[题解] [AGC005F] Many Easy Problems

Face questions

answer

Quality problems of seniors lectures and sure enough I usually look for those that are not the same

Still thinking it is somewhat clever

We calculated taking into account not good for all of size \ (I \) of the set point, it can contain a minimum block size communicated

Conversion topic

This time we should think of the target into a single point \ (i \) to select \ (k \) contribution at points

Then his contribution to the total number is subtracted scheme did not choose the right program

The program did not choose how count it

Note \ (sz [i] \) is to \ (I \) is the subtree rooted at points

Analysis found that when \ (K \) point simultaneously with its sons in the root of a subtree will not count \ (I \)

The son of significance here is that in order to \ (i \) is the root of the whole tree when its son

That is, to its parent is the root, of size \ (n - sz [i] \) of a subtree to be regarded as \ (I \) subtree

Then a great force equation
\ [\ displaystyle \ begin {aligned } f (k) = \ sum_ {u = 1} ^ n (C_n ^ k- \ sum_ {v \ in sonu} C_ {sz [v]} ^ K) \ the aligned End {} \]
\ (C_n ^ K \) is the total number of programs that is not back calculate \ (I \) the number of programs

But we also found that this thing is not a good count

Then again conversion, we set \ (cnt_i \) of size \ (I \) number subtree

Then the contribution of the subtree has the same size imputation up
\ [\ displaystyle \ begin {aligned } f (k) & = n * C_n ^ k - \ sum_ {i = k} ^ ncnt_i * C_i ^ k \\ & = n * C_n ^ k- \ sum_ {i = k} ^ {n} cnt_i * \ frac {i!} {k! * (ik)!} \\ & = n * C_n ^ k- \ frac {1} {k!} \ sum_ {i
= k} ^ {n} \ frac {cnt_i * i!} {(ik)!} \ end {aligned} \] order \ (F_i = cnt_i * I! \) , \ ( = g_i \ FRAC. 1 {I} {!} \) , \ (G_ H_i = {Ni} \)

所以有
\[ \displaystyle\begin{aligned}f(k)&=n*C_n^k-\frac{1}{k!}\sum_{i=k}^{n}F_i*H_{n+k-i}\end{aligned} \]
NTT 即可

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
const int mod = 924844033;
const int N = 200005; 
using namespace std;

int m, lim, n, g, ig, p[105], cnt, f[N * 8], inv[2 * N], fac[2 * N], h[N * 8], sz[N], head[N], rev[N * 8];
struct edge { int to, nxt; } e[N << 1]; 

template < typename T >
inline T read()
{
    T x = 0, w = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * w; 
}

inline void adde(int u, int v) { e[++cnt] = (edge) { v, head[u] }, head[u] = cnt; }

int fpow(int x, int y)
{
    int res = 1;
    for( ; y; y >>= 1, x = 1ll * x * x % mod)
        if(y & 1) res = 1ll * res * x % mod;
    return res; 
}

int getroot(int x)
{
    int rem = x - 1, tmp = x - 1;
    for(int i = 2; i * i <= rem; i++)
        if(!(tmp % i))
        {
            p[++cnt] = i;
            while(!(tmp % i)) tmp /= i; 
        }
    if(tmp > 1) p[++cnt] = tmp;
    for(int flag = 0, i = 2; i <= x; i++, flag = 0)
    {
        for(int j = 1; j <= cnt; j++)
            if(fpow(i, rem / p[j]) == 1) flag = 1;
        if(!flag) return i; 
    }
}

void pre()
{
    for(int i = (fac[0] = 1); i <= 2 * n; i++)
        fac[i] = 1ll * fac[i - 1] * i % mod; 
    inv[2 * n] = fpow(fac[2 * n], mod - 2); 
    for(int i = 2 * n - 1; i >= 0; i--)
        inv[i] = 1ll * inv[i + 1] * (i + 1) % mod; 
    for(int i = 1; i <= n; i++)
        h[i] = inv[n - i]; 
}

int C(int n, int m) { return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod; }

void dfs(int u, int fa)
{
    sz[u] = 1; 
    for(int v, i = head[u]; i; i = e[i].nxt)
    {
        v = e[i].to; if(v == fa) continue; 
        dfs(v, u), sz[u] += sz[v]; 
    }
    f[sz[u]]++, f[n - sz[u]]++; 
}

void ntt(int *p, int opt)
{
    for(int i = 0; i < n; i++)
        if(i < rev[i]) swap(p[i], p[rev[i]]);
    for(int i = 1; i < n; i <<= 1)
    {
        int rt = fpow(opt == 1 ? g : ig, (mod - 1) / (i << 1));
        for(int j = 0; j < n; j += (i << 1))
        {
            int w = 1;
            for(int k = j; k < j + i; k++, w = 1ll * w * rt % mod)
            {
                int x = p[k], y = 1ll * w * p[k + i] % mod;
                p[k] = (x + y) % mod, p[k + i] = (x - y + mod) % mod; 
            }
        }
    }
}

int main()
{
    n = read <int> (), g = getroot(mod), ig = fpow(g, mod - 2); 
    cnt = 0, pre(); 
    for(int u, v, i = 1; i < n; i++)
    {
        u = read <int> (), v = read <int> (); 
        adde(u, v), adde(v, u); 
    }
    dfs(1, 0); 
    for(m = 3 * n, n = 1; n <= m; n <<= 1, lim++); lim--; 
    for(int i = 0; i < n; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lim);
    f[m / 3]--;
    for(int i = 0; i < m / 3; i++)
        f[i] = 1ll * f[i] * fac[i] % mod; 
    ntt(f, 1), ntt(h, 1); 
    for(int i = 0; i < n; i++)
        f[i] = 1ll * f[i] * h[i] % mod; 
    ntt(f, -1); 
    int tmp = fpow(n, mod - 2); 
    for(int i = 0; i < n; i++)
        f[i] = 1ll * f[i] * tmp % mod; 
    m /= 3; 
    for(int i = 1; i <= m; i++)
        printf("%lld\n", (1ll * m * C(m, i) % mod - 1ll * inv[i] * f[m + i] % mod + mod) % mod); 
    return 0; 
}