洛谷P4725 【模板】多项式对数函数(多项式ln)

题意

题目链接

Sol

这个不用背XD

前置知识：

\(f(x) = ln(x), f'(x) = \frac{1}{x}\)

\(f(g(x)) = f'(g(x)) g'(x)\)

我们要求的是\(G(x) = F(A(x)), F(x) = ln(x)\)

可以直接对两边求导\(G'(A(x)) = F'(A(x))A'(x) = \frac{A(x)}{A'(x)}\)

发现这个可以算，只要求个逆就行了。

那么就直接求导之后积分回去，复杂度\(O(nlogn)\)

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
#define ull unsigned long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 4e5 + 10, INF = 1e9 + 10;
const double eps = 1e-9, pi = acos(-1);
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int a[MAXN], b[MAXN];
namespace Poly {
    int rev[MAXN], GPow[MAXN], GiPow[MAXN], A[MAXN], B[MAXN], C[MAXN], D[MAXN], lim;
    const int G = 3, Gi = 332748118, mod = 998244353;
    template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
    template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
    template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
    template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
    int fp(int a, int p, int P = mod) {
        int base = 1;
        for(; p; p >>= 1, a = 1ll * a * a % P) if(p & 1) base = 1ll * base *  a % P;
        return base;
    }
    int GetLen(int x) {
        int lim = 1;
        while(lim < x) lim <<= 1;
        return lim;
    }
    int GetOrigin(int x) {//¼ÆËãÔ­¸ù
        static int q[MAXN]; int tot = 0, tp = x - 1;
        for(int i = 2; i * i <= tp; i++) if(!(tp % i)) {q[++tot] = i;while(!(tp % i)) tp /= i;}
        if(tp > 1) q[++tot] = tp;
        for(int i = 2, j; i <= x - 1; i++) {
            for(j = 1; j <= tot; j++) if(fp(i, (x - 1) / q[j], x) == 1) break;
            if(j == tot + 1) return i;
        }
    }
    void Init(/*int P,*/ int Lim) {
        //mod = P; G = GetOrigin(mod); Gi = fp(G, mod - 2);
        for(int i = 1; i <= Lim; i++) GPow[i] = fp(G, (mod - 1) / i), GiPow[i] = fp(Gi, (mod - 1) / i);
    }
    void NTT(int *A, int lim, int opt) {
        int len = 0; for(int N = 1; N < lim; N <<= 1) ++len;
        for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
        for(int i = 0; i <= lim; i++) if(i < rev[i]) swap(A[i], A[rev[i]]);
        for(int mid = 1; mid < lim; mid <<= 1) {
            int Wn = (opt == 1 ? GPow[mid << 1] : GiPow[mid << 1]);
            for(int i = 0; i < lim; i += (mid << 1)) {
                for(int j = 0, w = 1; j < mid; j++, w = mul(w, Wn)) {
                    int x = A[i + j], y = mul(w, A[i + j + mid]);
                    A[i + j] = add(x, y), A[i + j + mid] = add(x, -y);
                }
            }
        }
        if(opt == -1) {
            int Inv = fp(lim, mod - 2);
            for(int i = 0; i <= lim; i++) mul2(A[i], Inv);
        }
    }
    void Mul(int *a, int *b, int N, int M) {
        memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
        int lim = 1, len = 0;
        while(lim <= N + M) len++, lim <<= 1;
        for(int i = 0; i <= N; i++) A[i] = a[i];
        for(int i = 0; i <= M; i++) B[i] = b[i];
        NTT(A, lim, 1); NTT(B, lim, 1);
        for(int i = 0; i <= lim; i++) B[i] = mul(B[i], A[i]);
        NTT(B, lim, -1);
        for(int i = 0; i <= N + M; i++) b[i] = B[i];
    }
    void Inv(int *a, int *b, int len) {//B1 = 2B - A1 * B^2
        if(len == 1) {b[0] = fp(a[0], mod - 2); return ;}
        Inv(a, b, len >> 1);
        for(int i = 0; i < len; i++) A[i] = a[i], B[i] = b[i];
        NTT(A, len << 1, 1); NTT(B, len << 1, 1);
        for(int i = 0; i < (len << 1); i++) mul2(A[i], mul(B[i], B[i]));
        NTT(A, len << 1, -1);
        for(int i = 0; i < len; i++) add2(b[i], add(b[i], -A[i]));
        for(int i = 0; i < (len << 1); i++) A[i] = B[i] = 0;
    }
    void Dao(int *a, int *b, int len) {
    	for(int i = 1; i < len; i++) b[i - 1] = mul(i, a[i]);
    }
    void Ji(int *a, int *b, int len) {
        for(int i = 1; i < len; i++) b[i] = mul(a[i - 1], fp(i, mod - 2));
    }
    void Ln(int *a, int *b, int len) {
    	Dao(a, C, len);
        Inv(a, D, len);
    	NTT(C, len << 1, 1); NTT(D, len << 1, 1);
    	for(int i = 0; i < (len << 1); i++) D[i] = mul(C[i], D[i]);
    	NTT(D, len << 1, -1);
    	Ji(D, b, len << 1);
    }
    /*
    void PolySqrt(int *a, int *b, int len) {//B1 = \frac{1}{2} (B + A / B)
    	if(len == 1) {b[0] = };
    }
    */
};
using namespace Poly;
signed main() {
    int N = read();
    for(int i = 0; i < N; i++) a[i] = read();
    Init(4 * N);
    Ln(a, b, GetLen(N));
    for(int i = 0; i < N; i++) cout << b[i] << " ";
    return 0;
}