cf438E. The Child and Binary Tree(生成函数 多项式开根 多项式求逆)
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2022-10-06 14:58:54
题意 "链接" Sol 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$\su ......
题意
sol
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) {//¼æëãô¸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; }