标签：出现 freopen ref cos lse gpo 题意 break size
生成函数博大精深Orz
我们设\(f(i)\)表示权值为\(i\)的二叉树数量，转移的时候可以枚举一下根节点
\(f(n) = \sum_{w \in C_1 \dots C_n} \sum_{j=0}^{n-w} f(j) f(n-w-j)\)
设\(T =n-w\)，后半部分变为\(\sum_{j=0}^T f(j) f(T-j)\)，是个标准的卷积形式。
对于第一重循环我们可以设出现过的数的生成函数\(C(x)\)
可以得到\(f = C * f * f + 1\)，+1是因为\(f[0] = 1\)
可以解得\(f = \frac{1\pm\sqrt{1-4G}}{2G} = \frac{2}{1\pm\sqrt{1-4C}}\)
现在问题来了，我们是要取\(+\)还是取\(-\)。
结论是取\(+\)，因为当取\(-\)时，C中x的取值趋向于\(0\)时分母会无意义
举个例子(来自cf讨论区)
\(C = 2x - 4x^2\)，\(+\sqrt{1-4C} = 1 - 4x, -\sqrt{1-4C} = -1+4x\)
后者带入得到\(F = \frac{2}{4x}\)，这玩意儿显然是无解的，因为多项式有逆元的充要条件是常数项在模意义下有逆元，然而这玩意儿的常数项是0.。
感觉做这种题直接还是要先推一推暴力dp的式子吧，不然直接用生成函数推根本无从下手。。
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
#define ull unsigned long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 1;
const double eps = 1e-9, pi = acos(-1);
inline int read() {
char c = getchar(); int x = 0, f = 1;
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 N, M, a[MAXN], b[MAXN], c[MAXN], d[MAXN];
namespace Poly {
int rev[MAXN], GPow[MAXN], A[MAXN], B[MAXN], C[MAXN], lim, INV2;
const int G = 3, mod = 998244353;
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
int fp(int a, int p, int P = mod) {
int base = 1;
for(; p; p >>= 1, a = 1ll * a * a % P) if(p & 1) base = 1ll * base * a % P;
return base;
}
int GetLen(int x) {
int lim = 1;
while(lim <= x) lim <<= 1;
return lim;
}
int GetOrigin(int x) {//?????-?ù
static int q[MAXN]; int tot = 0, tp = x - 1;
for(int i = 2; i * i <= tp; i++) if(!(tp % i)) {q[++tot] = i;while(!(tp % i)) tp /= i;}
if(tp > 1) q[++tot] = tp;
for(int i = 2, j; i <= x - 1; i++) {
for(j = 1; j <= tot; j++) if(fp(i, (x - 1) / q[j], x) == 1) break;
if(j == tot + 1) return i;
}
}
void Init(/*int P,*/ int Lim) {
//mod = P; G = GetOrigin(mod); Gi = fp(G, mod - 2);
INV2 = fp(2, mod - 2);
for(int i = 1; i < Lim; i++) GPow[i] = fp(G, (mod - 1) / i);
}
void NTT(int *A, int lim, int opt) {
int len = 0; for(int N = 1; N < lim; N <<= 1) ++len;
for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
for(int i = 0; i <= lim; i++) if(i < rev[i]) swap(A[i], A[rev[i]]);
for(int mid = 1; mid < lim; mid <<= 1) {
int Wn = GPow[mid << 1];
for(int i = 0; i < lim; i += (mid << 1)) {
for(int j = 0, w = 1; j < mid; j++, w = mul(w, Wn)) {
int x = A[i + j], y = mul(w, A[i + j + mid]);
A[i + j] = add(x, y), A[i + j + mid] = add(x, -y);
}
}
}
if(opt == -1) {
reverse(A + 1, A + lim);
int Inv = fp(lim, mod - 2);
for(int i = 0; i <= lim; i++) mul2(A[i], Inv);
}
}
void Mul(int *a, int *b, int N, int M) {
memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
int lim = 1, len = 0;
while(lim <= N + M) len++, lim <<= 1;
for(int i = 0; i <= N; i++) A[i] = a[i];
for(int i = 0; i <= M; i++) B[i] = b[i];
NTT(A, lim, 1); NTT(B, lim, 1);
for(int i = 0; i <= lim; i++) B[i] = mul(B[i], A[i]);
NTT(B, lim, -1);
for(int i = 0; i <= N + M; i++) b[i] = B[i];
memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
}
void Inv(int *a, int *b, int len) {//B1 = 2B - A1 * B^2
if(len == 1) {b[0] = fp(a[0], mod - 2); return ;}
Inv(a, b, len >> 1);
for(int i = 0; i < len; i++) A[i] = a[i], B[i] = b[i];
NTT(A, len << 1, 1); NTT(B, len << 1, 1);
for(int i = 0; i < (len << 1); i++) mul2(A[i], mul(B[i], B[i]));
NTT(A, len << 1, -1);
for(int i = 0; i < len; i++) add2(b[i], add(b[i], -A[i]));
for(int i = 0; i < (len << 1); i++) A[i] = B[i] = 0;
}
void Dao(int *a, int *b, int len) {
for(int i = 1; i < len; i++) b[i - 1] = mul(i, a[i]); b[len - 1] = 0;
}
void Ji(int *a, int *b, int len) {
for(int i = 1; i < len; i++) b[i] = mul(a[i - 1], fp(i, mod - 2)); b[0] = 0;
}
void Ln(int *a, int *b, int len) {//G(A) = \frac{A}{A'} qiudao zhihou jifen
static int A[MAXN], B[MAXN];
Dao(a, A, len);
Inv(a, B, len);
NTT(A, len << 1, 1); NTT(B, len << 1, 1);
for(int i = 0; i < (len << 1); i++) B[i] = mul(A[i], B[i]);
NTT(B, len << 1, -1);
Ji(B, b, len << 1);
memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
}
void Exp(int *a, int *b, int len) {//F(x) = F_0 (1 - lnF_0 + A) but code ..why....
if(len == 1) return (void) (b[0] = 1);
Exp(a, b, len >> 1); Ln(b, C, len);
C[0] = add(a[0] + 1, -C[0]);
for(int i = 1; i < len; i++) C[i] = add(a[i], -C[i]);
NTT(C, len << 1, 1); NTT(b, len << 1, 1);
for(int i = 0; i < (len << 1); i++) mul2(b[i], C[i]);
NTT(b, len << 1, -1);
for(int i = len; i < (len << 1); i++) C[i] = b[i] = 0;
}
void Sqrt(int *a, int *b, int len) {
static int B[MAXN];
Ln(a, B, len);
for(int i = 0; i < len; i++) B[i] = mul(B[i], INV2);
Exp(B, b, len);
}
};
using namespace Poly;
signed main() {
N = read(); M = read(); int Lim = GetLen(M); Init(4 * Lim);
for(int i = 1; i <= N; i++) a[i] = read();
for(int i = 1; i <= N; i++) b[a[i]] = (-4 + mod); add2(b[0], 1);
Sqrt(b, c, Lim);
add2(c[0], 1);
Inv(c, d, Lim);
for(int i = 1; i <= M; i++) cout << mul(2, d[i]) << '\n';
return 0;
}cf438E. The Child and Binary Tree(NTT 多项式开根 多项式求逆)
标签：出现 freopen ref cos lse gpo 题意 break size
原文地址：https://www.cnblogs.com/zwfymqz/p/10523926.html
