【BZOJ3625】【CF438E】小朋友和二叉树
题目
思路&做法
我们可以用\(v_i\)表示\(i\)在\(c\)中出现了几次， 用\(f_i\)表示权值为\(i\)的神犇树的总数, 于是
\[f_x = \sum_{i = 0}^{x}v_i \bigg( \sum_{j = 0}^{x-i}f_jf_{x-i-j} \bigg)
\]
\[f_0 = 1
\]
然后我们设\(v\)的生成函数为\(V = \sum_{i = 0} ^{\infty}v_ix^i\),
设\(f\)的生成函数为\(F = \sum_{i = 0}^{\infty}f_ix^i\)
所以
\[F(x) = C(x) F(x)F(x) + 1
\]
然后移一下项
\[V(x)F^2(x) - F(x) + 1 = 0
\]
接着直接用求根公式
\[F(x) = {1 \pm \sqrt{1 - 4 \times V(x)} \over 2 \times V(x)}
\]
于是有两种情况
\((1)\)
\[F(x) = {1 + \sqrt{1 - 4 \times V(x)} \over 2 \times V(x)}
\]
要舍去\((\)原因？不知道，我太弱了\()\)
\((2)\)
\[F(x) = {1 - \sqrt{1 - 4 \times V(x)} \over 2 \times V(x)}
\]
可以把分子和分母同时乘上\(1 + \sqrt{1 + 4 \times V(x)}\)
所以
\[\begin {aligned}
F(x)&={1^2 - \big(\sqrt{1 - 4 \times V(x)}\big)^2 \over {2 \times V(x) \times \big(1 + \sqrt{1 - 4 \times V(x)} \big)}} \\\
&={4 \times V(x) \over {2 \times V(x) \times \big(1 + \sqrt{1 - 4 \times V(x)}} \big)} \\\
&={2 \over {1 + \sqrt{1 - 4 \times V(x)}}}
\end {aligned}
\]
多项式求逆和多项式开根可以看我的这篇blog : 多项式的一些东西
代码
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 998244353LL;
const int N = 400010;
char I[8000010], *pos, *End;
inline char GetChar()
{ if(pos == End)
{ End = (pos = I) + fread(I, 1, 8000000, stdin);
if(pos == End) return EOF;
}
return *pos++;
}
inline int rd_int()
{ int x = 0, f = 1;
char c = GetChar();
while(c < '0' || c > '9') { if(c == '-') f = -1; c = GetChar(); }
while(c >= '0' && c <= '9') { x = x*10 + c-'0'; c = GetChar(); }
return x*f;
}
int Bit;
inline LL power(LL a, LL n)
{ LL ans = 1;
while (n)
{ if (n & 1) ans = (ans * a) % mod;
a = (a * a) % mod;
n >>= 1;
}
return ans;
}
int rev[N];
void getReverse(int Len)
{ for (int i = 0; i < Len; i++)
rev[i] = (rev[i>>1] >> 1) | ((i&1) * (Len>>1));
}
LL w_n[N];
void NTT(LL * a, int opt, int Len)
{ getReverse(Len);
for (int i = 0; i < Len; i++)
if (i < rev[i]) swap(a[i], a[rev[i]]);
for (register int i = 2; i <= Len; i <<= 1)
{ LL w_n = power(3LL, (mod-1LL) / (LL)i);
if (opt == -1) w_n = power(w_n, mod-2);
for (register int j = 0; j < Len; j += i)
{ LL w = 1;
for (register int k = 0; k < (i>>1); k++)
{ LL x = a[j + k];
LL y = (w * a[j + k + (i>>1)]) % mod;
a[j + k] = (x + y) % mod;
a[j + k + (i>>1)] = (x - y + mod) % mod;
w = (w * w_n) % mod;
}
}
}
if (opt == -1)
{ LL num = power(Len, mod-2);
for (register int i = 0; i < Len; i++)
a[i] = (a[i] * num) % mod;
}
}
int getBit(int l)
{ Bit = 0;
for (; (1 << Bit) <= l; Bit++);
}
inline void cpy(LL * a, LL * b, int n, int Len)
{ for (register int i = 0; i < n; i++) a[i] = b[i];
for (register int i = n; i < Len; i++) a[i] = 0;
}
LL tmp1[N], tmp2[N];
void get_Inv(LL * a, LL * b, int l)
{ b[0] = power(a[0], mod-2);
for (register int i = 2; i <= l; i <<= 1)
{ int Len = i << 1;
cpy(tmp1, a, i, Len);
cpy(tmp2, b, i>>1, Len);
NTT(tmp1, 1, Len);
NTT(tmp2, 1, Len);
for (register int j = 0; j < Len; j++)
tmp1[j] = ((2LL*tmp2[j])%mod + mod - (((tmp2[j]*tmp2[j])%mod)*tmp1[j])%mod) % mod;
NTT(tmp1, -1, Len);
for (register int j = 0; j < i; j++)
b[j] = tmp1[j];
}
}
LL inv2;
LL tmp3[N], tmp4[N], tmp5[N];
void Sqrt(LL * a, LL * b, int l)
{ b[0] = 1;
for (register int i = 1; i <= l; i <<= 1)
{ int Len = i << 1;
for (register int j = 0; j < Len; j++)
if (j < i) tmp3[j] = (2LL * b[j]) % mod;
else tmp3[j] = 0;
get_Inv(tmp3, tmp4, i);
for (register int j = i; j < Len; j++) tmp4[j] = 0;
cpy(tmp5, a, i, Len);
NTT(tmp4, 1, Len);
NTT(tmp5, 1, Len);
for (register int j = 0; j < Len; j++)
tmp4[j] = (tmp4[j] * tmp5[j]) % mod;
NTT(tmp4, -1, Len);
for (register int j = 0; j < (i << 1); j++)
b[j] = (inv2 * b[j] + tmp4[j]) % mod;
}
}
LL a[N], b[N], ans[N];
LL v[N];
int main()
{ int n, m;
n = rd_int(), m = rd_int();
m++;
for (register int i = 1; i <= n; i++)
{ int x;
x = rd_int();
v[x]++;
}
inv2 = power(2, mod - 2);
getBit(m);
a[0] = 1;
for (register int i = 1; i < m; i++)
a[i] = (4 * (mod - v[i])) % mod;
Sqrt(a, b, (1 << Bit));
b[0] = (b[0] + 1) % mod;
get_Inv(b, ans, (1 << Bit));
for (register int i = 1; i < m; i++)
printf("%lld\n", (ans[i] * 2LL) % mod);
return 0;
}
