LOJ154 集合划分计数

即要求计算一个集合幂级数在子集卷积意义下的

$${\exp }_{\le k}f=\sum _{j=0}^k\frac{f^k}{k!}$$

第一步自然是 fmt，然后进行每个分量上的形式幂级数计算。
考虑到 $g(z)={\exp }_{\le k}f(z)$ 满足方程

$$g^{\prime }=g{f}^{\prime }-\frac{f^k}{k!}{f}^{\prime }$$

通过递推式，此问题可在 $\Theta (n^2{2}^n)$ 内解决。

#include 
#include 
#include 
#include 
#include 

#include 

#define LOG(FMT...) fprintf(stderr, FMT)

using namespace std;

typedef unsigned long long ll;

const int N = 21, P = 998244353;

int n, m, k, ifac = 1;
int f[1 << N][N + 1];
int a[N + 1], b[N + 1], c[N + 1], iinv[N + 1];

int norm(int x) { return x >= P ? x - P : x; }

void exGcd(int a, int b, int& x, int& y) {
    if (!b) {
        x = 1;
        y = 0;
        return;
    }
    exGcd(b, a % b, y, x);
    y -= a / b * x;
}

int inv(int a) {
    int x, y;
    exGcd(a, P, x, y);
    return norm(x + P);
}

int mpow(int x, int k) {
    int ret = 1;
    while (k) {
        if (k & 1)
            ret = ret * (ll)x % P;
        k >>= 1;
        x = x * (ll)x % P;
    }
    return ret;
}

int main() {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; ++i) iinv[i] = inv(i);
    for (int i = 1; i <= k; ++i) ifac = ifac * (ll)iinv[i] % P;
    while (m--) {
        int s;
        scanf("%d", &s);
        ++f[s][__builtin_popcount(s)];
    }
    for (int i = 0; i < n; ++i)
        for (int s = 0; s < 1 << n; ++s)
            if (!(s >> i & 1))
                for (int j = 0; j <= n; ++j) f[s | 1 << i][j] = norm(f[s | 1 << i][j] + f[s][j]);
    int ans = 0;
    for (int s = 0; s < 1 << n; ++s) {
        int lead = -1;
        for (int i = 0; i <= n; ++i)
            if (f[s][i]) {
                lead = i;
                break;
            }
        if (lead == -1 || lead * k > n)
            continue;
        int in = inv(f[s][lead]), pw = mpow(f[s][lead], k);
        memset(a, 0, sizeof(a));
        for (int i = lead; i <= n; ++i) a[i - lead] = f[s][i] * (ll)in % P;
        for (int i = 0; i < n - lead * k; ++i) b[i] = a[i + 1] * (ll)(i + 1) % P;
        b[n] = 0;
        for (int i = 0; i <= n - lead * k; ++i) {
            ll v = b[i];
            for (int j = 1; j <= i; ++j) {
                v += (P - a[j]) * (ll)c[i - j];
                if ((j & 15) == 15)
                    v %= P;
            }
            c[i] = v % P;
        }
        for (int i = 0; i <= n - lead * k; ++i) c[i] = c[i] * (ll)k % P;
        a[0] = 1;
        for (int i = 1; i <= n - lead * k; ++i) {
            ll v = 0;
            for (int j = 0; j < i; ++j) {
                v += a[j] * (ll)c[i - 1 - j];
                if ((j & 15) == 15)
                    v %= P;
            }
            a[i] = v % P * (ll)iinv[i] % P;
        }
        for (int i = lead * k; i <= n; ++i) b[i] = a[i - lead * k] * (ll)pw % P * ifac % P;
        memset(c, 0, sizeof(c));
        for (int i = 0; i < n; ++i) a[i] = f[s][i + 1] * (ll)(i + 1) % P;
        for (int i = lead * k; i <= n; ++i) {
            ll v = 0;
            for (int j = 0; j <= i - lead * k; ++j) {
                v += b[i - j] * (ll)a[j];
                if ((j & 15) == 15)
                    v %= P;
            }
            c[i] = v % P;
        }
        for (int i = 0; i < n; ++i) b[i] = f[s][i + 1] * (ll)(i + 1) % P;
        b[n] = 0;
        a[0] = 1;
        for (int i = 1; i <= n; ++i) {
            ll v = norm(P - c[i - 1]);
            for (int j = 0; j < i; ++j) {
                v += a[j] * (ll)b[i - 1 - j];
                if ((j & 15) == 15)
                    v %= P;
            }
            a[i] = v % P * iinv[i] % P;
        }
        if (__builtin_parity(s ^ ((1 << n) - 1)))
            ans = norm(ans - a[n] + P);
        else
            ans = norm(ans + a[n]);
    }
    printf("%d\n", ans);
    return 0;
}