手动博客搬家: 本文发表于20181125 13:25:03, 原地址https://blog.csdn.net/suncongbo/article/details/84487306

题目链接: https://www.luogu.org/problemnew/show/P4725

题目大意: 给定一个\(n\)次多项式\(A(x)\), 求一个\(n\)次多项式\(B(x)\)满足\(B(x)\equiv \ln A(x) (\mod x^n)\)

题解: 神数学模板题……
数学真奇妙！
前驱知识
导数、积分相关
幂函数的求导
\(f(x)=x^n, f'(x)=nx^{n-1}\)
和的导数等于导数的和
\((f+g)'(x)=f'(x)+g'(x)\)
一般多项式的求导
\(f(x)=\sum^{n-1}_{i=0} a_ix^i, f'(x)=\sum^{n-2}_{i=0} (i+1)a_{i+1}x^i\)
对数函数\(\ln\)的求导
\(f(x)=\ln(x), f'(x)=\frac{1}{x}\)
复合函数求导——链式法则
\(f(g(x))'=f'(g(x))g'(x)\)
求导的逆运算——积分

本题解法
\(g(x)\equiv \ln f(x) (\mod x^n)\)
两边同时求导可得
\(g'(x)\equiv \frac{f'(x)}{f(x)} (\mod x^n)\)
结束！
多项式求逆算\(\frac{1}{f(x)}\)，再和\(f'(x)\)相乘即可得到\(g'(x)\)。
(多项式求逆见蒟蒻一篇博客 https://blog.csdn.net/suncongbo/article/details/84485718)
\(g'(x)\)积个分得到\(g(x)\). 常数项，直接为\(0\).
时间复杂度\(O(n\log n)\)
常数，我写的大概\(9\)倍吧，求逆是\(6\)倍，再做个乘法就是\(3\)倍。
UPD: 仔细想了一下我这个常数好像是\(18\)倍（见我多项式求逆那篇博客）

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define llong long long
#define ldouble long double
#define uint unsigned int
#define ullong unsigned long long
#define udouble unsigned double
#define uldouble unsigned long double
#define modinc(x) {if(x>=P) x-=P;}
#define pii pair<int,int>
#define piii pair<pair<int,int>,int>
#define piiii pair<pair<int,int>,pair<int,int> >
#define pli pair<llong,int>
#define pll pair<llong,llong>
#define Memset(a,x) {memset(a,x,sizeof(a));}
using namespace std;const int N = 1<<19;
const int P = 998244353;
const int LGN = 19;
const int G = 3;
llong a[N+3];
llong b[N+3];
llong tmp1[N+3],tmp2[N+3],tmp3[N+3],tmp4[N+3]; //inv
llong tmp7[N+3],tmp8[N+3],tmp9[N+3],tmp10[N+3]; //ln
int id[N+2];
int n;void initid(int _len)
{id[0] = 0;for(int i=1; i<(1<<_len); i++) id[i] = (id[i>>1]>>1)|((i&1)<<(_len-1));
}llong quickpow(llong x,llong y)
{llong cur = x,ret = 1ll;for(int i=0; y; i++){if(y&(1ll<<i)){y-=(1ll<<i); ret = ret*cur%P;}cur = cur*cur%P;}return ret;
}
llong mulinv(llong x) {return quickpow(x,P-2);}void ntt(int dgr,int coe,llong poly[],llong ret[])
{int len = 0; for(int i=0; i<=LGN; i++) if((1<<i)==dgr) {len = i; break;}initid(len); for(int i=0; i<dgr; i++) ret[i] = 0ll;for(int i=0; i<dgr; i++) ret[i] = poly[i];for(int i=0; i<dgr; i++) if(i<id[i]) swap(ret[i],ret[id[i]]);for(int i=1; i<=(dgr>>1); i<<=1){llong tmp = quickpow(G,(P-1)/(i<<1));if(coe==-1) tmp = mulinv(tmp);for(int j=0; j<dgr; j+=(i<<1)){llong expn = 1ll;for(int k=0; k<i; k++){llong x = ret[j+k],y = (expn*ret[j+i+k])%P;ret[j+k] = x+y; modinc(ret[j+k]);ret[j+i+k] = x-y+P; modinc(ret[j+i+k]);expn = (expn*tmp)%P;}}}if(coe==-1){llong tmp = mulinv(dgr);for(int i=0; i<dgr; i++) ret[i] = ret[i]*tmp%P;}
}void polyinv(int dgr,llong poly[],llong ret[])
{for(int i=0; i<dgr; i++) ret[i] = 0ll;ret[0] = mulinv(poly[0]);for(int i=1; i<=(dgr>>1); i<<=1){for(int j=0; j<(i<<2); j++) tmp1[j] = j<i ? ret[j] : 0ll;for(int j=0; j<(i<<2); j++) tmp2[j] = j<(i<<1) ? poly[j] : 0ll;ntt((i<<2),1,tmp1,tmp3); ntt((i<<2),1,tmp2,tmp4);for(int j=0; j<(i<<2); j++) tmp3[j] = tmp3[j]*tmp3[j]%P*tmp4[j]%P;ntt((i<<2),-1,tmp3,tmp4);for(int j=0; j<(i<<1); j++) ret[j] = (tmp1[j]+tmp1[j]-tmp4[j]+P)%P;}for(int i=dgr; i<(dgr<<1); i++) ret[i] = 0ll;
}void polyder(int dgr,llong poly[],llong ret[])
{for(int i=0; i<dgr-1; i++) ret[i] = poly[i+1]*(i+1)%P;
}void polyint(int dgr,llong poly[],llong ret[])
{for(int i=1; i<=dgr; i++) ret[i] = poly[i-1]*mulinv(i)%P;
}void polyln(int dgr,llong poly[],llong ret[])
{polyder(dgr,poly,tmp7);polyinv(dgr,poly,tmp8);ntt((dgr<<1),1,tmp8,tmp9); ntt((dgr<<1),1,tmp7,tmp10);for(int i=0; i<(dgr<<1); i++) tmp9[i] = tmp9[i]*tmp10[i]%P;ntt((dgr<<1),-1,tmp9,tmp10);polyint(dgr,tmp10,ret);
}int main()
{scanf("%d",&n); int dgr = 1; while(dgr<=n) dgr<<=1;for(int i=0; i<n; i++) scanf("%lld",&a[i]);polyln(dgr,a,b);for(int i=0; i<n; i++) printf("%lld ",b[i]);return 0;
}