題解 下降冪多項式轉普通多項式
阿新 • • 發佈:2020-07-20
題目大意
給出一個下降冪多項式\(F(x)=\sum_{i=0}^{n-1} a_ix^{\underline{i}}\),求一個普通多項式\(G(x)\)使得\(G(x)=F(x)\)。
\(n\le 2\times 10^5\)
思路
一個非常\(\texttt {naive}\)的想法就是我們直接乘上\(e^x\)轉換成點值\(\texttt{EGF}\),然後再多項式快速插值插回去。
同時,我們發現點值是連續的,於是,我們就不需要多項式多點求值和多項式取模了。
最終的時間複雜度為\(\Theta(n\log^2n)\)。
\(\texttt{Code}\)
#include <bits/stdc++.h> using namespace std; #define SZ(x) ((int)x.size()) #define Int register int #define mod 998244353 #define ll long long #define MAXN 2000005 int mul (int a,int b){return 1ll * a * b % mod;} int dec (int a,int b){return a >= b ? a - b : a + mod - b;} int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;} int qkpow (int a,int k){ int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod; return res; } int inv (int x){return qkpow (x,mod - 2);} typedef vector <int> poly; int w[MAXN],rev[MAXN]; void init_ntt (){ int lim = 1 << 19; for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 18); int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1; for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn); for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1]; } void ntt (poly &a,int lim,int type){ #define G 3 #define Gi 332748118 static int d[MAXN]; for (Int i = 0,z = 19 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i]; for (Int i = 1;i < lim;i <<= 1) for (Int j = 0;j < lim;j += i << 1) for (Int k = 0;k < i;++ k){ int x = mul (w[i + k],d[i + j + k]); d[i + j + k] = dec (d[j + k],x),d[j + k] = add (d[j + k],x); } for (Int i = 0;i < lim;++ i) a[i] = d[i] % mod; if (type == -1){ reverse (a.begin() + 1,a.begin() + lim); for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv); } #undef G #undef Gi } poly operator + (poly a,poly b){ a.resize (max (SZ (a),SZ (b))); for (Int i = 0;i < SZ (b);++ i) a[i] = add (a[i],b[i]); return a; } poly operator * (poly a,poly b){ int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1; a.resize (lim),b.resize (lim); ntt (a,lim,1),ntt (b,lim,1); for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]); ntt (a,lim,-1),a.resize (d); return a; } template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;} template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);} template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');} int n,fac[MAXN],ifac[MAXN]; poly fuck,III,D[MAXN << 2],DR[MAXN << 2]; void divide (int i,int l,int r){ if (l == r){ DR[i].resize (2),DR[i][0] = mod - l,DR[i][1] = 1; D[i].resize (1),D[i][0] = (n - l) & 1 ? mod - mul (fuck[l],ifac[n - l]) : mul (fuck[l],ifac[n - l]); return ; } int mid = (l + r) >> 1;divide (i << 1,l,mid),divide (i << 1 | 1,mid + 1,r); D[i] = DR[i << 1 | 1] * D[i << 1] + DR[i << 1] * D[i << 1 | 1]; DR[i] = DR[i << 1] * DR[i << 1 | 1]; } signed main(){ init_ntt(),read (n),fuck.resize (n),III.resize (n + 1); fac[0] = 1;for (Int i = 1;i <= n;++ i) fac[i] = mul (fac[i - 1],i); ifac[n] = inv (fac[n]);for (Int i = n;i;-- i) ifac[i - 1] = mul (ifac[i],i); for (Int i = 0;i < n;++ i) read (fuck[i]),III[i] = ifac[i];III[n] = ifac[n]; fuck = fuck * III,fuck.resize (n + 1);divide (1,0,n); for (Int i = 0;i < n;++ i) write (D[1][i]),putchar (' ');putchar ('\n'); return 0; }