bzoj3625

fft

分治虽然是万能的，但是太慢了

分治是nlog^2n的，太慢了，于是我们用求逆和开根

设f(x)表示答案为x的方案数

c表示物品的生成函数

那么f=f*f*c+1

f*f表示左右儿子的方案数

c表示根的方案数

+1是空树，也就是+上t(x)=1这个生成函数

然后求根公式得出f=2/(1+sqrt(1-4*g))

为什么是+号呢？因为x=1时方案数=1，那么只能是+号

然后就是多项式开根和求逆

具体看picks博客

然后是卡常

所有数组开int

fft蝴蝶操作一定要开int

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int P = ;
ll power(ll x, int t)
{
    ll ret = ;
    for(; t; t >>= , x = x * x % P) if(t & ) ret = ret * x % P;
    return ret;
}
const int N = ( << ) + ;
int n, k, m;
int rev[N];
ll inv2;
void ntt(int *a, int n, int k, int f)
{
    for(int i = ; i < n; ++i) if(i < rev[i]) swap(a[i], a[rev[i]]);
    for(int l = ; l <= n; l <<= )
    {
        ll w = power(, f ==  ? (P - ) / l : P -  - (P - ) / l);
        int m = l >> ;
        for(int i = ; i < n; i += l)
        {
            ll t = ;
            for(int k = ; k < m; ++k, t = t * w % P)
            {
                int x = a[i + k], y = t * a[i + k + m] % P;
                a[i + k] = (x + y) % P;
                a[i + m + k] = (x - y + P) % P;
            }
        }
    }
    if(f == -)
    {
        ll inv = power(n, P - );
        for(int i = ; i < n; ++i) a[i] = a[i] * inv % P;
    }
}
int a[N], b[N], tmp[N], B[N], c[N];
void poly_inverse(int *a, int *b, int l)
{
    if(l == )
    {
        b[] = power(a[], P - );
        return;
    }
    poly_inverse(a, b, l >> );
    int n = , k = ;
    while(n <= l) n <<= , ++k;
    for(int i = ; i < n; ++i) rev[i] = (rev[i >> ] >> ) | ((i & ) << (k - ));
    for(int i = ; i < l; ++i) tmp[i] = a[i];
    for(int i = l; i < n; ++i) tmp[i] = ;
    ntt(tmp, n, k, );
    ntt(b, n, k, );
    for(int i = ; i < n; ++i) b[i] = (ll)b[i] * ( - (ll)tmp[i] * b[i] % P + P) % P;
    ntt(b, n, k, -);
    for(int i = l; i < n; ++i) b[i] = ;
}
void poly_sqrt(int *a, int *b, int l)
{
    if(l == )
    {
        b[] = ;
        return;
    }
    int n = , k = ;
    while(n <= l) n <<= , ++k;
    poly_sqrt(a, b, l >> );
    for(int i = ; i < n; ++i) B[i] = ;
    poly_inverse(b, B, l);
    for(int i = ; i < n; ++i) rev[i] = (rev[i >> ] >> ) | ((i & ) << (k - ));
    for(int i = ; i < l; ++i) tmp[i] = a[i];
    for(int i = l; i < n; ++i) tmp[i] = , B[i] = ;
    ntt(B, n, k, );
    ntt(tmp, n, k, );
    ntt(b, n, k, );
    for(int i = ; i < n; ++i) b[i] = (ll)inv2 * (b[i] + (ll)B[i] * tmp[i] % P) % P;
    ntt(b, n, k, -);
    for(int i = l; i < n; ++i) b[i] = ;
}
int main()
{
    scanf("%d%d", &n, &m);
    for(int i = ; i <= n; ++i)
    {
        int x;
        scanf("%d", &x);
        c[x] -= ;
    }
    inv2 = power(, P - );
    int len = ;
    while(len <= m) len <<= ;
    ++c[];
    for(int i = ; i <= m; ++i) if(c[i] < ) c[i] += P;
    poly_sqrt(c, b, len);
    ++b[];
    b[] %= P;
    poly_inverse(b, a, len);
    for(int i = ; i <= m; ++i) printf("%d\n", a[i] *  % P);
    return ;
}