【數學】多項式取 ln
阿新 • • 發佈:2021-12-09
多項式取 ln
給你 \(A\),讓你求 \(B(x) \equiv lnA(x) \pmod {x^n}\)
兩遍求導,得到 \(B'(x) \equiv \frac{A'(x)}{A(x)} \pmod {x^n}\)
然後積分即有 \(B(x) \equiv \int\frac{A'(x)}{A(x)}dx \pmod {x^n}\)
#include<bits/stdc++.h> using namespace std; #define debug(x) cerr << #x << ": " << (x) << endl #define rep(i,a,b) for(int i=(a);i<=(b);i++) #define dwn(i,a,b) for(int i=(a);i>=(b);i--) using pii = pair<int, int>; using ll = long long; #define int long long inline void read(int &x){ int s=0; x=1; char ch=getchar(); while(ch<'0' || ch>'9') {if(ch=='-')x=-1;ch=getchar();} while(ch>='0' && ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar(); x*=s; } const int N=3e5+5, rt=3, mod=998244353; int rev[N], tot=1, bit; ll fpow(ll x, int p, ll mod){ int res=1; for(; p; p>>=1, x=x*x%mod) if(p&1) res=res*x%mod; return res; } ll inv(ll x, ll mod){ return fpow(x, mod-2, mod); } ll mul(ll x, int p, ll mod){ ll res=0; for(; p; p>>=1, x=(x+x)%mod) if(p&1) res=(res+x)%mod; return res; } void NTT(ll *a, int type, int mod){ for(int i=0; i<tot; i++){ a[i]%=mod; if(i<rev[i]) swap(a[i], a[rev[i]]); } for(int mid=1; mid<tot; mid<<=1){ ll w1=fpow(rt, (type==1? (mod-1)/(mid<<1): mod-1-(mod-1)/(mid<<1)), mod); for(int i=0; i<tot; i+=mid*2){ ll wk=1; for(int j=0; j<mid; j++, wk=wk*w1%mod){ auto x=a[i+j], y=wk*a[i+j+mid]%mod; a[i+j]=(x+y)%mod, a[i+j+mid]=(x-y+mod)%mod; } } } if(type==-1){ for(int i=0; i<tot; i++) a[i]=a[i]*inv(tot, mod)%mod; } } int n; int A[N], B[N], C[N]; void poly_inv(int sz, int *a, int *b){ if(sz==1) return b[0]=inv(a[0], mod), void(); poly_inv(sz+1>>1, a, b); // init bit=0, tot=1; while(tot<=(sz-1<<1)) tot<<=1, bit++; for(int i=0; i<tot; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1)); rep(i,0,sz-1) C[i]=a[i]; rep(i,sz,tot-1) C[i]=0; NTT(C, 1, mod), NTT(b, 1, mod); rep(i,0,tot-1) b[i]=(2-C[i]*b[i]%mod+mod)%mod*b[i]%mod; NTT(b, -1, mod); rep(i,sz,tot-1) b[i]=0; } void poly_der(int sz, int *a, int *b){ rep(i,1,sz-1) b[i-1]=i*a[i]%mod; b[sz-1]=0; } void poly_int(int sz, int *a, int *b){ rep(i,1,sz-1) b[i]=a[i-1]*inv(i, mod)%mod; b[0]=0; } int inv_A[N], der_A[N]; void conv(int *a, int *b){ NTT(a, 1, mod), NTT(b, 1, mod); rep(i,0,tot-1) a[i]=a[i]*b[i]%mod; NTT(a, -1, mod); } void poly_ln(int sz, int *a, int *b){ poly_inv(sz, a, inv_A); poly_der(sz, a, der_A); // init bit=0, tot=1; while(tot<=sz-1+sz-2) tot<<=1, bit++; for(int i=0; i<tot; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1)); conv(inv_A, der_A); poly_int(sz, inv_A, b); } signed main(){ cin>>n; rep(i,0,n-1) read(A[i]); poly_ln(n, A, B); rep(i,0,n-1) cout<<B[i]<<' '; cout<<endl; return 0; }