Luogu 4091 [HEOI2016/TJOI2016]求和

BZOJ 4555

一道模板题。

第二类斯特林数有公式：

$$S(n, m) = \frac{1}{m!}\sum_{i = 0}^{m}(-1)^i\binom{m}{i}(m - i)^n$$

考虑它的组合意义：$S(n, m)$表示$n$个不相同的小球放到$m$个相同的盒子里而且不能有空盒的方案数。

我们枚举空盒有$i$个，然后进行容斥。因为盒子没有区别，所以最后得到的值还要除以$m!$。

本题要求：

$$\sum_{i = 0}^{n}\sum_{j = 0}^{i}S(i, j)*2^j*(j!)$$

$$=\sum_{j = 0}^{n}2^j*(j!)\sum_{i = 0}^{n}S(i, j)$$

$$=\sum_{j = 0}^{n}2^j*(j!)\sum_{i = 0}^{n}\frac{1}{j!}\sum_{k = 0}^{j}(-1)^k\binom{j}{k}(j - k) ^ i$$

$$=\sum_{j = 0}^{n}2^j\sum_{i = 0}^{n}\sum_{k = 0}^{j}(-1)^k\frac{j!}{k!(j - k)!}*(j - k)^i$$

$$=\sum_{j = 0}^{n}2^j*(j!)\sum_{k = 0}^{j}\frac{(-1^k)}{k!} * \frac{\sum_{i = 0}^{n}(j - k)^i}{(j - k)!}$$

记$f(i) = \frac{(-1^k)}{k!}$，$g(i) = \frac{\sum_{j = 0}^{n}i^j}{(i)!}$，

因为$S(0, 0) = (-1)^0 * 1 * 0^0 = 1$，所以记$g(0) = 1$，$g(1) = n + 1$，剩下代入等比数列求和公式。

那么原式化为

$$\sum_{i = 0}^{n}2^i*(i!)(f*g)(i)$$

做一遍$NTT$就好了。

时间复杂度$O(nlogn)$。

Code：

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const int N = 3e5 + 5;
const ll P = 998244353LL;

int n, lim = 1, pos[N];
ll f[N], g[N], fac[N], inv[N], bin[N];

template <typename T>
inline void swap(T &x, T &y) {
    T t = x; x = y; y = t;
}

template <typename T>
inline void inc(T &x, T y) {
    x += y;
    if(x >= P) x -= P;
}

inline ll fpow(ll x, ll y) {
    ll res = 1LL;
    for (; y > 0; y >>= 1) {
        if (y & 1) res = res * x % P;
        x = x * x % P;
    }
    return res;
}

inline void prework() {
    int l = 0;
    for (; lim <= n * 2; ++l, lim <<= 1);
    for (int i = 0; i < lim; i++)
        pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << (l - 1)); 
}

inline void ntt(ll *c, int opt) {
    for (int i = 0; i < lim; i++)
        if (i < pos[i]) swap(c[i], c[pos[i]]);
    for (int i = 1; i < lim; i <<= 1) {
        ll wn = fpow(3, (P - 1) / (i << 1));
        if(opt == -1) wn = fpow(wn, P - 2);
        for (int len = i << 1, j = 0; j < lim; j += len) {
            ll w = 1LL;
            for (int k = 0; k < i; k++, w = w * wn % P) {
                ll x = c[j + k], y = w * c[j + k + i] % P;
                c[j + k] = (x + y) % P, c[j + k + i] = (x - y + P) % P;
            }
        }
    }
    
    if (opt == -1) {
        ll invP = fpow(lim, P - 2);
        for (int i = 0; i < lim; i++)
            c[i] = c[i] * invP % P;
    }
}

int main() {
    scanf("%d", &n);
    
    bin[0] = fac[0] = 1LL;
    for (int i = 1; i <= n; i++) {
        fac[i] = fac[i - 1] * i % P;
        bin[i] = bin[i - 1] * 2LL % P;
    }
    inv[n] = fpow(fac[n], P - 2);
    for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % P;
    
/*    for (int i = 0; i <= n; i++)
        printf("%lld%c", inv[i] * fac[i] % P, i == n ? '\n' : ' ');   */
    
    for (int i = 0; i <= n; i++) {
        f[i] = ((i & 1) ? (-1LL) : (1LL)) * inv[i] % P;
        if(f[i] < 0) f[i] += P;
        if(i == 0) g[i] = 1LL;
        else if (i == 1) g[i] = n + 1;
        else g[i] = (fpow(i, n + 1) - 1 + P) % P * fpow(i - 1, P - 2) % P;
        g[i] = g[i] * inv[i] % P;
    }
    
    prework();
    ntt(f, 1), ntt(g, 1);
    for (int i = 0; i < lim; i++) f[i] = f[i] * g[i] % P;
    ntt(f, -1);
    
    ll ans = 0LL;
    for (int i = 0; i <= n; i++) 
        inc(ans, bin[i] * fac[i] % P * f[i] % P);
    
    printf("%lld\n", ans);
    return 0;
}