BZOJ 3684: 大朋友和多叉树 [拉格朗日反演 多项式k次幂 生成函数]

3684: 大朋友和多叉树

题意：

求有n个叶子结点，非叶节点的孩子数量(in S, a otin S)的有根树个数，无标号，孩子有序。

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鏼鏼鏼！

树的OGF：(T(x) = sum_{ige 0} t_ix^i)

[T(x) = x + sum_{k in S}T(x)^k ]

因为一个树是叶子结点或者其他树拼接成的“序列”

[x = T(x) -sum_{k in S}T(x)^k = G(T(x)) ]

所以(T(x))是(G(x) = x - sum_{k in S}x^k)的复合逆，直接上拉格朗日反演求第n项的值

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = (1<<18) + 5;
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 P = 950009857;
inline int Pow(ll a, int b) {
    ll ans = 1;
    for(; b; b>>=1, a=a*a%P)
        if(b&1) ans=ans*a%P;
    return ans;
}
 
ll inv[N];
namespace ntt {
    int g = 7, rev[N];
    void dft(int *a, int n, int flag) {
        int k = 0; while((1<<k) < n) k++;
        for(int i=0; i<n; i++) {
            rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
            if(i < rev[i]) swap(a[i], a[rev[i]]);
        }
        for(int l=2; l<=n; l<<=1) {
            int m = l>>1, wn = Pow(g, flag == 1 ? (P-1)/l : P-1-(P-1)/l);
            for(int *p = a; p != a+n; p += l)
                for(int k=0, w=1; k<m; k++, w = (ll)w*wn %P) {
                    int t = (ll) w * p[k+m] %P, r = p[k];
                    p[k+m] = (r - t + P) %P;
                    p[k] = (r + t) %P;
                }
        }
        if(flag == -1) {
            ll inv = Pow(n, P-2);
            for(int i=0; i<n; i++) a[i] = a[i] * inv %P;
        }
    }
     
    void inverse(int *a, int *b, int l) {
        static int t[N];
        if(l == 1) {b[0] = Pow(a[0], P-2); return;}
        inverse(a, b, l>>1);
        int n = l<<1;
        for(int i=0; i<l; i++) t[i] = a[i], t[i+l] = 0;
        dft(t, n, 1); dft(b, n, 1);
        for(int i=0; i<n; i++) b[i] = (ll) b[i] * (2 - (ll) t[i] * b[i] %P + P) %P;
        dft(b, n, -1); for(int i=l; i<n; i++) b[i] = 0;
    }
 
    void ln(int *a, int *b, int l) {
        static int da[N], ia[N];
        int n = l<<1;
        for(int i=0; i<n; i++) da[i] = ia[i] = 0;
        for(int i=0; i<l-1; i++) da[i] = (ll) (i+1) * a[i+1] %P;
        inverse(a, ia, l);
        dft(da, n, 1); dft(ia, n, 1);
        for(int i=0; i<n; i++) b[i] = (ll) da[i] * ia[i] %P;
        dft(b, n, -1);
        for(int i=l-1; i>0; i--) b[i] = (ll) inv[i] * b[i-1] %P; b[0] = 0;
        for(int i=l; i<n; i++) b[i] = 0;
    }
     
    void exp(int *a, int *b, int l) {
        static int t[N];
        if(l == 1) {b[0] = 1; return;}
        exp(a, b, l>>1);
        int n = l<<1;
        for(int i=0; i<n; i++) t[i] = 0;
        ln(b, t, l);
        for(int i=0; i<l; i++) t[i] = (a[i] - t[i] + P) %P; t[0] = (t[0] + 1) %P;
        dft(b, n, 1); dft(t, n, 1);
        for(int i=0; i<n; i++) b[i] = (ll) b[i] * t[i] %P;
        dft(b, n, -1); for(int i=l; i<n; i++) b[i] = 0;
    }
 
    void power(int *a, int k, int l) {
        static int t[N];
        int n = l<<1;
        for(int i=0; i<n; i++) t[i] = 0;
        ln(a, t, l);
        for(int i=0; i<l; i++) t[i] = (ll) k * t[i] %P;
        for(int i=0; i<n; i++) a[i] = 0;
        exp(t, a, l);
    }
}
 
int n, m, g[N], a[N];
int main() {
    //freopen("in", "r", stdin);
    n=read(); m=read();
    g[0] = 1;
    for(int i=1; i<=m; i++) g[read()-1] = P-1;
    int len = 1; while(len <= n) len <<= 1;
    inv[1] = 1; 
    for(int i=2; i < len<<1; i++) inv[i] = (P - P/i) * inv[P%i] %P;
    ntt::inverse(g, a, len);
 
    ntt::power(a, n, len);
    int ans = (ll) a[n-1] * inv[n] %P;
    printf("%d", ans);
}