【BZOJ3625】小朋友和二叉树

【题目链接】

点击打开链接

【思路要点】

令多项式\(F_i=f_i\)，\(C_i=[i\ exist\ in\ c]\)，则有\(F\equiv F*F*C+1(mod\ x^{m+1})\)。
解这个方程，有\(F\equiv \frac{1\pm\sqrt{1-4C}}{2C}(mod\ x^{m+1})\)。
分式上下同乘\(1\mp\sqrt{1-4C}\)，有\(F\equiv \frac{4C}{2C*(1\pm\sqrt{1-4C})}\equiv \frac{2}{1\pm\sqrt{1-4C}}(mod\ x^{m+1})\)。
注意到\(1-4C\)常数项为1，因此\(1-\sqrt{1-4C}\)常数项为0，因此它不存在逆元。
因此，取\(F\equiv \frac{2}{1+\sqrt{1-4C}}(mod\ x^{m+1})\)。
按该式计算\(F\)即可，时间复杂度\(O(MLogM)\)。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 524288;
const int P = 998244353;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
 x = 0; int f = 1;
 char c = getchar();
 for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
 for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
 x *= f;
}
template <typename T> void write(T x) {
 if (x < 0) x = -x, putchar('-');
 if (x > 9) write(x / 10);
 putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
 write(x);
 puts("");
}
namespace NTT {
 const int G = 3;
 int power(int x, int y) {
  if (y == 0) return 1;
  int tmp = power(x, y / 2);
  if (y % 2 == 0) return 1ll * tmp * tmp % P;
  else return 1ll * tmp * tmp % P * x % P;
 }
 int N, Log, home[MAXN];
 void NTTinit() {
  for (int i = 0; i < N; i++) {
   int ans = 0, tmp = i;
   for (int j = 1; j <= Log; j++) {
    ans <<= 1;
    ans += tmp & 1;
    tmp >>= 1;
   }
   home[i] = ans;
  }
 }
 void NTT(int *a, int mode) {
  for (int i = 0; i < N; i++)
   if (home[i] < i) swap(a[i], a[home[i]]);
  for (int len = 2; len <= N; len <<= 1) {
   int delta;
   if (mode == 1) delta = power(G, (P - 1) / len);
   else delta = power(G, P - 1 - (P - 1) / len);
   for (int i = 0; i < N; i += len) {
    int now = 1;
    for (int j = i, k = i + len / 2; k < i + len; j++, k++) {
     int tmp = a[j];
     int tnp = 1ll * a[k] * now % P;
     a[j] = (tmp + tnp) % P;
     a[k] = (tmp - tnp + P) % P;
     now = 1ll * now * delta % P;
    }
   }
  }
  if (mode == -1) {
   int inv = power(N, P - 2);
   for (int i = 0; i < N; i++)
    a[i] = 1ll * a[i] * inv % P;
  }
 }
 void times(int *a, int *b, int *c, int limit) {
  N = 1, Log = 0;
  while (N < 2 * limit) {
   N <<= 1;
   Log++;
  }
  for (int i = limit; i < N; i++)
   a[i] = b[i] = 0;
  NTTinit();
  NTT(a, 1);
  NTT(b, 1);
  for (int i = 0; i < N; i++)
   c[i] = 1ll * a[i] * b[i] % P;
  NTT(c, -1);
 }
 void timesabb(int *a, int *b, int *c, int limit) {
  N = 1, Log = 0;
  while (N < 2 * limit) {
   N <<= 1;
   Log++;
  }
  for (int i = limit; i < N; i++)
   a[i] = 0;
  for (int i = limit / 2; i <= N; i++)
   b[i] = 0;
  NTTinit();
  NTT(a, 1);
  NTT(b, 1);
  for (int i = 0; i < N; i++)
   c[i] = 1ll * a[i] * b[i] % P * b[i] % P;
  NTT(c, -1);
 }
 void inverse(int *a, int *b, int limit) {
  for (int i = 0; i < 2 * limit; i++) {
   if (i >= limit) a[i] = 0;
   b[i] = 0;
  }
  b[0] = power(a[0], P - 2);
  for (int now = 1; now < limit; now <<= 1) {
   static int c[MAXN], d[MAXN];
   for (int i = 0; i < now * 2; i++)
    c[i] = a[i], d[i] = b[i];
   timesabb(c, d, d, now * 2);
   for (int i = 0; i < now * 2; i++)
    b[i] = (2ll * b[i] - d[i] + P) % P;
  }
 }
 void getsqrt(int *a, int *b, int limit, int residue) {
  for (int i = 0; i < 2 * limit; i++) {
   if (i >= limit) a[i] = 0;
   b[i] = 0;
  }
  b[0] = residue; int inv = power(2, P - 2);
  for (int now = 1; now < limit; now <<= 1) {
   static int c[MAXN], d[MAXN];
   for (int i = 0; i < now * 2; i++)
    c[i] = a[i];
   inverse(b, d, now * 2);
   times(c, d, d, now * 2);
   for (int i = 0; i < now * 2; i++)
    b[i] = 1ll * (b[i] + d[i]) * inv % P;
  }
 }
}
int n, m;
int f[MAXN], c[MAXN], g[MAXN];
int main() {
 read(n), read(m);
 for (int i = 1; i <= n; i++) {
  int x; read(x);
  c[x] = P - 4;
 }
 c[0] = 1;
 NTT::getsqrt(c, g, m + 1, 1);
 g[0] = (g[0] + 1) % P;
 NTT::inverse(g, f, m + 1);
 for (int i = 1; i <= m; i++)
  writeln(f[i] * 2 % P);
 return 0;
}