标签：define 思路 get math 直接 not 解决 ack 答案
给出\(n\)个点\((x_i,y_i)\)，求出经过这\(n\)个点的一个\(n-1\)次多项式。
\(n\le 10^5\)
差点卡常死在这里。论多项式里面的调参（有\(\texttt{SA}\)内味了）（雾
我们发现这个东西我们显然可以使用拉格朗日插值法，我们可以求到答案为:
变一下我们就可以得到:
我们假设\(g(x)=\prod_{i=1}^{n}(x-x_i)\)，那么上面的式子就是:
我们发现\(\frac{g(x_i)}{x_i-x_i}\)可以使用洛必达法则求到，答案即为\(g^{‘}(x_i)\)。
所以式子就是:
我们发现这个式子如果我们直接求的话会炸成\(\Theta(n^2\log^2n)\)，但是我们发现可以使用分治解决这个问题。
我们假设当前区间为\(l,r\)，答案为\(f_{l,r}\)，中点为\(mid\)，那么可以得到:
我们从中间拆开就可以得到:
即:
于是我们就可以直接分治解决了。至于\(g^{‘}(x_i)\)可以使用多项式多点求值\(\Theta(n\log ^2n)\)预处理求出来，每一层的\(\prod_{k=l}^{r}(x-x_k)\)也可以使用分治求出来。总时间复杂度根据主定理为\(F(n)=F(n/2)+\Theta(n\log n)=\Theta(n\log^2n)\)。
但是这道题有点卡常，所以多项式多点求值的地方需要调参，我差点自闭了。。。
#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 1000005
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 << 18;
for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 17);
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 = 18 - __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){
a.resize (max (SZ (a),SZ (b)));
for (Int i = 0;i < SZ (b);++ i) a[i] = dec (a[i],b[i]);
return a;
}
poly operator * (poly a,int b){
for (Int i = 0;i < SZ (a);++ i) a[i] = mul (a[i],b);
return a;
}
poly der (poly a){
for (Int i = 0;i < SZ (a) - 1;++ i) a[i] = mul (a[i + 1],i + 1);
a.pop_back ();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;
}
poly inv (poly a,int n){
poly b(1,inv (a[0])),c;
for (Int l = 4;(l >> 2) < n;l <<= 1){
c.resize (l >> 1);
for (Int i = 0;i < (l >> 1);++ i) c[i] = i < n ? a[i] : 0;
c.resize (l),b.resize (l);
ntt (c,l,1),ntt (b,l,1);
for (Int i = 0;i < l;++ i) b[i] = mul (b[i],dec (2,mul (b[i],c[i])));
ntt (b,l,-1),b.resize (l >> 1);
}
b.resize (n);
return b;
}
poly inv (poly a){return inv (a,SZ (a));}
poly Mod (poly F,poly G){
int n = SZ (F) - 1,m = SZ (G) - 1;poly Q;Q.resize (m + 1);for (Int i = 0;i <= m;++ i) Q[i] = G[i];
reverse (F.begin(),F.end()),reverse (G.begin(),G.end()),reverse (Q.begin(),Q.end()),Q.resize (n - m + 1),Q = inv (Q) * F,Q.resize (n - m + 1),reverse (Q.begin(),Q.end());
reverse (F.begin(),F.end()),reverse (G.begin(),G.end()),Q = G * Q,Q.resize (m),Q = F - Q,Q.resize (m);
return Q;
}
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,a[MAXN],bb[MAXN],AAns[MAXN];
poly AAA,D[MAXN << 2],DR[MAXN << 2];
void divide1 (int i,int l,int r){
if (l == r) return DR[i].resize (2),DR[i][0] = mod - a[l],DR[i][1] = 1,void ();
int mid = (l + r) >> 1;divide1 (i << 1,l,mid),divide1 (i << 1 | 1,mid + 1,r);
DR[i] = DR[i << 1] * DR[i << 1 | 1];
}
ll c1,c2,c3,c4,b[17];
void divide2 (int i,int l,int r,poly AA){
if (r - l <= 512){
for (Int i = l;i <= r;++ i){
int x = a[i];b[0] = 1;int now = r - l;
for (Int j = 1;j <= 16;++ j) b[j] = b[j - 1] * x % mod;
ll Ans = AA[now];
for (Int j = now - 1;j - 15 >= 0;j -= 16){
c1 = Ans * b[16] + AA[j] * b[15] + AA[j - 1] * b[14] + AA[j - 2] * b[13],c1 %= mod;
c2 = AA[j - 3] * b[12] + AA[j - 4] * b[11] + AA[j - 5] * b[10] + AA[j - 6] * b[9],c2 %= mod;
c3 = AA[j - 7] * b[8] + AA[j - 8] * b[7] + AA[j - 9] * b[6] + AA[j - 10] * b[5],c3 %= mod;
c4 = AA[j - 11] * b[4] + AA[j - 12] * b[3] + AA[j - 13] * b[2] + AA[j - 14] * b[1],c4 %= mod;
Ans = (c1 + c2 + c3 + c4 + AA[j - 15]) % mod;
}
for (Int j = now % 16 - 1;~j;-- j) Ans = (Ans * x + AA[j]) % mod;
AAns[i] = Ans;
}
return ;
}
if (l == r) return write (AA[0]),putchar (‘\n‘),void ();
poly B = Mod (AA,DR[i << 1]);int mid = (l + r) >> 1;divide2 (i << 1,l,mid,B);B = Mod (AA,DR[i << 1 | 1]);divide2 (i << 1 | 1,mid + 1,r,B);
}
void divide3 (int i,int l,int r){
if (l == r) return D[i].resize (1),D[i][0] = mul (bb[l],inv (AAns[l])),void ();
int mid = (l + r) >> 1;divide3 (i << 1,l,mid),divide3 (i << 1 | 1,mid + 1,r);
D[i] = DR[i << 1 | 1] * D[i << 1] + DR[i << 1] * D[i << 1 | 1];
}
signed main(){
init_ntt(),read (n);
for (Int i = 1;i <= n;++ i) read (a[i],bb[i]);
divide1 (1,1,n),AAA = der (DR[1]),divide2 (1,1,n,AAA),divide3 (1,1,n);
for (Int i = 0;i < n;++ i) write (D[1][i]),putchar (‘ ‘);putchar (‘\n‘);
return 0;
}
标签：define 思路 get math 直接 not 解决 ack 答案
原文地址：https://www.cnblogs.com/Dark-Romance/p/13344207.html
