dfrac{F'}{1+F^2}dx$$ 套一下多项式求导，积分，开根，乘法，求逆即可。

#include <bits/stdc++.h>
using namespace std;
namespace IO {
    //read and write
} using namespace IO;
const int N = 2.7e5 + 10;
namespace Polynomial {
    const int mod = 998244353, G = 3, Gi = 332748118, inv2 = 499122177;
    int lim, rev[N], a[N], b[N], c[N];
    int qpow(int n, int k) {
        int res = 1;
        for(; k; n = 1ll * n * n % mod, k >>= 1)
            if(k & 1) res = 1ll * res * n % mod;
        return res;
    }
    void NTT(int *f, int T) {
        for(int i = 0; i < lim; i++)
            if(i < rev[i]) 
                swap(f[i], f[rev[i]]);
        for(int mid = 1; mid < lim; mid <<= 1) {
            int wn = qpow(T == 1 ? G : Gi, (mod - 1) / (mid << 1));
            int len = mid << 1;
            for(int i = 0; i < lim; i += (mid << 1)) {
                int w = 1;
                for(int j = 0; j < mid; j++, w = 1ll * w * wn % mod) {
                    int x = f[i + j], y = 1ll * w * f[i + j + mid] % mod;
                    f[i + j] = (x + y) % mod;
                    f[i + j + mid] = (x - y + mod) % mod;
                }
            }
        }
        if(T == -1) {
            int inv = qpow(lim, mod - 2);
            for(int i = 0; i < lim; i++) 
                f[i] = 1ll * f[i] * inv % mod;
        }
    }
    void init(int n) {
        for(lim = 1; lim < n; lim <<= 1);
        for(int i = 0; i < lim; i++)
            rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (lim >> 1));
    } 
    void mul(int *f, int *g, int *h, int n, int m) {
        static int a[N], b[N];
        init(n + m - 1);
        memset(a, 0, lim << 2);
        memcpy(a, f, n << 2);
        memset(b, 0, lim << 2);
        memcpy(b, g, m << 2);
        NTT(a, 1), NTT(b, 1);
        for(int i = 0; i < lim; i++) 
            h[i] = 1ll * a[i] * b[i] % mod;
        NTT(h, -1);
    }
    void inv(int *f, int *g, int n) {
        if(n == 1) { g[0] = qpow(f[0], mod - 2); return; }
        inv(f, g, n + 1 >> 1);
        init(n << 1);
        copy(f, f + n, a);
        fill(a + n, a + lim, 0);
        NTT(a, 1), NTT(g, 1);
        for(int i = 0; i < lim; i++)
            g[i] = (2 - 1ll * a[i] * g[i] % mod + mod) % mod * g[i] % mod;
        NTT(g, -1);
        fill(g + n, g + lim, 0);
    }
    void sqrt(int *f, int *g, int n) {
        if(n == 1) { g[0] = 1; return; }
        sqrt(f, g, n + 1 >> 1);
        memset(b, 0, n << 2);
        inv(g, b, n);
        mul(f, b, b, n, n);
        for(int i = 0; i < n; i++)
            g[i] = 1ll * (g[i] + b[i]) * inv2 % mod;
    }
    void dev(int *f, int *g, int n) {
        for(int i = 1; i < n; i++)
            g[i - 1] = 1ll * i * f[i] % mod;
        g[n - 1] = 0;
    }
    void invdev(int *f, int *g, int n) {
        for(int i = n - 1; i; i--)
            g[i] = 1ll * f[i - 1] * qpow(i, mod - 2) % mod;
        g[0] = 0;
    }
    void arcsin(int *f, int *g, int n) {
        static int a[N], b[N], c[N];
        dev(f, a, n); 
        mul(f, f, b, n, n);
        for(int i = 0; i < n; i++)
            b[i] = (mod - b[i]) % mod;
        b[0] = (b[0] + 1) % mod;
        sqrt(b, c, n);
        memset(b, 0, lim << 2);
        inv(c, b, n);
        memset(c, 0, lim << 2);
        mul(a, b, c, n, n);
        invdev(c, g, n);
    }
    void arctan(int *f, int *g, int n) {
        static int a[N], b[N], c[N];
        dev(f, a, n); 
        mul(f, f, b, n, n);
        b[0] = (b[0] + 1) % mod;
        inv(b, c, n);
        memset(b, 0, lim << 2);
        mul(a, c, b, n, n);
        invdev(b, g, n);
    }
} using namespace Polynomial;
int n, type, f[N], g[N];
int main() {
    n = read(), type = read();
    for(int i = 0; i < n; i++)
        f[i] = read();
    if(!type) arcsin(f, g, n);
    else arctan(f, g, n);
    for(int i = 0; i < n; i++)
        write(g[i]), putc(' ');
    flush();
    return 0;
}